Logo
Jeeify
MCQMediumJEE 2026Photoelectric Effect

JEE Physics 2026 Question with Solution

A light wave described by E=60[sin⁡(3×1015t)+sin⁡(12×1015t)]E = 60[\sin(3 \times 10^{15}t) + \sin(12 \times 10^{15}t)] (in SI units) falls on a metal surface of work function 2.8 eV2.8 \, \text{eV}. The maximum kinetic energy of ejected photoelectron is (approximately) _____ eV\text{eV}. (h=6.6×10−34 J.sh = 6.6 \times 10^{-34} \, \text{J.s} and e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C})

  • A

    3.83.8

  • B

    5.15.1

  • C

    6.06.0

  • D

    7.87.8

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The incident light contains two angular frequencies, ω1=3×1015 rad/s\omega_1 = 3 \times 10^{15} \, \text{rad/s} and ω2=12×1015 rad/s\omega_2 = 12 \times 10^{15} \, \text{rad/s}. The work function is ϕ=2.8 eV\phi = 2.8 \, \text{eV}.

Find: The maximum kinetic energy of the emitted photoelectron.

In photoelectric effect, one photon interacts with one electron. Therefore, the maximum kinetic energy is determined by the photon of highest frequency.

So we take

ωmax⁡=12×1015 rad/s\omega_{\max} = 12 \times 10^{15} \, \text{rad/s}

Photon energy is

Ephoton=hf=hω2πE_{\text{photon}} = hf = \frac{h\omega}{2\pi}

Substituting the given values,

Ephoton=(6.6×10−34)(12×1015)2πE_{\text{photon}} = \frac{(6.6 \times 10^{-34})(12 \times 10^{15})}{2\pi} Ephoton≈79.2×10−196.283≈12.605×10−19 JE_{\text{photon}} \approx \frac{79.2 \times 10^{-19}}{6.283} \approx 12.605 \times 10^{-19} \, \text{J}

Converting to electron-volt,

Ephoton=12.605×10−191.6×10−19≈7.88 eVE_{\text{photon}} = \frac{12.605 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 7.88 \, \text{eV}

Now apply Einstein's photoelectric equation:

Kmax⁡=Ephoton−ϕK_{\max} = E_{\text{photon}} - \phi Kmax⁡=7.88−2.8=5.08 eVK_{\max} = 7.88 - 2.8 = 5.08 \, \text{eV}

Hence, the maximum kinetic energy is approximately 5.1 eV5.1 \, \text{eV}. Therefore, the correct option is B.

Common mistakes

  • Using the sum of the two frequencies or assuming both photons combine to eject one electron. This is wrong because photoelectric effect involves absorption of one photon by one electron. Use only the highest frequency photon to find the maximum kinetic energy.

  • Treating 3×10153 \times 10^{15} and 12×101512 \times 10^{15} as ordinary frequencies ff instead of angular frequencies ω\omega. This is wrong because the given wave is of the form sin⁡(ωt)\sin(\omega t). Convert using f=ω/(2π)f = \omega/(2\pi).

  • Subtracting the work function in joules from photon energy in electron-volts, or mixing units midway. This is wrong because both quantities must be in the same unit before subtraction. Convert photon energy fully into eV first, then apply Kmax⁡=E−ϕK_{\max} = E - \phi.

Practice more Photoelectric Effect questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Photoelectric Effect

Related questions