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Dual Nature of Matter and Radiation — JEE Previous Year Questions

Every Dual Nature of Matter and Radiation question asked in JEE Main and JEE Advanced across the last 186 papers — 174 questions, each with its correct answer. Free to read, no account needed.

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174

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83%

All 174 Dual Nature of Matter and Radiation questions

Most recent papers first.

Q1·PhysicsSingle correctJEE Main 2026
A monochromatic source of light operating at 15 kW emits 2.5×10222.5 \times 10^{22}2.5×1022 photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take h=6.6×10−34h = 6.6 \times 10^{-34}h=6.6×10−34 J·s and c=3×108c = 3 \times 10^{8}c=3×108 m/s).
  1. (A)Microwave
  2. (B)Infrared
  3. (C)Visible
  4. (D)Ultraviolet

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correctJEE Main 2026
K1K_1K1​ and K2K_2K2​ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength λ1\lambda_1λ1​ and λ2\lambda_2λ2​, respectively. If λ1=2λ2\lambda_1 = 2\lambda_2λ1​=2λ2​ then the work function of material is given by:
  1. (A)K2+2K1K_2 + 2K_1K2​+2K1​
  2. (B)2K2−K12K_2 - K_12K2​−K1​
  3. (C)K1−2K2K_1 - 2K_2K1​−2K2​
  4. (D)K2−2K1K_2 - 2K_1K2​−2K1​

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsNumericalJEE Main 2026
The de Broglie wavelength for an electron accelerated through the potential difference of V1V_{1}V1​ volt is λ1\lambda_{1}λ1​. When the potential difference is changed to V2V_{2}V2​ volt, the associated de Broglie wavelength is increased by 50%. If (V1/V2)=(9/α)(V_{1}/V_{2}) = (9/\alpha)(V1​/V2​)=(9/α), then the value of α\alphaα is __________.

Correct answer: 4

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
An electron is travelling with a velocity vvv in free space and when it enters a medium, its velocity is reduced by 20%. The de Broglie wavelength of electron in the medium is αλ0\alpha\lambda_0αλ0​, where λ0\lambda_0λ0​ is its de Broglie wavelength in free space. The value of α\alphaα is ________.
  1. (A)1.20
  2. (B)1.0
  3. (C)1.25
  4. (D)0.75

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19\alpha \times 10^{-19}α×10−19 J. The value of α\alphaα is _____. (h=6.62×10−34h = 6.62 \times 10^{-34}h=6.62×10−34 J s, e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C and c=3×108c = 3 \times 10^{8}c=3×108 m/s)
  1. (A)3.68
  2. (B)4.68
  3. (C)5.68
  4. (D)2.68

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
An electron of mass mmm is moving in an electric field E⃗=−2E0i^\vec{E} = -2E_0\hat{i}E=−2E0​i^ (E0=E_0 =E0​= constant >0> 0>0), with an initial velocity V⃗=v0i^\vec{V} = v_0\hat{i}V=v0​i^ (v0=v_0 =v0​= constant >0> 0>0). If λ0=h4mv0\lambda_0 = \frac{h}{4mv_0}λ0​=4mv0​h​, its de Broglie wavelength at time ttt is __________. (e=e =e= charge of electron)
  1. (A)4λ0[1−E0e2mtv0]\frac{4\lambda_0}{\left[1 - \frac{E_0e}{2m}\frac{t}{v_0}\right]}[1−2mE0​e​v0​t​]4λ0​​
  2. (B)4λ0[1+E0e2mtv0]\frac{4\lambda_0}{\left[1 + \frac{E_0e}{2m}\frac{t}{v_0}\right]}[1+2mE0​e​v0​t​]4λ0​​
  3. (C)4λ0[1+2E0emtv0]\frac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}[1+m2E0​e​v0​t​]4λ0​​
  4. (D)4λ0[1−2E0emtv0]\frac{4\lambda_0}{\left[1 - \frac{2E_0e}{m}\frac{t}{v_0}\right]}[1−m2E0​e​v0​t​]4λ0​​

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
The de Broglie wavelength associated with an electron accelerated through a potential difference V is λe\lambda_{e}λe​ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is λp\lambda_{p}λp​. If their corresponding masses are mem_{e}me​ and mpm_{p}mp​, respectively, then the ratio of their de Broglie wavelengths (λeλp)\left(\frac{\lambda_{e}}{\lambda_{p}}\right)(λp​λe​​) is ______.
  1. (A)mpme\sqrt{\frac{m_{p}}{m_{e}}}me​mp​​​
  2. (B)memp\sqrt{\frac{m_{e}}{m_{p}}}mp​me​​​
  3. (C)mpme\frac{m_{p}}{m_{e}}me​mp​​
  4. (D)(mpme)2\left(\frac{m_{p}}{m_{e}}\right)^{2}(me​mp​​)2

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
The graph shows variation of stopping potential VoV_oVo​ with the frequency ν\nuν of the incident radiation for three photosensitive metals X1X_1X1​, X2X_2X2​ and X3X_3X3​. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?
  1. (A)X1X_1X1​
  2. (B)X2X_2X2​
  3. (C)X3X_3X3​
  4. (D)All the metals will give out photo electrons with same kinetic energies.

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
For a certain metal, when monochromatic light of wavelength λ is incident, the stopping potential for photoelectrons is 3V03V_{0}3V0​. When the same metal is illuminated by light of wavelength 2λ, then the stopping potential becomes V0V_{0}V0​. The threshold wavelength for photoelectric emission for the given metal is αλ. The value of α is ________.
  1. (A)1
  2. (B)4
  3. (C)2
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsNumericalJEE Main 2026
The ratio of de Broglie wavelength of a deutron with kinetic energy EEE to that of an alpha particle with kinetic energy 2E2E2E, is nnn : 1. The value of nnn is _________. (Assume mass of proton = mass of neutron) :

Correct answer: 2

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is _____. (Given : h = 6.63×10−346.63 \times 10^{-34}6.63×10−34 J.s and c = 3×1083 \times 10^{8}3×108 m/s)
  1. (A)5×10165 \times 10^{16}5×1016
  2. (B)5×10155 \times 10^{15}5×1015
  3. (C)10×101510 \times 10^{15}10×1015
  4. (D)2×10162 \times 10^{16}2×1016

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is__________m. (h=6.63×10−34 J.s,e=1.6×10−19 C,c=3×108 m/s)(h = 6.63 \times 10^{-34}\,\text{J.s}, e = 1.6 \times 10^{-19}\,\text{C}, c = 3 \times 10^{8}\,\text{m/s})(h=6.63×10−34J.s,e=1.6×10−19C,c=3×108m/s)
  1. (A)2.9×10−82.9 \times 10^{-8}2.9×10−8
  2. (B)2.2×10−82.2 \times 10^{-8}2.2×10−8
  3. (C)3.1×10−73.1 \times 10^{-7}3.1×10−7
  4. (D)2.5×10−72.5 \times 10^{-7}2.5×10−7

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
The de Broglie wavelength of an oxygen molecule at 27°C is x×10−12x \times 10^{-12}x×10−12 m. The value of x is (take Planck's constant =6.63×10−34= 6.63 \times 10^{-34}=6.63×10−34 J.s, Boltzmann constant =1.38×10−23= 1.38 \times 10^{-23}=1.38×10−23 J/K, mass of oxygen. Molecule =5.31×10−26= 5.31 \times 10^{-26}=5.31×10−26 kg).
  1. (A)26
  2. (B)24
  3. (C)30
  4. (D)20

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
Light is incident on a metallic plate having work function 110×10−20110 \times 10^{-20}110×10−20 J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is __________ rad/s. (h =6.63×10−34= 6.63 \times 10^{-34}=6.63×10−34 J.s)
  1. (A)1.04×10161.04 \times 10^{16}1.04×1016
  2. (B)1.04×10131.04 \times 10^{13}1.04×1013
  3. (C)1.66×10161.66 \times 10^{16}1.66×1016
  4. (D)1.66×10151.66 \times 10^{15}1.66×1015

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2026
A light wave described by E = 60sin⁡(3×1015)t+sin⁡(12×1015)t]60\sin(3 \times 10^{15})t + \sin(12 \times 10^{15})t]60sin(3×1015)t+sin(12×1015)t] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) __________ eV. (h = 6.6×10−346.6 \times 10^{-34}6.6×10−34 J-s. and e = 1.6×10−191.6 \times 10^{-19}1.6×10−19C)
  1. (A)5.1
  2. (B)3.8
  3. (C)6.0
  4. (D)7.8

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsNumericalJEE Main 2026
A particle having electric charge 3×10−193\times10^{-19}3×10−19 C and mass 6×10−276\times10^{-27}6×10−27 kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α×10−12\alpha\times10^{-12}α×10−12 m. The value of α\alphaα is __________. (Take Planck's constant =6.6×10−34= 6.6\times10^{-34}=6.6×10−34 J.s)

Correct answer: 10

Step-by-step solution →
Q17·PhysicsIntegerJEE Main 2025
An electron is released from rest near an infinite non-conducting sheet of uniform charge density '−σ-\sigma−σ'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nthn^{th}nth power of time. The numerical value of nnn is ___

Correct answer: 2

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2025
A photo-emissive substance is illuminated with a radiation of wavelength λi\lambda_iλi​ so that it releases electrons with de-Broglie wavelength λe\lambda_eλe​. The longest wavelength of radiation that can emit photoelectron is λ0\lambda_0λ0​. Expression for de-Broglie wavelength is given by: (mmm = mass of the electron, hhh = Planck's constant and ccc = speed of light)
  1. (A)λe=h2mc(1λi−1λ0)\lambda_e=\sqrt{\dfrac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}}λe​=2mc(λi​1​−λ0​1​)h​​
  2. (B)λe=hλ02mc\lambda_e=\sqrt{\dfrac{h\lambda_0}{2mc}}λe​=2mchλ0​​​
  3. (C)λe=h2mc(1λi−1λ0)\lambda_e=\dfrac{h}{\sqrt{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}}λe​=2mc(λi​1​−λ0​1​)​h​
  4. (D)λe=hλi2mc\lambda_e=\sqrt{\dfrac{h\lambda_i}{2mc}}λe​=2mchλi​​​

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both A and R are true but R is NOT the correct explanation of A
  2. (B)A is false but R is true
  3. (C)A is true but R is false
  4. (D)Both A and R are true and R is the correct explanation of A

