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 the process is

Then wavelength

\(=\operatorname{Rcz}^{2} \frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}=\operatorname{Rc}(1)^{2}\left[1-\frac{1}{4}\right]\)

\(\lambda=\operatorname{Rc}\left[\frac{3}{4}\right]\)   ..... \((1)\)

Forionized lithium

\(\lambda=\operatorname{Rc}(3)^{2}\left[\frac{1}{n^{2}}\right]=\operatorname{Rc} 9\left[\frac{1}{n^{2}}\right]\)    ..... \((2)\)

\(\operatorname{Rc}\left[\frac{3}{4}\right]=\operatorname{Rc9}\left[\frac{1}{n^{2}}\right]\)

\(\Rightarrow \frac{3}{4}=\frac{9}{n^{2}} \Rightarrow n=\sqrt{12}=2 \sqrt{3}\)

The least quantum number must be \(4\).