A hydrogen atom changes its state from \(n=3\) to \(\mathrm{n}=2\). Due to recoil, the percentage change in the wave length of emitted light is approximately \(1 \times 10^{-\mathrm{n}}\). The value of \(\mathrm{n}\) is _______. [Given \(\mathrm{Rhc}=13.6 \mathrm{eV}, \mathrm{hc}=1242 \mathrm{eV} \mathrm{nm}\), \(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\), mass of the hydrogen atom \(\left.=1.6 \times 10^{-27} \mathrm{~kg}\right]\)

\(\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}\)

\(\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}\)

\(\mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{f}}\)

\(0=-\mathrm{mv}+\frac{\mathrm{h}}{\lambda^{\prime}}\)

\(\Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\)

\(\Delta \mathrm{E}=\frac{1}{2} \mathrm{mv^{2 }}+\frac{\mathrm{hc}}{\lambda^{\prime}}\)

\(=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\right)^2+\frac{\mathrm{hc}}{\lambda^{\prime}}\)

Now

\(\Delta \mathrm{E}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^{\prime 2}}+\frac{\mathrm{hc}}{\lambda^{\prime}}\)

\(\lambda^{\prime 2} \Delta \mathrm{E}-\mathrm{hc} \lambda^{\prime}-\frac{\mathrm{h}^2}{2 \mathrm{~m}}=0\)

\(\lambda^{\prime}=\frac{\mathrm{hc} \pm \sqrt{\mathrm{h}^2 \mathrm{c}^2+\frac{4 \Delta \mathrm{Eh}^2}{2 \mathrm{~m}}}}{2 \Delta \mathrm{E}}\)

\(\lambda^{\prime}=\frac{\mathrm{hc} \pm \mathrm{hc} \sqrt{1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^2}}}{2 \Delta \mathrm{E}}\)

\(\frac{\lambda^{\prime}}{\lambda}=\frac{1+\left(1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^2}\right)^{\frac{1}{2}}}{2}=\frac{1+1+\frac{\Delta \mathrm{E}}{\mathrm{mc}^2}}{2}\)

\(\frac{\lambda^{\prime}}{\lambda}=1+\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^2}\)

\(\frac{\lambda^{\prime}-\lambda}{\lambda}=\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^2}=\frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 9 \times 10^{16}}=10^{-9}\)

\(\therefore \% \text { change } \approx 10^{-7}\)

Correct answer \(7\)