Physics - Modern Physics Question with Solution | TestHub

We have, $d\sin \phi =n\lambda$ For $\varphi =90^{\circ }$ and $\mathrm{n}=1$, we get $\mathrm{d}=\lambda$ $\begin{array}{l}\text{ But }\lambda =\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{meV}}}=\sqrt{\frac{{\mathrm{h}}^2}{2\mathrm{meV}}}\\ =\sqrt{\left(\frac{{\left(6.63\times 10^{-34}\right)}^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times \mathrm{V}}\right)}\\ =\sqrt{\frac{1.5}{\mathrm{V}}\times 10^{-9}\mathrm{m}}\therefore \mathrm{d}=\sqrt{\frac{1.5}{\mathrm{V}}}\times 10^{-9}\end{array}$ or $5\times 10^{-10}=\sqrt{\frac{1.5}{\mathrm{V}}}\times 10^{-9}$ or $0.5=\sqrt{\frac{1.5}{\mathrm{V}}}$ or $0.5\times 0.5=\frac{1.5}{\mathrm{V}}\Rightarrow \mathrm{V}=\frac{1.5}{0.5\times 0.5}=6\mathrm{V}$ No option is matching with the exact answer. But $5\mathrm{V}$ is approximately equal to the exact potential. Therefore, option (b) should be the correct option.