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2025
The work function of a metal is 3 eV. The colour of the visible light that is required to cause emission of photoelectrons is:
  1. (A)Green
  2. (B)Blue
  3. (C)Red
  4. (D)Yellow

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2025
An electron with mass 'm' with an initial velocity v⃗=v0i^\vec{v}=v_0\hat{i}v=v0​i^ (v0>0)(v_0>0)(v0​>0) enters a magnetic field B⃗=B0j^\vec{B}=B_0\hat{j}B=B0​j^​. If the initial de-Broglie wavelength at t=0t=0t=0 is λ0\lambda_0λ0​, then its value after time ttt would be:
  1. (A)λ01−e2B02t2m2\dfrac{\lambda_0}{\sqrt{1-\frac{e^2B_0^2t^2}{m^2}}}1−m2e2B02​t2​​λ0​​
  2. (B)λ01+e2B02t2m2\dfrac{\lambda_0}{\sqrt{1+\frac{e^2B_0^2t^2}{m^2}}}1+m2e2B02​t2​​λ0​​
  3. (C)λ01+e2B02t2m2\lambda_0\sqrt{1+\frac{e^2B_0^2t^2}{m^2}}λ0​1+m2e2B02​t2​​
  4. (D)λ0\lambda_0λ0​

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2025
A monochromatic light is incident on a metallic plate having work function ϕ\phiϕ. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of the electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is: (Given: the magnitude of charge of an electron is eee and mass is mmm, hhh is Planck’s constant, ccc is velocity of light, magnetic field BBB exists throughout the path of the electron)
  1. (A)2m(hcλ−ϕ)eB\dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eB2m(λhc​−ϕ)​​
  2. (B)m(hcλ−ϕ)eB\dfrac{\sqrt{m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eBm(λhc​−ϕ)​​
  3. (C)8m(hcλ−ϕ)eB\dfrac{\sqrt{8m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eB8m(λhc​−ϕ)​​
  4. (D)2m(hcλ−ϕ)eB\dfrac{2\sqrt{m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eB2m(λhc​−ϕ)​​

Correct answer: (C)

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2025
If λ\lambdaλ and K are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be (graphs of 1K\dfrac{1}{K}K1​ versus λ\lambdaλ):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)(A) is false but (R) is true.
  2. (B)(A) is true but (R) is false.
  3. (C)Both (A) and (R) are true and (R) is the correct explanation of (A).
  4. (D)Both (A) and (R) are true but (R) is not the correct explanation of (A).

Correct answer: (D)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2025
In an experiment with photoelectric effect, the stopping potential,
  1. (A)increases with increase in the wavelength of the incident light
  2. (B)increases with increase in the intensity of the incident light
  3. (C)is (1e)\left(\dfrac{1}{e}\right)(e1​) times the maximum kinetic energy of the emitted photoelectrons
  4. (D)decreases with increase in the intensity of the incident light

Correct answer: (C)

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2025
A proton of mass "mpm_pmp​" has same energy as that of a photon of wavelength "λ\lambdaλ". If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.
  1. (A)1c2Emp\frac{1}{c}\sqrt{\frac{2E}{m_p}}c1​mp​2E​​
  2. (B)1cEmp\frac{1}{c}\sqrt{\frac{E}{m_p}}c1​mp​E​​
  3. (C)1cE2mp\frac{1}{c}\sqrt{\frac{E}{2m_p}}c1​2mp​E​​
  4. (D)12cEmp\frac{1}{2c}\sqrt{\frac{E}{m_p}}2c1​mp​E​​

Correct answer: (C)

Step-by-step solution →
Q27·PhysicsSingle correctJEE Main 2025
Which of the following phenomena can not be explained by wave theory of light?
  1. (A)Reflection of light
  2. (B)Diffraction of light
  3. (C)Refraction of light
  4. (D)Compton effect

Correct answer: (D)

Step-by-step solution →
Q28·PhysicsIntegerJEE Main 2025
The ratio of the power of a light source S1S_1S1​ to the light source S2S_2S2​ is 2. S1S_1S1​ is emitting 2×10152\times10^{15}2×1015 photons per second at 600 nm. If the wavelength of the source S2S_2S2​ is 300 nm, then the number of photons per second emitted by S2S_2S2​ is __________ ×1014\times10^{14}×1014.

Correct answer: 5

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2025
In photoelectric effect, the stopping potential (V0)(V_0)(V0​) v/s frequency (ν)(\nu)(ν) curve is plotted. (hhh is the Planck's constant and ϕ0\phi_0ϕ0​ is work function of metal) (A) V0V_0V0​ v/s ν\nuν is linear. (B) The slope of V0V_0V0​ v/s ν\nuν curve =ϕ0h=\dfrac{\phi_0}{h}=hϕ0​​. (C) hhh is constant is related to the slope of V0V_0V0​ v/s ν\nuν line. (D) The value of electric charge of electron is not required to determine hhh using the V0V_0V0​ v/s ν\nuν curve. (E) The work function can be estimated without knowing the value of hhh. Choose the correct answer from the options given below :
  1. (A)(A), (B) and (C) only
  2. (B)(C) and (D) only
  3. (C)(A), (C) and (E) only
  4. (D)(D) and (E) only

Correct answer: (C)

Step-by-step solution →
Q30·PhysicsSingle correctJEE Main 2025
An electron of mass ‘m’ with an initial velocity v⃗=v0i^ (v0>0)\vec v=v_0\hat i\ (v_0>0)v=v0​i^ (v0​>0) enters an electric field E⃗=−E0k^\vec E=-E_0\hat kE=−E0​k^. If the initial de Broglie wavelength is λ0\lambda_0λ0​, the value after time t would be :-
  1. (A)λ01+e2E02t2m2v02\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}1+m2v02​e2E02​t2​​λ0​​
  2. (B)λ01−e2E02t2m2v02\dfrac{\lambda_0}{\sqrt{1-\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}1−m2v02​e2E02​t2​​λ0​​
  3. (C)λ0\lambda_0λ0​
  4. (D)λ01+e2E02t2m2v02\lambda_0\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}λ0​1+m2v02​e2E02​t2​​

Correct answer: (A)

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2025
A sub-atomic particle of mass 10−3010^{-30}10−30 kg is moving with a velocity 2.21×1052.21\times10^52.21×105 m/s. Under the matter wave consideration, the particle will behave closely like _______ (h=6.63×10−34h=6.63\times10^{-34}h=6.63×10−34 J.s)
  1. (A)Infra-red radiation
  2. (B)X-rays
  3. (C)Gamma rays
  4. (D)Visible radiation

Correct answer: (B)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2025
In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.142.142.14 eV and stopping potential is 222V, what is the wavelength of the em-wave? (Given hc=1242hc=1242hc=1242 eVnm where hhh is the Planck's constant and ccc is the speed of light in vacuum.)
  1. (A)400400400 nm
  2. (B)600600600 nm
  3. (C)200200200 nm
  4. (D)300300300 nm

Correct answer: (D)

Step-by-step solution →
Q33·PhysicsSingle correctJEE Main 2025
The work functions of caesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV respectively. For light of wavelength 550 nm, the photoelectric effect is possible for the case of:
  1. (A)Li only
  2. (B)Cs only
  3. (C)Neither Cs nor Li
  4. (D)Both Cs and Li

Correct answer: (B)

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2025
A light source of wavelength λ\lambdaλ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength λ2\frac{\lambda}{2}2λ​, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV)
  1. (A)2 eV
  2. (B)6 eV
  3. (C)5 eV
  4. (D)4 eV

Correct answer: (C)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2024
A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:
  1. (A)λe>λα>λp\lambda_e > \lambda_\alpha > \lambda_pλe​>λα​>λp​
  2. (B)λα<λp<λe\lambda_\alpha < \lambda_p < \lambda_eλα​<λp​<λe​
  3. (C)λp<λe<λα\lambda_p < \lambda_e < \lambda_\alphaλp​<λe​<λα​
  4. (D)λp>λe>λα\lambda_p > \lambda_e > \lambda_\alphaλp​>λe​>λα​

Correct answer: (B)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2024
UV light of 4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV. The maximum kinetic energy of ejected photoelectrons will be:
  1. (A)1.13 eV
  2. (B)1 eV
  3. (C)3.13 eV
  4. (D)7.26 eV

Correct answer: (B)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2024
A proton and an electron have the same de Broglie wavelength. If KpK_pKp​ and KeK_eKe​ be the kinetic energies of proton and electron respectively, then choose the correct relation:
  1. (A)Kp>KeK_p>K_eKp​>Ke​
  2. (B)Kp=KeK_p=K_eKp​=Ke​
  3. (C)Kp=Ke2K_p=\dfrac{K_e}{2}Kp​=2Ke​​
  4. (D)Kp<KeK_p<K_eKp​<Ke​

Correct answer: (D)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2024
A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h=6.63×10−34h=6.63\times10^{-34}h=6.63×10−34 J s, me=9.0×10−31m_e=9.0\times10^{-31}me​=9.0×10−31 kg and mp=1836m_p=1836mp​=1836 times mem_eme​)
  1. (A)1:18361:18361:1836
  2. (B)1:118361:\dfrac{1}{1836}1:18361​
  3. (C)1:118361:\dfrac{1}{\sqrt{1836}}1:1836​1​
  4. (D)1:18361:\sqrt{1836}1:1836​

Correct answer: (A)

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2024
When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is : (Given hc=1240hc = 1240hc=1240 eV nm)
  1. (A)4 V
  2. (B)4.1 V
  3. (C)2 V
  4. (D)1.5 V

Correct answer: (C)

Step-by-step solution →
Q40·PhysicsSingle correctJEE Main 2024
In a photoelectric experiment, light of energy 2.48 eV irradiates a photosensitive material. The stopping potential was measured to be 0.5 V. The work function of the photosensitive material is:
  1. (A)0.50.50.5 eV
  2. (B)1.681.681.68 eV
  3. (C)2.482.482.48 eV
  4. (D)1.981.981.98 eV

Correct answer: (D)

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2024
Which of the following phenomena is not explained by the wave nature of light? (A) reflection (B) diffraction (C) photoelectric effect (D) interference (E) polarization. Choose the most appropriate answer from the options given below:
  1. (A)(E) only
  2. (B)(C) only
  3. (C)(B), (D) only
  4. (D)(A), (C) only

Correct answer: (B)

Step-by-step solution →
Q42·PhysicsSingle correctJEE Main 2024
Given below are two statements: Statement-I: The figure shows the variation of stopping potential with frequency (ν\nuν) for two photosensitive materials M1M_1M1​ and M2M_2M2​. The slope gives the value of he\dfrac{h}{e}eh​, where hhh is Planck’s constant and eee is the charge of the electron. Statement-II: M2M_2M2​ will emit photoelectrons of greater kinetic energy for incident radiation having the same frequency. In the light of the above statements, choose the most appropriate answer from the options given below.
  1. (A)Statement-I is correct and Statement-II is incorrect.
  2. (B)Statement-I is incorrect but Statement-II is correct.
  3. (C)Both Statement-I and Statement-II are incorrect.
  4. (D)Both Statement-I and Statement-II are correct.

