A hydrogen atom changes its state from $n=3$ to $n=2$. Due to recoil, the percentage change in the wave…

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 $\mathrm{Rhc}=13.6 \mathrm{eV}, \mathrm{hc}=1242 \mathrm{eV} \mathrm{nm}, \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ mass of the hydrogenatom $=1.6 \times 10^{-27} \mathrm{~kg}$ ]

Solution

$\begin{aligned} & \Delta \mathrm{E}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.9 \mathrm{eV} \\ & \Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \\ & \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}} \\ & \mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{t}} \\ & 0=-\mathrm{mv}+\frac{\mathrm{h}}{\lambda^{\prime \prime}} \\ & \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}} \\ & \Delta \mathrm{E}=\frac{1}{2} \mathrm{mv^{2 }}+\frac{\mathrm{hc}}{\lambda^{\prime \prime}} \\ & =\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\right)^2+\frac{\mathrm{hc}}{\lambda^{\prime}} \end{aligned}$
Now $\begin{aligned} & \Delta \mathrm{E}=\frac{h^2}{2 \mathrm{~m} \lambda^{\prime^2}}+\frac{\mathrm{hc}}{\lambda^{\prime}} \\ & \lambda^{\prime 2} \Delta \mathrm{E}-\mathrm{hc} \lambda^{\prime}-\frac{\mathrm{h}^2}{2 \mathrm{~m}}=0 \\ & \lambda^{\prime}=\frac{\mathrm{hc} \pm \sqrt{\mathrm{h}^2 \mathrm{c}^2+\frac{4 \Delta \mathrm{Eh}^2}{2 \mathrm{~m}}}}{2 \Delta \mathrm{E}} \\ & \lambda^{\prime}=\frac{\mathrm{hc} \pm \mathrm{hc} \sqrt{1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}}}}{2 \Delta \mathrm{E}} \end{aligned}$ $\begin{aligned} & \frac{\lambda^{\prime}}{\lambda}=\frac{1+\left(1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^2}\right)^{\frac{1}{2}}}{2}=\frac{1+1+\frac{\Delta \mathrm{E}}{\mathrm{mc}^2}}{2} \\ & \frac{\lambda^{\prime}}{\lambda}=1+\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^2} \\ & \frac{\lambda^{\prime}-\lambda}{\lambda}=\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^2}=\frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 9 \times 10^{15}}=10^{-9} \\ & \therefore \% \text { change } \approx 10^{-7} \end{aligned}$
Correct answer 7

Asked in: JEE Main 2024 (04 Apr Shift 1)

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