Question

(Marking scheme for all questions will be: correct answer +3 and wrong answer -1)

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 $h/\pi$ , are made to fall incidently on sodium metal (work function, $W=2.3\text{eV})$. The maximum possible kinetic energy of emitted photoelectrons is:

• A. $7.9\text{eV}$
• B. $0.25\text{eV}$
• C. $10.45\text{eV}$
• D. $9.79\text{eV}$

Answer 1

The correct option is D $9.79\text{eV}$

$\text{Difference in angular momentum}=\frac{h}{\pi }$
$\therefore (n_2-n_1)\frac{h}{2\pi }=\frac{h}{\pi }$
$\therefore n_2-n_1=2\text{(Difference in shell no.)}$
For photoelectric effect to be observed,
$\text{Energy of photon > Work function}\left(2.3\text{eV}\right)$
$\therefore$Two photons are possible in H-atom where difference in shell number is 2 and energy $>2.3\text{eV}$
$\therefore E_{photon}=12.09\text{eV}$ (From 3 $\to$ 1 transition) and $2.55\text{eV}$ (From 4 $\to$ 2 transition)
Maximum KE of photoelectron will correspond to maximum energy of the incident photon.
$\therefore (KE{)}_{max}=12.09–2.3=9.79\text{eV}$