An electron and a proton are accelerated through the same potential...

Question

An electron and a proton are accelerated through the same potential difference. Let e and p be the de Broglie wavelength associated with electron and proton respectively. Then e/p is 1 (A) me/mp (B) mp/me me mp mp me

Accepted Answer

$\lambda=\dfrac{h}{\sqrt{2mqV}}$. For the same $V$ and same magnitude of charge $e$ for both electron and proton:

$$\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}$$

(Since $m_p \gg m_e$, this ratio is much greater than 1 — the electron has a much longer de Broglie wavelength.)

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