An electron and an alpha particle are accelerated by the same potential difference. Let and denote the de Broglie wavelengths of the electron and the alpha particle, respectively, then [2024]

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Solution

Q.
An electron and an alpha particle are accelerated by the same potential difference. Let $λ_{e}$ and $λ_{α}$ denote the de Broglie wavelengths of the electron and the alpha particle, respectively, then [2024]

1 $λ_{e} > λ_{α}$

2 $λ_{e} = 4 λ_{α}$

3 $λ_{e} = λ_{α}$

4 $λ_{e} < λ_{α}$

Ans.
(1)

The de-Broglie wavelength in terms of potential difference is given by,

$λ = \frac{h}{\sqrt{2 m q v}}$

For same potential difference,

$λ \propto \frac{1}{\sqrt{m}} and λ \propto \frac{1}{\sqrt{q}}$

So, $λ_{e} > λ_{α} (∵ m_{α} and q_{α} are greater than m_{e} and q_{e})$

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