A hydrogen atom makes a transition from $n=2$ to $n=1$ and emits a photon. This photon strikes a dou

Question

A hydrogen atom makes a transition from $n=2$ to $n=1$ and emits a photon. This photon strikes a doubly ionized lithium atom ( $Z=3$ ) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for this process is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(4) : Energy of emitted photon $ \ E=\left[\frac{1}{1^2}-\frac{1}{2^2}ight] 	imes 13.6 \mathrm{eV}=\frac{3}{4} 	imes 13.6 \mathrm{eV} \ $ Energy required to completely remove the electron from $n^{	ext {th }}$ excited state of doubly ionized lithium, As $E \geq E^{\prime} \Rightarrow \frac{3}{4} 	imes 13.6 \geq \frac{13.6 	imes 9}{n^2}$ \ $ \Rightarrow n^2 \geq 3 	imes 4 	ext { or } n \geq \sqrt{12}=3.46 $ \ $	herefore \quad$ Least quantum number for the excited state $=4$.