Correct answer: (A)

Step-by-step solution →
Q43·PhysicsSingle correctJEE Main 2024
Which of the following statements is not true about stopping potential (V0)(V_0)(V0​)?
  1. (A)It depends on the nature of emitter material.
  2. (B)It depends upon the frequency of the incident light.
  3. (C)It increases with increase in intensity of the incident light.
  4. (D)It is 1e\tfrac{1}{e}e1​ times the maximum kinetic energy of electrons emitted.

Correct answer: (C)

Step-by-step solution →
Q44·PhysicsSingle correctJEE Main 2024
Which figure shows the correct variation of applied potential difference (V)(V)(V) with photoelectric current (I)(I)(I) at two different intensities of light (I1<I2)(I_1<I_2)(I1​<I2​) of same wavelengths:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2024
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Number of photons increases with increase in frequency of light. Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both A and R are correct and R is NOT the correct explanation of A.
  2. (B)A is correct but R is not correct.
  3. (C)Both A and R are correct and R is the correct explanation of A.
  4. (D)A is not correct but R is correct.

Correct answer: (D)

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2024
The de Broglie wavelengths of a proton and an α\alphaα particle are λ\lambdaλ and 2λ2\lambda2λ respectively. The ratio of the velocities of proton and α\alphaα particle will be:
  1. (A)1:81:81:8
  2. (B)1:21:21:2
  3. (C)4:14:14:1
  4. (D)8:18:18:1

Correct answer: (D)

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2024
Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of photodiode is found to be (X/8)(X/8)(X/8) eV. The value of X is: (Given, h=6.6×10−34h=6.6\times 10^{-34}h=6.6×10−34 Js, e=1.6×10−19e=1.6\times 10^{-19}e=1.6×10−19 C)
  1. (A)15
  2. (B)11
  3. (C)13
  4. (D)21

Correct answer: (A)

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2024
Monochromatic light of frequency 6×10146\times 10^{14}6×1014 Hz is produced by a laser. The power emitted is 2×10−32\times 10^{-3}2×10−3 W. How many photons per second on an average, are emitted by the source? (Given h=6.63×10−34h=6.63\times 10^{-34}h=6.63×10−34 Js)
  1. (A)9×10189\times 10^{18}9×1018
  2. (B)6×10156\times 10^{15}6×1015
  3. (C)5×10155\times 10^{15}5×1015
  4. (D)7×10167\times 10^{16}7×1016

Correct answer: (C)

Step-by-step solution →
Q49·PhysicsSingle correctJEE Main 2024
When a metal surface is illuminated by light of wavelength λ\lambdaλ, the stopping potential is 8V. When the same surface is illuminated by light of wavelength 3λ3\lambda3λ, stopping potential is 2V. The threshold wavelength for this surface is :
  1. (A)5λ5\lambda5λ
  2. (B)3λ3\lambda3λ
  3. (C)9λ9\lambda9λ
  4. (D)4.5λ4.5\lambda4.5λ

Correct answer: (C)

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2024
In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:
  1. (A)Doubled
  2. (B)Quadrupled
  3. (C)Zero
  4. (D)Halved

Correct answer: (C)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Main 2024
The work function of a substance is 3.0 eV3.0\,eV3.0eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:
  1. (A)215 nm215\,nm215nm
  2. (B)414 nm414\,nm414nm
  3. (C)400 nm400\,nm400nm
  4. (D)200 nm200\,nm200nm

Correct answer: (B)

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2024
For the photoelectric effect, the maximum kinetic energy (Ek)(E_k)(Ek​) of the photoelectrons is plotted against the frequency (ν)(\nu)(ν) of the incident photons as shown in figure. The slope of the graph gives:
  1. (A)Ratio of Planck's constant to electric charge
  2. (B)Work function of the metal
  3. (C)Charge of electron
  4. (D)Planck's constant

Correct answer: (D)

Step-by-step solution →
Q53·PhysicsSingle correctJEE Main 2024
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
  1. (A)11\dfrac{1}{1}11​
  2. (B)18\dfrac{1}{8}81​
  3. (C)81\dfrac{8}{1}18​
  4. (D)14\dfrac{1}{4}41​

Correct answer: (B)

Step-by-step solution →
Q54·PhysicsSingle correctJEE Main 2024
Two sources of light emit with a power of 200 W. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be:
  1. (A)1 : 5
  2. (B)1 : 3
  3. (C)5 : 3
  4. (D)3 : 5

Correct answer: (D)

Step-by-step solution →
Q55·PhysicsSingle correctJEE Main 2024
A convex lens of focal length 40 cm forms an image of an extended source of light on a photo-electric cell. A current III is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm. The photoelectric current now is:
  1. (A)I/2I/2I/2
  2. (B)4I4I4I
  3. (C)2I2I2I
  4. (D)III

Correct answer: (D)

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2024
The threshold frequency of a metal with work function 6.63 eV is :
  1. (A)16×101516\times10^{15}16×1015 Hz
  2. (B)16×101216\times10^{12}16×1012 Hz
  3. (C)1.6×10121.6\times10^{12}1.6×1012 Hz
  4. (D)1.6×10151.6\times10^{15}1.6×1015 Hz

Correct answer: (D)

Step-by-step solution →
Q57·PhysicsSingle correctJEE Main 2023
The de Broglie wavelength of an electron having kinetic energy EEE is λ\lambdaλ. If the kinetic energy of electron becomes E4\frac{E}{4}4E​, then its de-Broglie wavelength will be :
  1. (A)λ2\frac{\lambda}{\sqrt{2}}2​λ​
  2. (B)λ2\frac{\lambda}{2}2λ​
  3. (C)2λ2\lambda2λ
  4. (D)2λ\sqrt{2}\lambda2​λ

Correct answer: (C)

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2023
The difference between threshold wavelengths for two metal surfaces AAA and BBB having work function φA=9 eV\varphi_{A}=9\,eVφA​=9eV and φB=4.5 eV\varphi_{B}=4.5\,eVφB​=4.5eV in nm is: (Given, hc=1242 eV nmhc=1242\,eV\,nmhc=1242eVnm)
  1. (A)264264264
  2. (B)138138138
  3. (C)276276276
  4. (D)540540540

Correct answer: (B)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface. Statement II: Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light. In the light of above statements, choose the correct answer from the options given below
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are false

Correct answer: (A)

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2023
A proton and an α\alphaα-particle are accelerated from rest by 2V and 4V potentials, respectively. The ratio of their de-Broglie wavelength is:
  1. (A)4:1
  2. (B)2:1
  3. (C)8:1
  4. (D)16:1

Correct answer: (A)

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2023
The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume mp=me×1849m_p=m_e\times1849mp​=me​×1849)
  1. (A)1:431:431:43
  2. (B)1:301:301:30
  3. (C)1:621:621:62
  4. (D)2:432:432:43

Correct answer: (A)

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2023
A metallic surface is illuminated with radiation of wavelength λ\lambdaλ, the stopping potential is V0V_0V0​. If the same surface is illuminated with radiation of wavelength 2λ2\lambda2λ, the stopping potential becomes V04\dfrac{V_0}{4}4V0​​. The threshold wavelength for this metallic surface will be -
  1. (A)λ4\dfrac{\lambda}{4}4λ​
  2. (B)4λ4\lambda4λ
  3. (C)32λ\dfrac{3}{2}\lambda23​λ
  4. (D)3λ3\lambda3λ

Correct answer: (D)

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2023
The variation of stopping potential (V0)(V_0)(V0​) as a function of the frequency (ν)(\nu)(ν) of the incident light for a metal is shown in the figure. The work function of the surface is:
  1. (A)18.618.618.6 eV
  2. (B)2.982.982.98 eV
  3. (C)2.072.072.07 eV
  4. (D)1.361.361.36 eV

Correct answer: (C)

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2023
The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ1\lambda_1λ1​. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes
  1. (A)12λ1\dfrac{1}{\sqrt2}\lambda_12​1​λ1​
  2. (B)2λ12\lambda_12λ1​
  3. (C)12λ1\dfrac12\lambda_121​λ1​
  4. (D)2λ1\sqrt2\lambda_12​λ1​

Correct answer: (A)

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2023
In photoelectric effect: A. The photocurrent is proportional to the intensity of the incident radiation. B. Maximum kinetic energy with which photoelectrons are emitted depends on the intensity of incident light. C. Maximum K.E. with which photoelectrons are emitted depends on the frequency of incident light. D. The emission of photoelectrons requires a minimum threshold intensity of incident radiation. E. Maximum K.E. of the photoelectrons is independent of the frequency of the incident light. Choose the correct answer from the options below:
  1. (A)A and C only
  2. (B)A and E only
  3. (C)B and C only
  4. (D)A and B only

Correct answer: (A)

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2023
Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, mp=1849 mem_{p}=1849\,m_{e}mp​=1849me​)
  1. (A)1:431:431:43
  2. (B)43:143:143:1
  3. (C)1:18491:18491:1849
  4. (D)1:11:11:1

Correct answer: (D)

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2023
The kinetic energy of an electron, α\alphaα-particle and a proton are given as 4K, 2K4K,\,2K4K,2K and KKK respectively. The de-Broglie wavelength associated with electron (λe)(\lambda e)(λe), α\alphaα-particle (λα)(\lambda\alpha)(λα) and the proton (λp)(\lambda p)(λp) are as follows:
  1. (A)λα=λp=λe\lambda\alpha=\lambda p=\lambda eλα=λp=λe
  2. (B)λα>λp>λe\lambda\alpha>\lambda p>\lambda eλα>λp>λe
  3. (C)λα<λp<λe\lambda\alpha<\lambda p<\lambda eλα<λp<λe
  4. (D)λα<λe<λp\lambda\alpha<\lambda e<\lambda pλα<λe<λp

Correct answer: (C)

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2023
The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is
  1. (A)1.241.241.24
  2. (B)222
  3. (C)111
  4. (D)1.51.51.5

Correct answer: (C)

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2023
The threshold frequency of a metal is f0f_0f0​. When the light of frequency 2f02f_02f0​ is incident on the metal plate, the maximum velocity of photoelectrons is v1v_1v1​. When the frequency of incident radiation is increased to 5f05f_05f0​, the maximum velocity of photoelectrons emitted is v2v_2v2​. The ratio of v1v_1v1​ to v2v_2v2​ is:
  1. (A)v1v2=18\dfrac{v_1}{v_2}=\dfrac18v2​v1​​=81​
  2. (B)v1v2=14\dfrac{v_1}{v_2}=\dfrac14v2​v1​​=41​
  3. (C)v1v2=116\dfrac{v_1}{v_2}=\dfrac{1}{16}v2​v1​​=161​
  4. (D)v1v2=12\dfrac{v_1}{v_2}=\dfrac12v2​v1​​=21​

Correct answer: (D)

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2023
A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of λ\lambdaλ. An alpha particle having certain kinetic energy has the same de-Broglie wavelength λ\lambdaλ. The ratio of kinetic energy of proton and that of alpha particle is:
  1. (A)2:12:12:1
  2. (B)1:21:21:2
  3. (C)1:41:41:4
  4. (D)4:14:14:1

Correct answer: (D)

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2023
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A: The beam of electrons show wave nature and exhibit interference and diffraction. Reason R: Davisson Germer Experimentally verified the wave nature of electrons. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both A and R are correct and R is the correct explanation of A
  2. (B)A is not correct but R is correct
  3. (C)A is correct but R is not correct
  4. (D)Both A and R are correct but R is Not the correct explanation of A

Correct answer: (A)

Step-by-step solution →
Q72·PhysicsSingle correctJEE Main 2023
If a source of electromagnetic radiation having power 151515 kW produces 101610^{16}1016 photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant h=6×10−34h = 6 \times 10^{-34}h=6×10−34 Js)
  1. (A)Micro waves
  2. (B)Ultraviolet rays
  3. (C)Gamma rays
  4. (D)Radio waves

Correct answer: (C)

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2023
If the two metals AAA and BBB are exposed to radiation of wavelength 350350350 nm. The work functions of metals AAA and BBB are 4.84.84.8 eV and 2.22.22.2 eV. Then choose the correct option.
  1. (A)Both metals A and B will emit photo-electrons
  2. (B)Metal A will not emit photo-electrons
  3. (C)Metal B will not emit photo-electrons
  4. (D)Both metals A and B will not emit photo-electrons

Correct answer: (B)

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2023
A small object at rest, absorbs a light pulse of power 202020 mW and duration 300300300 ns. Assuming speed of light as 3×1083\times10^83×108 m/s, the momentum of the object becomes equal to:
  1. (A)3×10−173\times10^{-17}3×10−17 kg m/s
  2. (B)2×10−172\times10^{-17}2×10−17 kg m/s
  3. (C)1×10−171\times10^{-17}1×10−17 kg m/s
  4. (D)0.5×10−170.5\times10^{-17}0.5×10−17 kg m/s

Correct answer: (B)

Step-by-step solution →
Q75·PhysicsNumericalJEE Main 2023
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 242424 W. The radius of curvature of hemisphere is 101010 cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is _________ ×10−8\times10^{-8}×10−8 N

Correct answer: 4

Step-by-step solution →
Q76·PhysicsSingle correctJEE Main 2023
An electron accelerated through a potential difference V1V_1V1​ has a de-Broglie wavelength of λ\lambdaλ. When the potential is changed to V2V_2V2​, its de-Broglie wavelength increases by 50%. The value of V1V2\dfrac{V_1}{V_2}V2​V1​​ is equal to:
  1. (A)333
  2. (B)32\dfrac{3}{2}23​
  3. (C)444
  4. (D)94\dfrac{9}{4}49​

Correct answer: (D)

Step-by-step solution →
Q77·PhysicsSingle correctJEE Main 2023
The ratio of de-Broglie wavelength of an α\alphaα particle and a proton accelerated from rest by the same potential is 1m\dfrac{1}{\sqrt{m}}m​1​, the value of mmm is:
  1. (A)161616
  2. (B)444
  3. (C)222
  4. (D)888

Correct answer: (D)

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2023
The threshold wavelength for photoelectric emission from a material is 5500 Å. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a: (A) 75 W infra-red lamp (B) 10 W infra-red lamp (C) 75 W ultra-violet lamp (D) 10 W ultra-violet lamp. Choose the correct answer from the options given below:
  1. (A)B and C only
  2. (B)A and D only
  3. (C)C only
  4. (D)C and D only

Correct answer: (D)

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source. Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light. In the light of above statements, choose the most appropriate answer from the options given below
  1. (A)Statement I is correct but statement II is incorrect
  2. (B)Statement I is incorrect but statement II is correct
  3. (C)Both Statement I and Statement II are correct
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (C)

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2023
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of λ0\lambda_0λ0​. If the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be:
  1. (A)3λ03\lambda_03λ0​
  2. (B)λ02\dfrac{\lambda_0}{2}2λ0​​
  3. (C)λ02\dfrac{\lambda_0}{\sqrt{2}}2​λ0​​
  4. (D)9λ09\lambda_09λ0​

Correct answer: (C)

Step-by-step solution →
Q81·PhysicsSingle correctJEE Main 2023
An α\alphaα-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:
  1. (A)λα<λp<λe\lambda_\alpha<\lambda_p<\lambda_eλα​<λp​<λe​
  2. (B)λα=λp=λe\lambda_\alpha=\lambda_p=\lambda_eλα​=λp​=λe​
  3. (C)λα>λp>λe\lambda_\alpha>\lambda_p>\lambda_eλα​>λp​>λe​
  4. (D)λα>λp<λe\lambda_\alpha>\lambda_p<\lambda_eλα​>λp​<λe​

Correct answer: (A)

Step-by-step solution →
Q82·PhysicsSingle correctJEE Main 2023
From the photoelectric effect experiment, following observations are made. Identify which of these are correct. A. The stopping potential depends only on the work function of the metal. B. The saturation current increases as the intensity of incident light increases. C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light. D. Photoelectric effect can be explained using wave theory of light. Choose the correct answer from the options given below:
  1. (A)A, C, D only
  2. (B)B, C only
  3. (C)B only
  4. (D)A, B, D only

Correct answer: (C)

Step-by-step solution →
Q83·PhysicsSingle correctJEE Advanced 2022
When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0V . This potential drops to 0.6V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take hc/e =1.24×10−6 JmC−1= 1.24 \times 10^{-6}\,\mathrm{JmC}^{-1}=1.24×10−6JmC−1.]
  1. (A)1.72×10−71.72 \times 10^{-7}1.72×10−7m, 1.20eV
  2. (B)1.72×10−71.72 \times 10^{-7}1.72×10−7m, 5.60eV
  3. (C)3.78×10−73.78 \times 10^{-7}3.78×10−7m, 5.60eV
  4. (D)3.78×10−73.78 \times 10^{-7}3.78×10−7m, 1.20eV

Correct answer: (A)

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2022
An α particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be :
  1. (A)2:1\sqrt{2} : 12​:1
  2. (B)22:12\sqrt{2} : 122​:1
  3. (C)42:14\sqrt{2} : 142​:1
  4. (D)8 : 1

Correct answer: (B)

Step-by-step solution →
Q85·PhysicsSingle correctJEE Main 2022
The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ\lambdaλ. To double the kinetic energy, the incident light must have wavelength :
  1. (A)hcEλ−hc\frac{hc}{E\lambda - hc}Eλ−hchc​
  2. (B)hcλEλ+hc\frac{hc\lambda}{E\lambda + hc}Eλ+hchcλ​
  3. (C)hλEλ+hc\frac{h\lambda}{E\lambda + hc}Eλ+hchλ​
  4. (D)hcλEλ−hc\frac{hc\lambda}{E\lambda - hc}Eλ−hchcλ​

Correct answer: (B)

Step-by-step solution →
Q86·PhysicsSingle correctJEE Main 2022
The equation λ=1.227x\lambda = \frac{1.227}{x}λ=x1.227​ nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for : Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V = Accelerating potential in volts for electron
  1. (A)mK\sqrt{mK}mK​
  2. (B)P\sqrt{P}P​
  3. (C)K\sqrt{K}K​
  4. (D)V\sqrt{V}V​

Correct answer: (D)

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2022
Two streams of photons, possessing energies to five and ten times the work function of metal are incident on the metal surface successively. The ratio of the maximum velocities of the photoelectron emitted, in the two cases respectively, will be
  1. (A)1:21 : 21:2
  2. (B)1:31 : 31:3
  3. (C)2:32 : 32:3
  4. (D)3:23 : 23:2

Correct answer: (C)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2022
An electron (mass m) with an initial velocity v⃗=v0i^ (v0>0)\vec{v} = v_0\hat{i}\,(v_0 > 0)v=v0​i^(v0​>0) is moving in an electric field E⃗=−E0i^ (E0>0)\vec{E} = -E_0\hat{i}\,(E_0 > 0)E=−E0​i^(E0​>0) where E0E_0E0​ is constant. If at t = 0 de Broglie wavelength is λ0=hmv0\lambda_0 = \frac{h}{mv_0}λ0​=mv0​h​, then its de Broglie wavelength after time t is given by
  1. (A)λ0\lambda_0λ0​
  2. (B)λ0(1+eE0tmv0)\lambda_0\left(1 + \frac{eE_0t}{mv_0}\right)λ0​(1+mv0​eE0​t​)
  3. (C)λ0t\lambda_0 tλ0​t
  4. (D)λ0(1+eE0tmv0)\dfrac{\lambda_0}{\left(1 + \frac{eE_0t}{mv_0}\right)}(1+mv0​eE0​t​)λ0​​

Correct answer: (D)

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2022
With reference to the observations in photo-electric effect, identify the correct statements from below: A. The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. B. The value of saturation current increases on moving the source of light away from the metal surface. C. The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. D. The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. E. Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves. Choose the correct answer from the options given below:
  1. (A)A and B only
  2. (B)A and E only
  3. (C)C and E only
  4. (D)D and E only

Correct answer: (B)

