Calculate the ratio of the accelerating potential required to accelerate (i) a proton and (ii) an $α$ -particle to have the same de-Broglie wavelength associated with them. [Given : Mass of proton = 1.6 x 10^-27 kg; Mass of $α$ -particle = 6.4 x 10^-27 kg]

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8:1

4:3

3:1

1:2

answer is D.

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Detailed Solution

If a particle is accelerated through a potential difference V, then

$qV = \frac{1}{2} {mv}^{2} = \frac{p^{2}}{2 m} ⇒ p = \sqrt{2 mqV}$

de Broglie wavelength

$\begin{array}{l} λ = \frac{h}{p} = \frac{h}{\sqrt{2 mqV}} \\ As λ_{p} = λ_{α}, ⇒ \frac{h}{\sqrt{2 m_{p} q_{p} V_{p}}} = \frac{h}{\sqrt{2 m_{α} q_{α} V_{α}}} \\ or, \frac{V_{p}}{V_{α}} = \frac{m_{α} q_{α}}{m_{p} q_{p}} = \frac{(6.4 \times 10^{- 27}) \times 2 e}{(1.6 \times 10^{- 27}) \times e} = 8 \end{array}$