Work function, \(W=2.75 \mathrm{eV}\)

According to Einstein's photoelectric equation

\(e V_{0}=h v-W \quad\) or \(\quad h v=e V_{0}+W\)

\(=10 \mathrm{eV}+2.75 \mathrm{eV}=12.75 \mathrm{eV}\)    ..... \((i)\)

When an electron in the hydrogen atom makes transition from excited state \(n\) to the gound state \((n=1), \text { then the frequency ( } v)\) of the emitted photor is given by

\(hv = {E_n} - {E_1} \Rightarrow \) \(hv =  - \frac{{13.6}}{{{n^2}}} - \left( { - \frac{{13.6}}{{{1^2}}}} \right)\)

\(\left[ {\because {\text{ For hydrogen atom, }}{E_n} =  - \frac{{13.6}}{{{n^2}}}{\text{eV}}} \right]\)

According to given problem

\({-\frac{13.6}{n^{2}}+13.6=12.75 \quad(\mathrm{U} \sin \mathrm{g}(\mathrm{i}))}\)

\({\frac{13.6}{n^{2}}=0.85 \Rightarrow n^{2}=\frac{13.6}{0.85}=16}\)

or  \({n=4}\)