Question: Q. 7. A proton and an $\alpha$-particle have the same de Broglie wavelength. Determine the ratio of (i) their accelerating potentials and (ii) their speeds.
U [Delhi I, II, III 2014]
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Solution:
Ans. (i)
$$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 m q V}} \Rightarrow V=\frac{h^{2}}{2 m q \lambda^{2}}{ }^{1 / 2} \ m_{\alpha} & =4 m_{p^{\prime}} q_{\alpha}=2 q_{p} \ \frac{V_{p}}{V_{\alpha}} & =\frac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}=\frac{4 m_{p} \times 2 q_{p}}{m_{p} q_{p}} \ & =8: 1 \end{aligned} $$
$$ \Rightarrow \quad \frac{V_{p}}{V_{\alpha}}=\frac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}=\frac{4 m_{p} \times 2 q_{p}}{m_{p} q_{p}} \quad 1 / 2 $$
(ii)
$$ \lambda=\frac{h}{m v} \Rightarrow v=\frac{h}{m \lambda} \quad 1 / 2 $$
$$ \Rightarrow \quad \frac{v_{p}}{v_{\alpha}}=\frac{m_{\alpha}}{m_{p}}=4 $$
[CBSE Marking Scheme 2014]
