JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Modern Physics

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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 \overset{˚}{A}

. What is the maximum kinetic energy of the emitted photoelectron?

(A)1.41 eV✓ Correct
(B)7.61 eV
(C)3.3 eV
(D)No photoelectron would be emitted

Answer: (A)

Solution

The electron arrives with

3

eV of kinetic energy from rest at infinity and settles in the second excited state,

n = 3

, where its energy is

E_{3} = - \frac{13.6}{3 ^{2}} = - 1.51 eV

The photon carries away all the energy lost:

E_{photon} = 3 + 1.51 = 4.51 eV

The threshold wavelength fixes the work function of the metal:

ϕ = \frac{12400 eV A ˚}{4000 A ˚} = 3.1 eV

K_{m a x} = E_{photon} - ϕ = 4.51 - 3.1 = 1.41 eV

The maximum kinetic energy of the emitted photoelectron is

1.41

eV.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Modern Physics

Solution

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