Question

Photons emitted when electrons in a hydrogen atom make a transition from a higher energy state to a lower energy state, whose difference in angular momentum is $\frac{h}{\pi }$, are made to be 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

The correct option is D 9.79 $eV$

Angular momentum = $\frac{nh}{2\pi }$
For two successive states $n_1,n_2$ the difference in angular momentum;
$(n_2-n_1)\times \frac{h}{2\pi }=\frac{h}{2\pi }$
$(n_2-n_1)=2$

For photons to be emitted from sodium,
Energy of incident photons from the hydrogen atom > Work function of sodium

Therefore;
Two photons are possible in H-atom where difference in shell number is 2 and energy > 2.3 $eV$
$E{}_{\text{photon}}=12.90eV(3\to 1\text{transition})$
$E{}_{\text{photon}}=2.55eV(4\to 2\text{transition})$
The maximum kinetic energy of the photoelectrons will correspond to the maximum energy of incident photon.
Therefore;
$(K.E{)}_{max}=12.09-2.3=9.79eV$