Q86 / 120 · VITEEE 2008 · Physics
An electron microscope is used to probe the atomic arrangements to a resolution of $5 Å$. What should be the electric potential to which the electrons need to be accelerated?
- A$2.5 \mathrm{~V}$
- Correct answerB$5 \mathrm{~V}$
- C$2.5 \mathrm{kV}$
- D$5 \mathrm{kV}$
Solution
We have, $\quad d \sin \varphi=n \lambda$
For $\phi=90^{\circ}$ 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.
Hence, the correct answer is option: 'B'.