JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

A light wave described by $E = 60 sin (3 \times 1 0^{15}) t + sin (12 \times 1 0^{15}) t]$ (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 = $6.6 \times 1 0^{- 34}$ J-s. and e = $1.6 \times 1 0^{- 19}$ C)

A
5.1
B
3.8
C
6.0
D
7.8

Solution

The maximum-KE photoelectrons come from the higher-frequency component: $ω = 12 \times 1 0^{15}$ rad/s, $ν = ω / (2 π) \approx 1.91 \times 1 0^{15}$ Hz.

Photon energy: $E_{γ} = h ν = (6.6 \times 1 0^{- 34}) (1.91 \times 1 0^{15}) \approx 1.26 \times 1 0^{- 18}$ J $= 1.26 \times 1 0^{- 18} / (1.6 \times 1 0^{- 19}) \approx 7.9$ eV.

Einstein's photoelectric equation: $K_{m a x} = E_{γ} - ϕ_{0} = 7.9 - 2.8 = 5.1$ eV.

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →