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 $ Å.
(1997 (C), 2M)
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Answer:
Correct Answer: 76. $(63.56$ $ V)$
Solution:
- If accelerated by potential difference of $V$ volt, then
$\begin{aligned} \dfrac{1}{2} m v^{2} & =e V \\ \Rightarrow \quad \dfrac{p^{2}}{2 m} & =e V, \text { here } p=\operatorname{momentum}(m v) \end{aligned}$
Using de-Broglie equation, $\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m e V}}$
$\Rightarrow \quad 1.54 \times 10^{-10}=\dfrac{6.625 \times 10^{-34}}{\left(2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} V\right)^{1 / 2}}$
Solving for $V$ gives : $V=63.56$ $ V$.
BETA
