Physics - Modern Physics Question with Solution | TestHub

$$\begin{array}{rl}& \Delta \mathrm{E}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.9\mathrm{eV}\\ & \Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda }\\ & \lambda =\frac{\mathrm{hc}}{\Delta \mathrm{E}}\\ & {\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{t}}\\ & 0=-\mathrm{mv}+\frac{\mathrm{h}}{{\lambda }^{\prime \prime }}\\ & \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m}{\lambda }^{\prime }}\\ & \Delta \mathrm{E}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2+\frac{\mathrm{hc}}{{\lambda }^{\prime \prime }}\\ & =\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{h}}{\mathrm{m}{\lambda }^{\prime }}\right)}^2+\frac{\mathrm{hc}}{{\lambda }^{\prime }}\end{array}$$

Now $$\begin{array}{rl}& \Delta \mathrm{E}=\frac{h^2}{2\mathrm{m}{\lambda }^{{\prime }^2}}+\frac{\mathrm{hc}}{{\lambda }^{\prime }}\\ & {\lambda }^{\prime 2}\Delta \mathrm{E}-\mathrm{hc}{\lambda }^{\prime }-\frac{{\mathrm{h}}^2}{2\mathrm{m}}=0\\ & {\lambda }^{\prime }=\frac{\mathrm{hc}\pm \sqrt{{\mathrm{h}}^2{\mathrm{c}}^2+\frac{4\Delta {\mathrm{Eh}}^2}{2\mathrm{m}}}}{2\Delta \mathrm{E}}\\ & {\lambda }^{\prime }=\frac{\mathrm{hc}\pm \mathrm{hc}\sqrt{1+\frac{2\Delta \mathrm{E}}{\mathrm{mc}}}}{2\Delta \mathrm{E}}\end{array}$$ $$\begin{array}{rl}& \frac{{\lambda }^{\prime }}{\lambda }=\frac{1+{\left(1+\frac{2\Delta \mathrm{E}}{{\mathrm{mc}}^2}\right)}^{\frac{1}{2}}}{2}=\frac{1+1+\frac{\Delta \mathrm{E}}{{\mathrm{mc}}^2}}{2}\\ & \frac{{\lambda }^{\prime }}{\lambda }=1+\frac{\Delta \mathrm{E}}{2{\mathrm{mc}}^2}\\ & \frac{{\lambda }^{\prime }-\lambda }{\lambda }=\frac{\Delta \mathrm{E}}{2{\mathrm{mc}}^2}=\frac{1.9\times 1.6\times 10^{-19}}{2\times 1.67\times 10^{-27}\times 9\times 10^{15}}=10^{-9}\\ & \therefore \%\text{ change }\approx 10^{-7}\end{array}$$Correct answer 7