A proton and an electron are accelerated by the same potential di…

Question

A proton and an electron are accelerated by the same potential difference. Let $\lambda_\text{e}$ and $\lambda_\text{p}$ denote the de Broglie wavelengths of the electron and the proton respectively.

$\lambda_\text{e}=\lambda_\text{p}.$
$\lambda_\text{e}<\lambda_\text{p}.$
$\lambda _\text{e}>\lambda_\text{p}.$
The relation between $\lambda_\text{e}$, and $\lambda_\text{p}$ depends on the accelerating potential difference.

✓

Answer

$\lambda _\text{e}>\lambda_\text{p}.$

Explanation:

Let m_e and m_p be the masses of electron and proton, respectively.

Let the applied potential difference be V.

Thus, the de-Broglie wavelength of the electron,

$\lambda=\frac{\text{h}}{\sqrt{2\text{m}_\text{e}\text{eV}}}\dots(1)$

And de-Broglie wavelength of the proton,

$\lambda=\frac{\text{h}}{\sqrt{2\text{m}_\text{p}\text{eV}}}\dots(2)$

Dividing equation (2) by equation (1), we get:

$\frac{\lambda_\text{p}}{\lambda_\text{e}}=\frac{\sqrt{\text{m}_\text{e}}}{\sqrt{\text{m}_\text{p}}}$

$\text{m}_\text{e}<\text{m}_\text{p}$

$\therefore\frac{\lambda_\text{p}}{\lambda_\text{e}}<1$

$\Rightarrow\lambda_ \text{p}<\lambda_\text{e}$

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