Question

The number of 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/t ,
are made to fall incidently on sodium metal (work function, W = 2.3 eV). The maximum possible kinetic energy of emitted photoelectrons is:​

Answer 1

Answer:

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 Hñatom 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