JEE Main 2024 — 4 April, Shift 1 — Question 50

$\Delta \mathrm{E}=13.6\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.9 \mathrm{eV}$

$\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$

$\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$

$\mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{f}}$

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

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

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

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

Now

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

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

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

$\lambda^{\prime}=\frac{h c \pm h c \sqrt{1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^{2}}}}{2 \Delta \mathrm{E}}$

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

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

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

$\therefore \%$ change $\approx 10^{-7}$