Dual Nature of Matter and Radiation — JEE Main practice

37 questions

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Sample questions with solutions

Q1 · 2026

A light wave described by E=60[sin⁡(3×1015)t+sin⁡(12×1015)t]E=60\left[\sin \left(3 \times 10^{15}\right) t+\sin \left(12 \times 10^{15}\right) t\right] (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=6.6×10−34\left(h=6.6 \times 10^{-34}\right. J.s. and e=1.6×10−19C)\left.e=1.6 \times 10^{-19} \mathrm{C}\right)

  • A.

    3.8

  • B.

    7.8

  • C.

    6.0

  • D.

    5.1

Answer: D
  1. Einstein's photoelectric equation says maximum kinetic energy equals photon energy minus work function.

Kmax⁡=E−ϕK_{\max} = E - \phi

  1. Photon energy relates to angular frequency by E=hν=hω2πE = h\nu = \dfrac{h\omega}{2\pi}.

  2. The given field has two angular frequencies mixed together: ω1=3×1015 rad/s\omega_1 = 3\times10^{15}\ \mathrm{rad/s} and ω2=12×1015 rad/s\omega_2 = 12\times10^{15}\ \mathrm{rad/s}. Since higher frequency means higher photon energy, and Kmax⁡K_{\max} depends on the most energetic photons present, we use ωmax⁡=12×1015 rad/s\omega_{\max} = 12\times10^{15}\ \mathrm{rad/s}.

  3. Converting this photon energy to eV:

E(in eV)=h⋅ωmax⁡2π⋅e=(6.6×10−34)(12×1015)2×3.14×(1.6×10−19)≈7.88 eVE(\text{in eV}) = \frac{h\cdot\omega_{\max}}{2\pi\cdot e} = \frac{(6.6\times10^{-34})(12\times10^{15})}{2\times3.14\times(1.6\times10^{-19})} \approx 7.88\ \mathrm{eV}

  1. Therefore:

Kmax⁡=7.88−2.8=5.08 eV≈5.1 eVK_{\max} = 7.88 - 2.8 = 5.08\ \mathrm{eV} \approx 5.1\ \mathrm{eV}

Hence, the answer is Option D: 5.1 eV.

Q2 · 2026

Light is incident on a metallic plate having work function 110×10−20 J110 \times 10^{-20} \mathrm{~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 J.s)\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right).

  • A.

    1.04×10131.04 \times 10^{13}

  • B.

    1.66×10161.66 \times 10^{16}

  • C.

    1.66×10151.66 \times 10^{15}

  • D.

    1.04×10161.04 \times 10^{16}

Answer: D
  1. When photoelectrons come out with zero kinetic energy, the incident light frequency must equal the threshold frequency, meaning the photon energy exactly equals the work function.

hν=ϕh\nu = \phi

  1. Given ϕ=110×10−20 J=1.10×10−18 J\phi = 110\times10^{-20}\ \mathrm{J} = 1.10\times10^{-18}\ \mathrm{J}, solve for ν\nu.

ν=ϕh=1.10×10−186.63×10−34≈1.66×1015 Hz\nu = \frac{\phi}{h} = \frac{1.10\times10^{-18}}{6.63\times10^{-34}} \approx 1.66\times10^{15}\ \mathrm{Hz}

  1. Angular frequency relates to ordinary frequency through ω=2πν\omega = 2\pi\nu, so:

ω=2π×1.66×1015≈1.04×1016 rad/s\omega = 2\pi \times 1.66\times10^{15} \approx 1.04\times10^{16}\ \mathrm{rad/s}

Hence, the answer is Option D: 1.04×1016 rad/s1.04 \times 10^{16}\ \mathrm{rad/s}.

Q3 · 2026

The de Broglie wavelength of an oxygen molecule at 27∘C27^{\circ} \mathrm{C} is x×10−12 mx \times 10^{-12} \mathrm{~m}. The value of xx is (take Planck's constant =6.63×10−34 J.s=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, Boltzmann constant =1.38×10−23 J/K=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}, mass of oxygen molecule =5.31×10−26 kg=5.31 \times 10^{-26} \mathrm{~kg} )

  • A.

    24

  • B.

    30

  • C.

    20

  • D.

    26

Answer: D
  1. For a gas molecule, kinetic theory gives its average kinetic energy in terms of temperature.

K=32kBTK = \frac{3}{2}k_BT

  1. Momentum relates to kinetic energy as p=2mKp = \sqrt{2mK}, so substituting KK gives:

p=2m(32kBT)=3mkBTp = \sqrt{2m\left(\frac{3}{2}k_BT\right)} = \sqrt{3mk_BT}

  1. The de Broglie wavelength is λ=h/p\lambda = h/p, so combining with the momentum expression:

λ=h3mkBT\lambda = \frac{h}{\sqrt{3mk_BT}}

  1. Convert the temperature to Kelvin:

T=27+273=300 KT = 27 + 273 = 300\ \mathrm{K}

  1. Substituting all values:

λ=6.63×10−343×(5.31×10−26)×300≈2.58×10−11 m=25.8×10−12 m\lambda = \frac{6.63\times10^{-34}}{\sqrt{3\times(5.31\times10^{-26})\times300}} \approx 2.58\times10^{-11}\ \mathrm{m} = 25.8\times10^{-12}\ \mathrm{m}

So x≈26x \approx 26.

Hence, the answer is Option D: 26.

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