Step-by-step solution →
Q90·PhysicsSingle correctJEE Main 2022
A parallel beam of light of wavelength 900 nm and intensity 100 Wm−2^{-2}−2 is incident on a surface perpendicular to the beam. Tire number of photons crossing 1 cm2^{2}2 area perpendicular to the beam in one second is :
  1. (A)3 × 1016^{16}16
  2. (B)4.5 × 1016^{16}16
  3. (C)4.5 × 1017^{17}17
  4. (D)4.5 × 1020^{20}20

Correct answer: (B)

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2022
A nucleus of mass M at rest splits into two parts having masses M′3\frac{M'}{3}3M′​ and 2M′3(M′<M)\frac{2M'}{3}\left(M' < M\right)32M′​(M′<M). The ratio of de Broglie wavelength of two parts will be :
  1. (A)1 : 2
  2. (B)2 : 1
  3. (C)1 : 1
  4. (D)2 : 3

Correct answer: (C)

Step-by-step solution →
Q92·PhysicsSingle correctJEE Main 2022
A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take hc = 1230 eV-nm).
  1. (A)1.537 eV
  2. (B)2.46 eV
  3. (C)0.615 eV
  4. (D)1.23 eV

Correct answer: (C)

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2022
The electric field at the point associated with a light wave is given by E=200[sin⁡(6×1015)t+sin⁡(9×1015)t] Vm−1E = 200\left[\sin\left(6 \times 10^{15}\right)t + \sin\left(9 \times 10^{15}\right)t\right]\ \mathrm{Vm}^{-1}E=200[sin(6×1015)t+sin(9×1015)t] Vm−1 Given : h=4.14×10−15h = 4.14 \times 10^{-15}h=4.14×10−15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be :
  1. (A)1.90 eV
  2. (B)3.27 eV
  3. (C)3.60 eV
  4. (D)3.42 eV

Correct answer: (D)

Step-by-step solution →
Q94·PhysicsSingle correctJEE Main 2022
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The photoelectric effect does not take place, if the energy of the incident radiation is less than the work function of a metal. Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. In the light of the above statements, choose the most appropriate answer from the options given below.
  1. (A)Both A and R are correct and R is the correct explanation of A
  2. (B)Both A and R are correct but R is not the correct explanation of A
  3. (C)A is correct but R is not correct
  4. (D)A is not correct but R is correct

Correct answer: (B)

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2022
Let K1_11​ and K2_22​ be the maximum kinetic energies of photo–electrons emitted when two monochromatic beams of wavelength λ1\lambda_1λ1​ and λ2\lambda_2λ2​, respectively are incident on a metallic surface. If λ1=3λ2\lambda_1 = 3\lambda_2λ1​=3λ2​ then:
  1. (A)K1>K23K_1 > \frac{K_2}{3}K1​>3K2​​
  2. (B)K1<K23K_1 < \frac{K_2}{3}K1​<3K2​​
  3. (C)K1=K23K_1 = \frac{K_2}{3}K1​=3K2​​
  4. (D)K2=K13K_2 = \frac{K_1}{3}K2​=3K1​​

Correct answer: (B)

Step-by-step solution →
Q96·PhysicsSingle correctJEE Main 2022
The de Brogue wavelengths for an electron and a photon are λe\lambda_{e}λe​ and λp\lambda_{p}λp​ respectively. For the same kinetic energy of electron and photon. which of the following presents the correct relation between the de Brogue wavelengths of two ?
  1. (A)λp∝λe2\lambda_{p} \propto \lambda_{e}^{2}λp​∝λe2​
  2. (B)λp∝λe\lambda_{p} \propto \lambda_{e}λp​∝λe​
  3. (C)λp∝λe\lambda_{p} \propto \sqrt{\lambda_{e}}λp​∝λe​​
  4. (D)λp∝1λe\lambda_{p} \propto \sqrt{\frac{1}{\lambda_{e}}}λp​∝λe​1​​

Correct answer: (A)

Step-by-step solution →
Q97·PhysicsSingle correctJEE Main 2022
An α particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelength (λa:λC12)(\lambda_{a} : \lambda_{C12})(λa​:λC12​) is :
  1. (A)1:31:\sqrt{3}1:3​
  2. (B)3:1\sqrt{3}:13​:1
  3. (C)3:13:13:1
  4. (D)2:32:\sqrt{3}2:3​

Correct answer: (B)

Step-by-step solution →
Q98·PhysicsSingle correctJEE Main 2022
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are EeE_{e}Ee​ and pep_{e}pe​ and that of photon are EphE_{ph}Eph​ and pphp_{ph}pph​ respectively. Which of the following is correct?
  1. (A)EeEph=2cv\frac{E_{e}}{E_{ph}}=\frac{2c}{v}Eph​Ee​​=v2c​
  2. (B)EeEph=v2c\frac{E_{e}}{E_{ph}}=\frac{v}{2c}Eph​Ee​​=2cv​
  3. (C)pepph=2cv\frac{p_{e}}{p_{ph}}=\frac{2c}{v}pph​pe​​=v2c​
  4. (D)pepph=v2c\frac{p_{e}}{p_{ph}}=\frac{v}{2c}pph​pe​​=2cv​

Correct answer: (B)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2022
A metal surface is illuminated by a radiation of wavelength 4500 Å. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :
  1. (A)1.36 eV
  2. (B)1.69 eV
  3. (C)2.78 eV
  4. (D)2.23 eV

Correct answer: (A)

Step-by-step solution →
Q100·PhysicsNumericalJEE Main 2022
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0.42 V. If the threshold frequency is x × 101310^{13}1013/s, where x is _______ (nearest integer). (Given, speed light = 3 × 10810^{8}108 m/s, Planck's constant = 6.63 × 10−3410^{-34}10−34 Js)

Correct answer: 35

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2022
Given below are two statements :- Statement I : Davisson-Germer experiment establishes the wave nature of electrons. Statement II : If electrons have wave nature, they can interfere and show diffraction. In the light of the above statements choose the correct answer from the options given below:-
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2022
A proton, a neutron, an electron and an α-particle have same energy. If λp,λn,λe\lambda_p, \lambda_n, \lambda_eλp​,λn​,λe​ and λα\lambda_\alphaλα​ are the de Broglie's wavelengths of proton, neutron, electron and α particle respectively, then choose the correct relation from the following :
  1. (A)λp=λn>λe>λα\lambda_p = \lambda_n > \lambda_e > \lambda_\alphaλp​=λn​>λe​>λα​
  2. (B)λα<λn<λp<λe\lambda_\alpha < \lambda_n < \lambda_p < \lambda_eλα​<λn​<λp​<λe​
  3. (C)λe<λp=λn>λα\lambda_e < \lambda_p = \lambda_n > \lambda_\alphaλe​<λp​=λn​>λα​
  4. (D)λe=λp=λn=λα\lambda_e = \lambda_p = \lambda_n = \lambda_\alphaλe​=λp​=λn​=λα​

Correct answer: (B)

Step-by-step solution →
Q103·PhysicsNumericalJEE Main 2022
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted election is v1v_{1}v1​. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2v_{2}v2​. If v2v_{2}v2​ = x v1v_{1}v1​, the value of x will be ______.

Correct answer: 2

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2022
The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectivly will be :
  1. (A)1 : 1
  2. (B)2 : 1
  3. (C)4 : 1
  4. (D)1 : 4

Correct answer: (B)

Step-by-step solution →
Q105·PhysicsIntegerJEE Advanced 2021
In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals PPP, QQQ and RRR are EPE_PEP​, EQE_QEQ​ and ERE_RER​, respectively, and they are related by EP=2EQ=2ERE_P = 2E_Q = 2E_REP​=2EQ​=2ER​. In this experiment, the same source of monochromatic light is used for metals PPP and QQQ while a different source of monochromatic light is used for the metal RRR. The work functions for metals PPP, QQQ and RRR are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal RRR, in eV, is ___.

Correct answer: 6

Step-by-step solution →
Q106·PhysicsSingle correctJEE Main 2021
The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is : [me_{e}e​ = mass of electron = 9 × 10−31^{-31}−31 kg h = Planck constant = 6.6 × 10−34^{-34}−34 Js kB_{B}B​ = Boltzmann constant = 1.38 × 10−23^{-23}−23 JK−1^{-1}−1]
  1. (A)6.26 nm
  2. (B)8.46 nm
  3. (C)2.26 nm
  4. (D)3.25 nm

Correct answer: (A)

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2021
Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-
  1. (A)(mpme)3/2\left(\frac{m_p}{m_e}\right)^{3/2}(me​mp​​)3/2
  2. (B)memp\sqrt{\frac{m_e}{m_p}}mp​me​​​
  3. (C)mpme\sqrt{\frac{m_p}{m_e}}me​mp​​​
  4. (D)mpme\frac{m_p}{m_e}me​mp​​

Correct answer: (C)

Step-by-step solution →
Q108·PhysicsSingle correctJEE Main 2021
A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :
  1. (A)Kp<KeK_{p}<K_{e}Kp​<Ke​ and Pp=PeP_{p}=P_{e}Pp​=Pe​
  2. (B)Kp=KeK_{p}=K_{e}Kp​=Ke​ and Pp=PeP_{p}=P_{e}Pp​=Pe​
  3. (C)Kp<KeK_{p}<K_{e}Kp​<Ke​ and Pp<PeP_{p}<P_{e}Pp​<Pe​
  4. (D)Kp>KeK_{p}>K_{e}Kp​>Ke​ and Pp=PeP_{p}=P_{e}Pp​=Pe​

Correct answer: (A)

Step-by-step solution →
Q109·PhysicsSingle correctJEE Main 2021
A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?
  1. (A)0.96 V
  2. (B)1.25 V
  3. (C)0.24 V
  4. (D)1.5 V

Correct answer: (B)

Step-by-step solution →
Q110·PhysicsSingle correctJEE Main 2021
In a photoelectric experiment, increasing the intensity of incident light :
  1. (A)increases the number of photons incident and also increases the K.E. of the ejected electrons
  2. (B)increases the frequency of photons incident and increases the K.E. of the ejected electrons.
  3. (C)increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
  4. (D)increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

Correct answer: (D)

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2021
The de-Broglie wavelength of a particle having kinetic energy E is λ\lambdaλ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value ?
  1. (A)19E\frac{1}{9}\mathrm{E}91​E
  2. (B)79E\frac{7}{9}\mathrm{E}97​E
  3. (C)E
  4. (D)169E\frac{16}{9}\mathrm{E}916​E

Correct answer: (B)

Step-by-step solution →
Q112·PhysicsSingle correctJEE Main 2021
In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ\phiϕ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 ×\times× 10−3410^{-34}10−34 Js, c = 3 ×\times× 10810^{8}108 ms−1ms^{-1}ms−1)
  1. (A)1.3 V
  2. (B)1.1 V
  3. (C)1.9 V
  4. (D)0.6 V

Correct answer: (A)

Step-by-step solution →
Q113·PhysicsNumericalJEE Main 2021
A particle of mass 9.1×10−3110^{-31}10−31 kg travels in a medium with a speed of 10610^{6}106 m /s and a photon of a radiation of linear momentum 10−2710^{-27}10−27 kg m /s travels in vacuum. The wavelength of photon is ____ times the wavelength of the particle.

Correct answer: 910

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2021
An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron ?
  1. (A)1.41 eV
  2. (B)7.61 eV .
  3. (C)3.3 eV
  4. (D)No photoelectron would be emitted

Correct answer: (A)

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2021
A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be :
  1. (A)3 : 1
  2. (B)1 : 1
  3. (C)1 : 3
  4. (D)1:31:\sqrt{3}1:3​

Correct answer: (B)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2021
When radiation of wavelength λ\lambdaλ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is:
  1. (A)6λ6\lambda6λ
  2. (B)8λ8\lambda8λ
  3. (C)2λ2\lambda2λ
  4. (D)4λ4\lambda4λ

Correct answer: (D)

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2021
An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :
  1. (A)v2c\frac{v}{2c}2cv​
  2. (B)2cv\frac{2c}{v}v2c​
  3. (C)v3c\frac{v}{3c}3cv​
  4. (D)3cv\frac{3c}{v}v3c​

Correct answer: (A)

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2021
What should be the order of arrangement of de-Broglie wavelength of electron (λe)\left(\lambda_e\right)(λe​), an α−\alpha-α− particle (λα)\left(\lambda_\alpha\right)(λα​) and proton (λp)\left(\lambda_p\right)(λp​) given that all have the same kinetic energy ?
  1. (A)λe=λp=λα\lambda_e = \lambda_p = \lambda_\alphaλe​=λp​=λα​
  2. (B)λe<λp<λα\lambda_e < \lambda_p < \lambda_\alphaλe​<λp​<λα​
  3. (C)λe=λp>λα\lambda_e = \lambda_p > \lambda_\alphaλe​=λp​>λα​
  4. (D)λe>λp>λα\lambda_e > \lambda_p > \lambda_\alphaλe​>λp​>λα​

Correct answer: (D)

Step-by-step solution →
Q119·PhysicsNumericalJEE Main 2021
A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular arc of radius 30cm. The value of B is ________ ×10−7\times 10^{-7}×10−7 T. Given hc = 20×10−2620 \times 10^{-26}20×10−26 J − m, mass of electron = 9×10−319 \times 10^{-31}9×10−31 kg

Correct answer: 125

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2021
An electron of mass mem_eme​ and a proton of mass mpm_pmp​ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :
  1. (A)mpme\frac{m_p}{m_e}me​mp​​
  2. (B)1
  3. (C)memp\frac{m_e}{m_p}mp​me​​
  4. (D)mpme\sqrt{\frac{m_p}{m_e}}me​mp​​​

Correct answer: (D)

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2021
The radiation corresponding to 3→23 \rightarrow 23→2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5×10−45 \times 10^{-4}5×10−4 T. Assume that the radius of the largest circular circular path followed by these electrons is 7mm, the work function of the metal is : (Mass of electron = 9.1×10−319.1 \times 10^{-31}9.1×10−31 kg )
  1. (A)0.82 eV
  2. (B)1.88 eV
  3. (C)1.36 eV
  4. (D)0.16 eV

Correct answer: (A)

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2021
An electron having de-Broglie wavelength λ\lambdaλ is indent on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:
  1. (A)0
  2. (B)2mc λ2h\dfrac{2mc\,\lambda^{2}}{h}h2mcλ2​
  3. (C)hcmc\dfrac{hc}{mc}mchc​
  4. (D)2m2c2λ2h2\dfrac{2m^{2}c^{2}\lambda^{2}}{h^{2}}h22m2c2λ2​

Correct answer: (B)

Step-by-step solution →
Q123·PhysicsNumericalJEE Main 2021
A certain metallic surface is illuminated by monochromatic radiation of wavelength λ\lambdaλ. The stopping potential for photoelectric current for this radiation is 3V03V_{0}3V0​. If the same surface is illuminated with a radiation of wavelength 2λ2\lambda2λ, the stopping potential is V0V_{0}V0​. The threshold wavelength of this surface for photoelectric effect is _________ λ\lambdaλ.

Correct answer: 4

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2021
The speed of electrons in a scanning electron microscope is 1 × 107^{7}7 ms−1^{-1}−1. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:
  1. (A)1837
  2. (B)11837\frac{1}{1837}18371​
  3. (C)1837\sqrt{1837}1837​
  4. (D)11837\frac{1}{\sqrt{1837}}1837​1​

Correct answer: (A)

Step-by-step solution →
Q125·PhysicsSingle correctJEE Main 2021
A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is :-
  1. (A)116\dfrac{1}{16}161​ times the mass of e−^{-}−
  2. (B)8 times the mass of e−^{-}−
  3. (C)16 times the mass of e−^{-}−
  4. (D)18\dfrac{1}{8}81​ times the mass of e−^{-}−

Correct answer: (D)

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2021
Two identical photocathodes receive the light of frequencies f1_{1}1​ and f2_{2}2​ respectively. If the velocities of the photo-electrons coming out are v1_{1}1​ and v2_{2}2​ respectively, then
  1. (A)v12−v22=2hm[f1−f2]v_{1}^{2} - v_{2}^{2} = \frac{2h}{m}\left[f_{1} - f_{2}\right]v12​−v22​=m2h​[f1​−f2​]
  2. (B)v12+v22=2hm[f1+f2]v_{1}^{2} + v_{2}^{2} = \frac{2h}{m}\left[f_{1} + f_{2}\right]v12​+v22​=m2h​[f1​+f2​]
  3. (C)v1+v2=[2hm(f1+f2)]12v_{1} + v_{2} = \left[\frac{2h}{m}\left(f_{1} + f_{2}\right)\right]^{\frac{1}{2}}v1​+v2​=[m2h​(f1​+f2​)]21​
  4. (D)v1−v2=[2hm(f1−f2)]1/2v_{1} - v_{2} = \left[\frac{2h}{m}\left(f_{1} - f_{2}\right)\right]^{1/2}v1​−v2​=[m2h​(f1​−f2​)]1/2

Correct answer: (A)

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2021
An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)
  1. (A)1c(2mE)1/2\frac{1}{c}\left(\frac{2m}{E}\right)^{1/2}c1​(E2m​)1/2
  2. (B)1c(E2m)1/2\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}c1​(2mE​)1/2
  3. (C)(E2m)1/2\left(\frac{E}{2m}\right)^{1/2}(2mE​)1/2
  4. (D)c (2mE)1/2c\,(2mE)^{1/2}c(2mE)1/2

Correct answer: (B)

Step-by-step solution →
Q128·PhysicsSingle correctJEE Main 2021
The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation :
  1. (A)Phase
  2. (B)Intensity
  3. (C)Amplitude
  4. (D)Frequency

Correct answer: (D)

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2021
The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths ? (mP_{P}P​ = 1.00727 u, me_{e}e​ = 0.00055u)
  1. (A)1860 : 1
  2. (B)(1860)2^{2}2 : 1
  3. (C)41.4 : 1
  4. (D)43 : 1

Correct answer: (D)

Step-by-step solution →
Q130·PhysicsNumericalJEE Main 2021
Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is ________.

Correct answer: 1

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2021
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : An electron microscope can achieve better resolving power than an optical microscope. Reason R : The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)A is true but R is false.
  2. (B)Both A and R are true but R is NOT the correct explanation of A.
  3. (C)Both A and R are true and R is the correct explanation of A.
  4. (D)A is false but R is true.

Correct answer: (C)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2021
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43V. The new wavelength is:
  1. (A)400 NM
  2. (B)382 nm
  3. (C)309 nm
  4. (D)329 nm

Correct answer: (B)

Step-by-step solution →
Q133·PhysicsSingle correctJEE Main 2021
An α\alphaα particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα\lambda_\alphaλα​ and λp\lambda_pλp​ respectively. The ratio λpλα\dfrac{\lambda_p}{\lambda_\alpha}λα​λp​​ is :
  1. (A)8
  2. (B)2.8
  3. (C)3.8
  4. (D)7.8

Correct answer: (B)

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2021
An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength λelectron^{electron}electron will be: λProton 1
  1. (A)918918918
  2. (B)183618361836
  3. (C)11836\dfrac{1}{1836}18361​
  4. (D)111

Correct answer: (B)

Step-by-step solution →
Q135·PhysicsNumericalJEE Main 2021
The wavelength of an X-ray beam is 10 Å. The mass of a fictitious particle having the same x energy as that of the X – ray photons is h kg. The value of x is _______. 3

Correct answer: 10

Step-by-step solution →
Q136·PhysicsSingle correctJEE Main 2021
Given below are two statements : Statement – I: Two photons having equal linear momenta have equal wavelengths. Statement-II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below.
  1. (A)Statemnet-I is false but Statement-II is true
  2. (B)Both Statement-I and Statement-II are true
  3. (C)Both Statement-I and Statement-II are false
  4. (D)Statement-I is true but Statement-II is false

Correct answer: (D)

Step-by-step solution →
Q137·PhysicsSingle correctJEE Main 2021
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :
  1. (A)10−210^{-2}10−2 nm
  2. (B)10−310^{-3}10−3 nm
  3. (C)10−410^{-4}10−4 nm
  4. (D)10−110^{-1}10−1 nm

Correct answer: (B)

Step-by-step solution →
Q138·PhysicsSingle correctJEE Main 2021
The de Broglie wavelength of a proton andα\alphaα-particle are equal. The ratio of their velocities is :
  1. (A)4:2
  2. (B)4:1
  3. (C)1:4
  4. (D)4:3

Correct answer: (B)

Step-by-step solution →
Q139·PhysicsSingle correctJEE Main 2020
Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to: (Given: nitrogen molecule weight: 4.64×10−264.64 × 10^{-26}4.64×10−26 kg, Boltzman constant: 1.38×10−231.38 × 10^{-23}1.38×10−23 J/K, Planck constant: 6.63×10−346.63 × 10^{-34}6.63×10−34 Js)
  1. (A)0.24 Å
  2. (B)0.20 Å
  3. (C)0.34 Å
  4. (D)0.44 Å

Correct answer: (A)

Step-by-step solution →
Q140·PhysicsSingle correctJEE Main 2020
An electron, a doubly ionized helium ion (He++)\left(He^{++}\right)(He++) and a proton are having the same kinetic energy. The relation between their respective de – Broglie wavelength λe,λHe++\lambda_e, \lambda_{He^{++}}λe​,λHe++​ and λp\lambda_pλp​ is:
  1. (A)λe>λHe++>λP\lambda_e > \lambda_{He^{++}} > \lambda_Pλe​>λHe++​>λP​
  2. (B)λe<λHe++=λP\lambda_e < \lambda_{He^{++}} = \lambda_Pλe​<λHe++​=λP​
  3. (C)λe>λP>λHe++\lambda_e > \lambda_P > \lambda_{He^{++}}λe​>λP​>λHe++​
  4. (D)λe<λP<λHe++\lambda_e < \lambda_P < \lambda_{He^{++}}λe​<λP​<λHe++​

Correct answer: (C)

Step-by-step solution →
Q141·PhysicsNumericalJEE Main 2020
The surface of a metal is illuminated alternately with photons of energies E1_11​ = 4 eV and E2_22​ = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is __________.

Correct answer: 2.00

Step-by-step solution →
Q142·PhysicsSingle correctJEE Main 2020
Particle A of mass mA=m2m_{A} = \frac{m}{2}mA​=2m​ moving along the x-axis with velocity v0v_{0}v0​ collides elastically with another particle B at rest having mass mB=m3m_{B} = \frac{m}{3}mB​=3m​. If both particles move along the x-axis after the collision, the change Δλ\Delta\lambdaΔλ in de-Broglie wavelength of particle A, in terms of its de-Broglie wavelength (λ0)(\lambda_{0})(λ0​) before collision is:
  1. (A)Δλ=32λ0\Delta\lambda = \frac{3}{2}\lambda_{0}Δλ=23​λ0​
  2. (B)Δλ=52λ0\Delta\lambda = \frac{5}{2}\lambda_{0}Δλ=25​λ0​
  3. (C)Δλ=2λ0\Delta\lambda = 2\lambda_{0}Δλ=2λ0​
  4. (D)Δλ=4λ0\Delta\lambda = 4\lambda_{0}Δλ=4λ0​

Correct answer: (D)

Step-by-step solution →
Q143·PhysicsSingle correctJEE Main 2020
In photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:
  1. (A)Slope of the straight line get more steep
  2. (B)Straight line shifts to left
  3. (C)Straight line shifts to right
  4. (D)Graph does not change

Correct answer: (D)

Step-by-step solution →
Q144·PhysicsSingle correctJEE Main 2020
Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant h=6.62×10−34h = 6.62 \times 10^{-34}h=6.62×10−34 J.s)
  1. (A)1.93 eV
  2. (B)2.10 eV
  3. (C)2.59 eV
  4. (D)2.27 eV

Correct answer: (D)

Step-by-step solution →
Q145·PhysicsSingle correctJEE Main 2020
When the wavelength of radiation falling on a metal is charged from 500 nm to 200 nm the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metals is close to:
  1. (A)1.02 eV
  2. (B)0.61 eV
  3. (C)0.52 eV
  4. (D)0.81 eV

Correct answer: (B)

Step-by-step solution →
Q146·PhysicsSingle correctJEE Main 2020
Two sources of light emit X − rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X − rays to the number density of photons of the visible light of the given wavelengths is:
  1. (A)1500\dfrac{1}{500}5001​
  2. (B)500
  3. (C)250
  4. (D)1250\dfrac{1}{250}2501​

Correct answer: (A)

Step-by-step solution →
Q147·PhysicsSingle correctJEE Main 2020
An electron of mass m and magnitude of charge ∣e∣|e|∣e∣ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is
  1. (A)∣e∣Eth\frac{|e|Et}{h}h∣e∣Et​
  2. (B)−h∣e∣Et-\frac{h}{|e|Et}−∣e∣Eth​
  3. (C)−h∣e∣Et-\frac{h}{|e|E\sqrt{t}}−∣e∣Et​h​
  4. (D)−h∣e∣Et2-\frac{h}{|e|Et^{2}}−∣e∣Et2h​

Correct answer: (D)

Step-by-step solution →
Q148·PhysicsSingle correctJEE Main 2020
A particle moving with kinetic energy E has de Broglie wavelength λ\lambdaλ. If energy ΔE\Delta EΔE ;is added to its energy, the wavelength become λ/2\lambda/2λ/2. Value of ΔE\Delta EΔE, is:
  1. (A)E
  2. (B)3E
  3. (C)2E
  4. (D)4E

Correct answer: (B)

Step-by-step solution →
Q149·PhysicsSingle correctJEE Main 2020
Radiation, with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3×10−43 \times 10^{-4}3×10−4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to:
  1. (A)0.8 eV
  2. (B)1.6 eV
  3. (C)1.8 eV
  4. (D)1.1 eV

Correct answer: (D)

Step-by-step solution →
Q150·PhysicsSingle correctJEE Main 2020
When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAT_ATA​ eV and de-Broglie wavelength λA\lambda_AλA​. The maximum kinetic energy of photoelectrons liberated form another metal B by photon of energy 4.50 eV is TB=(TA−1.5)T_B = (T_A - 1.5)TB​=(TA​−1.5)eV. If the de-Broglie wavelength of these photoelectrons λB=2λA\lambda_B = 2\lambda_AλB​=2λA​, then the work function of metal B is
  1. (A)4 eV
  2. (B)1.5 eV
  3. (C)2 eV
  4. (D)3 eV

Correct answer: (A)

Step-by-step solution →
Q151·PhysicsSingle correctJEE Main 2020
An electron (mass m) with initial velocity v⃗=v0i^+v0j^\vec{v} = v_0\hat{i} + v_0\hat{j}v=v0​i^+v0​j^​ is in an electric field E⃗=−E0k^\vec{E} = -E_0\hat{k}E=−E0​k^. If λ0\lambda_0λ0​ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by
  1. (A)λ01+e2E02t2m2v02\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^{2}E_0^{2}t^{2}}{m^{2}v_0^{2}}}}1+m2v02​e2E02​t2​​λ0​​
  2. (B)λ02+e2E2t2m2v02\dfrac{\lambda_0}{\sqrt{2+\dfrac{e^{2}E^{2}t^{2}}{m^{2}v_0^{2}}}}2+m2v02​e2E2t2​​λ0​​
  3. (C)λ021+e2E2t2m2v02\dfrac{\lambda_0\sqrt{2}}{\sqrt{1+\dfrac{e^{2}E^{2}t^{2}}{m^{2}v_0^{2}}}}1+m2v02​e2E2t2​​λ0​2​​
  4. (D)λ01+e2E2t22m2v02\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^{2}E^{2}t^{2}}{2m^{2}v_0^{2}}}}1+2m2v02​e2E2t2​​λ0​​

Correct answer: (D)

Step-by-step solution →
Q152·PhysicsNumericalJEE Main 2020
A beam of electromagnetic radiation of intensity 6.4×10−56.4 \times 10^{-5}6.4×10−5 W/cm2^{2}2 is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function φ = 2 eV) of surface area 1 cm2^{2}2. If one in 10310^{3}103 photons ejects an electron, total number of electrons ejected in is 10x10^{x}10x. (hch_chc​ = 1240 eVnm, 1 eV = 1.6×10−191.6 \times 10^{-19}1.6×10−19 J), then x is _________.

Correct answer: 11

Step-by-step solution →
Q153·PhysicsSingle correctJEE Main 2020
An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)
  1. (A)c(2mE)1/2^{1/2}1/2
  2. (B)1c(E2m)1/2\dfrac{1}{c}\left(\dfrac{E}{2m}\right)^{1/2}c1​(2mE​)1/2
  3. (C)1c(2Em)1/2\dfrac{1}{c}\left(\dfrac{2E}{m}\right)^{1/2}c1​(m2E​)1/2
  4. (D)(E2m)1/2\left(\dfrac{E}{2m}\right)^{1/2}(2mE​)1/2

Correct answer: (B)

Step-by-step solution →
Q154·PhysicsNumericalJEE Advanced 2019
A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency Ω\OmegaΩ such that 4πMΩh=1024 m−2\frac{4\pi M\Omega}{h} = 10^{24}\ \mathrm{m}^{-2}h4πMΩ​=1024 m−2 with h as Planck's constant. N photons of wavelength λ=8π×10−6\lambda = 8\pi \times 10^{-6}λ=8π×10−6 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 μ\muμm. If the value of N is x×1012x \times 10^{12}x×1012, then the value of x is __________. [Consider the spring as massless]

Correct answer: 1.00

Step-by-step solution →
Q155·PhysicsSingle correctJEE Main 2019
The stopping potential V0V_0V0​ (in volt) as a function of frequency (ν\nuν) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be : (Given : Planck's constant (h) = 6.63 ×\times× 10−34^{-34}−34 Js, electron charge e = 1.6 ×\times× 10−19^{-19}−19 C)
  1. (A)1.82 eV
  2. (B)1.66 eV
  3. (C)2.12 eV
  4. (D)1.95 eV

Correct answer: (B)

Step-by-step solution →
Q156·PhysicsSingle correctJEE Main 2019
A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is [Given Planck's constant h = 6.6×10−346.6 \times 10^{-34}6.6×10−34 Js, speed of light c = 3.0×1083.0 \times 10^{8}3.0×108 m/s]
  1. (A)1×10161 \times 10^{16}1×1016
  2. (B)1.5×10161.5 \times 10^{16}1.5×1016
  3. (C)2×10162 \times 10^{16}2×1016
  4. (D)5×10155 \times 10^{15}5×1015

Correct answer: (D)

Step-by-step solution →
Q157·PhysicsSingle correctJEE Main 2019
A particle P is formed due to a completely inelastic collision of particles x and y having de – Broglie wavelengths λx\lambda_{x}λx​ and λy\lambda_{y}λy​ respectively. If x and y were moving in opposite directions, then the de – Broglie wavelength of P is:
  1. (A)λx+λy\lambda_{x} + \lambda_{y}λx​+λy​
  2. (B)λxλyλx+λy\dfrac{\lambda_{x}\lambda_{y}}{\lambda_{x}+\lambda_{y}}λx​+λy​λx​λy​​
  3. (C)λxλy∣λx−λy∣\dfrac{\lambda_{x}\lambda_{y}}{|\lambda_{x}-\lambda_{y}|}∣λx​−λy​∣λx​λy​​
  4. (D)λx−λy\lambda_{x} - \lambda_{y}λx​−λy​

