A hydrogen atom changes its state from $n=3$ to $n=2$. Due to recoil, the percentage change in the wave length of emitted light is approximately $1 \times 10^{-n}$. The value of $n$ is _______.

[Given 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 hydrogenatom $=1.6 \times 10^{-27} \mathrm{~kg}$]

<p>$$\begin{aligned} \Delta E & =13.6 \mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}\right) \\ & =\frac{68}{36} \mathrm{eV}=1.89 \mathrm{eV} \end{aligned}$$</p> <p>Due to recoil of hydrogen atom, the energy of emitted photon will decrease by very small amount.</p> <p>So for approximate calculations,</p> <p>$$\begin{aligned} & \% \text { charge }= \frac{\Delta E_{\text {atom }}}{\Delta E} \times 100 \\ &=\frac{\frac{\left(\frac{\Delta E}{C}\right)^2}{2 m}}{\Delta E} \times 100 \\ &=\frac{\Delta E}{C^2 \times 2 m} \times 100 \\ &=\frac{1.89 \times 1.6 \times 10^{-19} \times 100}{\left(3 \times 10^8\right)^2 \times 2 \times 1.6 \times 10^{-27}} \\ &=1.05 \times 10^{-7} \% \\ & \therefore n=7 \end{aligned}$$</p>