Question 108 of 239
A hydrogen atom changes its state from to . Due to recoil, the percentage change in the wave length of emitted light is approximately . The value of is_____.
[Given mass of the hydrogenatom ]
Show answer and solution
Answer: 7
$\begin{aligned}
& \Delta \mathrm{E}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.9 \mathrm{eV} \\
& \Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \\
& \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}} \\
& \mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{t}} \\
& 0=-\mathrm{mv}+\frac{\mathrm{h}}{\lambda^{\prime \prime}} \\
& \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}} \\
& \Delta \mathrm{E}=\frac{1}{2} \mathrm{mv^{2 }}+\frac{\mathrm{hc}}{\lambda^{\prime \prime}} \\
& =\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\right)^2+\frac{\mathrm{hc}}{\lambda^{\prime}}
\end{aligned}$
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
& \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