Correct answer: (C)

Step-by-step solution →
Q158·PhysicsSingle correctJEE Main 2019
The electric field of light wave is given as E⃗=10−3cos⁡(2πx5×10−7−2π×6×1014t)x^NC\vec{E}=10^{-3}\cos\left(\frac{2\pi x}{5\times10^{-7}}-2\pi\times6\times10^{14}t\right)\hat{x}\frac{N}{C}E=10−3cos(5×10−72πx​−2π×6×1014t)x^CN​. This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is:
  1. (A)0.48 V
  2. (B)2.48 V
  3. (C)0.72 V
  4. (D)2.0 V

Correct answer: (A)

Step-by-step solution →
Q159·PhysicsSingle correctJEE Main 2019
Two particles move at right angle to each other. Their de Broglie wavelengths are λ1\lambda_1λ1​ and λ2\lambda_2λ2​ respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength λ\lambdaλ, of the final particle, is given by:
  1. (A)λ=λ1λ2\lambda = \sqrt{\lambda_1 \lambda_2}λ=λ1​λ2​​
  2. (B)λ=λ1+λ22\lambda = \dfrac{\lambda_1 + \lambda_2}{2}λ=2λ1​+λ2​​
  3. (C)2λ=1λ1+1λ2\dfrac{2}{\lambda} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2}λ2​=λ1​1​+λ2​1​
  4. (D)1λ2=1λ12+1λ22\dfrac{1}{\lambda^{2}} = \dfrac{1}{\lambda_1^{2}} + \dfrac{1}{\lambda_2^{2}}λ21​=λ12​1​+λ22​1​

Correct answer: (D)

Step-by-step solution →
Q160·PhysicsSingle correctJEE Main 2019
A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelength λAλB\dfrac{\lambda_A}{\lambda_B}λB​λA​​ is close to:
  1. (A)10.00
  2. (B)0.07
  3. (C)14.14
  4. (D)4.47

Correct answer: (C)

Step-by-step solution →
Q161·PhysicsSingle correctJEE Main 2019
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is −V0_{0}0​/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is −V0_{0}0​. The threshold frequency for photoelectric emission is :
  1. (A)5v3\frac{5v}{3}35v​
  2. (B)43v\frac{4}{3}v34​v
  3. (C)2 v
  4. (D)3v2\frac{3v}{2}23v​

Correct answer: (D)

Step-by-step solution →
Q162·PhysicsSingle correctJEE Main 2019
If the deBroglie wavelength of an electron is equal to 10−310^{-3}10−3 times the wavelength of a photon of frequency 6×10146 \times 10^{14}6×1014 Hz, then the speed of electron is equal to: (Speed of light = 3 x 10 = 8 = m/s ; Planck's constant = 6.63×10−346.63 \times 10^{-34}6.63×10−34 J.s ; Mass of electron = 9.1×10−319.1 \times 10^{-31}9.1×10−31 kg)
  1. (A)1.1×1061.1 \times 10^{6}1.1×106 m/s
  2. (B)1.7×1061.7 \times 10^{6}1.7×106 m/s
  3. (C)1.8×1061.8 \times 10^{6}1.8×106 m/s
  4. (D)1.45×1061.45 \times 10^{6}1.45×106 m/s

Correct answer: (D)

Step-by-step solution →
Q163·PhysicsSingle correctJEE Main 2019
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to (hce=1240 nm−V)\left(\frac{hc}{e}=1240\text{ nm}-\text{V}\right)(ehc​=1240 nm−V)
  1. (A)0.5 V
  2. (B)1.5 V
  3. (C)1.0 V
  4. (D)2.0 V

Correct answer: (C)

Step-by-step solution →
Q164·PhysicsSingle correctJEE Main 2019
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5×10−127.5 \times 10^{-12}7.5×10−12 m, the minimum electron energy required is close to:
  1. (A)500 keV
  2. (B)100 keV
  3. (C)1 keV
  4. (D)25 keV

Correct answer: (D)

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2019
The magnetic field associated with a light wave is given, at the origin, by B=B0[sin⁡(3.14×107)ct+sin⁡(6.28×107)ct]B = B_0[\sin(3.14 \times 10^{7})ct + \sin(6.28 \times 10^{7})ct]B=B0​[sin(3.14×107)ct+sin(6.28×107)ct] If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons?
  1. (A)6.82 eV
  2. (B)12.5 eV
  3. (C)8.52 eV
  4. (D)7.72 eV

Correct answer: (D)

Step-by-step solution →
Q166·PhysicsSingle correctJEE Main 2019
Surface of certain metal is first illuminated with light of wavelength λ1\lambda_{1}λ1​ = 350 nm and then, the light of wavelength λ2\lambda_{2}λ2​ = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to: (Energy of photon = 1240λ(in nm)\dfrac{1240}{\lambda(\text{in nm})}λ(in nm)1240​eV)
  1. (A)1.8
  2. (B)2.5
  3. (C)5.6
  4. (D)1.4

Correct answer: (A)

Step-by-step solution →
Q167·PhysicsNumericalJEE Advanced 2018
In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F=n×10−4F = n \times 10^{-4}F=n×10−4 N due to the impact of the electrons. The value of n is ___________. Mass of the electron me=9×10−31m_{e} = 9 \times 10^{-31}me​=9×10−31 kg and 1.0 eV =1.6×10−19= 1.6 \times 10^{-19}=1.6×10−19 J.

Correct answer: 24.00

Step-by-step solution →
Q168·PhysicsSingle correctJEE Advanced 2017
A photoelectric material having work-function ϕ0\phi_{0}ϕ0​ is illuminated with light of wavelength λ\lambdaλ (λ<hcϕ0)\left(\lambda < \dfrac{hc}{\phi_{0}}\right)(λ<ϕ0​hc​). The fastest photoelectron has a de Broglie wavelength λd\lambda_{d}λd​. A change in wavelength of the incident light by Δλ\Delta\lambdaΔλ results in change Δλd\Delta\lambda_{d}Δλd​ in λd\lambda_{d}λd​. then the ratio Δλd/Δλ\Delta\lambda_{d}/\Delta\lambdaΔλd​/Δλ is proportional to
  1. (A)λd/λ\lambda_{d}/\lambdaλd​/λ
  2. (B)λd2/λ\lambda_{d}^{2}/\lambdaλd2​/λ
  3. (C)λd3/λ\lambda_{d}^{3}/\lambdaλd3​/λ
  4. (D)λd3/λ2\lambda_{d}^{3}/\lambda^{2}λd3​/λ2

Correct answer: (D)

Step-by-step solution →
Q169·PhysicsSingle correctJEE Advanced 2016
In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ\lambdaλ) of incident light and the corresponding stopping potential (V0V_0V0​) are given below: λ\lambdaλ (μm) | V0V_0V0​ (Volt) 0.3 | 2.0 0.4 | 1.0 0.5 | 0.4 Given that c=3×108 ms−1c = 3 \times 10^{8}\ \text{ms}^{-1}c=3×108 ms−1 and e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}e=1.6×10−19 C, Planck's constant (in units of J s) found from such an experiment is
  1. (A)6.0×10−346.0 \times 10^{-34}6.0×10−34
  2. (B)6.4×10−346.4 \times 10^{-34}6.4×10−34
  3. (C)6.6×10−346.6 \times 10^{-34}6.6×10−34
  4. (D)6.8×10−346.8 \times 10^{-34}6.8×10−34

Correct answer: (B)

Step-by-step solution →
Q170·PhysicsMultiple correctJEE Advanced 2016
Light of wavelength λph\lambda_{ph}λph​ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is ϕ\phiϕ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe\lambda_{e}λe​, which of the following statement(s) is(are) true?
  1. (A)For large potential difference (V >> ϕ/e\phi/eϕ/e), λe\lambda_{e}λe​ is approximately halved if V is made four times
  2. (B)λe\lambda_{e}λe​ increases at the same rate as λph\lambda_{ph}λph​ for λph<hc/ϕ\lambda_{ph} < hc/\phiλph​<hc/ϕ
  3. (C)λe\lambda_{e}λe​ is approximately halved, if d is doubled
  4. (D)λe\lambda_{e}λe​ decreases with increase in ϕ\phiϕ and λph\lambda_{ph}λph​

Correct answer: (A)

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Q171·PhysicsMultiple correctJEE Advanced 2015
For photo-electric effect with incident photon wavelength λ\lambdaλ, the stopping potential is V0V_{0}V0​. Identify the correct variation(s) of V0V_{0}V0​ with λ\lambdaλ and 1/λ1/\lambda1/λ.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A), (C)

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Q172·PhysicsSingle correctJEE Advanced 2014
A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1u_{1}u1​ and u2u_{2}u2​, respectively. If the ratio u1:u2=2:1u_{1} : u_{2} = 2 : 1u1​:u2​=2:1 and hc=1240hc = 1240hc=1240 eV nm, the work function of the metal is nearly
  1. (A)3.7 eV
  2. (B)3.2 eV
  3. (C)2.8 eV
  4. (D)2.5 eV

Correct answer: (A)

Step-by-step solution →
Q173·PhysicsIntegerJEE Advanced 2013
The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is

Correct answer: 1

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Q174·PhysicsSingle correctJEE Advanced 2013
A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3×1083\times 10^{8}3×108 m/s. The final momentum of the object is
  1. (A)0.3×10−170.3\times 10^{-17}0.3×10−17 kg ms−1^{-1}−1
  2. (B)1.0×10−171.0\times 10^{-17}1.0×10−17 kg ms−1^{-1}−1
  3. (C)3.0×10−173.0\times 10^{-17}3.0×10−17 kg ms−1^{-1}−1
  4. (D)9.0×10−179.0\times 10^{-17}9.0×10−17 kg ms−1^{-1}−1

Correct answer: (B)

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Dual Nature of Matter and Radiation — frequently asked

How many questions from Dual Nature of Matter and Radiation appear in JEE?

Dual Nature of Matter and Radiation has appeared in 155 of the last 186 JEE Main and JEE Advanced papers — about 83% of them — contributing 174 questions in total across those papers.

Is Dual Nature of Matter and Radiation an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 83% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Dual Nature of Matter and Radiation questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

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