Atomic Structure - Result Question 100
####76. An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54 \AA$.
(1997 (C), 2M)
Show Answer
Answer:
Correct Answer: 76. $(471 nm)$
Solution:
- If accelerated by potential difference of $V$ volt, then
$$ \begin{aligned} \frac{1}{2} m v^{2} & =e V \ \Rightarrow \quad \frac{p^{2}}{2 m} & =e V, \text { here } p=\operatorname{momentum}(m v) \end{aligned} $$
Using de-Broglie equation, $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e V}}$
$$ \Rightarrow \quad 1.54 \times 10^{-10}=\frac{6.625 \times 10^{-34}}{\left(2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} V\right)^{1 / 2}} $$
Solving for $V$ gives : $V=63.56 V$.
