Question

Photons emitted when electrons in a H-atom make transition from a higher energy state to lower energy state, whose difference in angular momentum is h/π , are made to incident on sodium metal (work function, W = 2.3 eV). The maximum possible kinetic energy of emitted photoelectrons is :

(A) 7.9 eV

(B) 0.25 eV

(C) 10.45 eV

(D) 9.79 eV

Answer 1

Correct option (D) 9.79 eV

Explanation:

Difference in angular momentum = h/π

∴ (n2 - n1)h/2π = h/π

∴ n2 - n1 = 2 (Difference in shell no.)

For photoelectric effect to be observed,

Energy of photon > Work function (2.3 eV)

∴ Two photons are possible in Hatom where difference in shell number is 2 and energy > 2.3 eV

∴ Ephoton = 12.09 eV (From 3 → 1 transition)

& 2.55 eV (From 4 → 2 transition)

Max KE of photoelectron will correspond to max energy of incident photon.

∴ (KE)max = 12.09 - 2.3 = 9.79 eV