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.
✓
Answer
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$:$
$\frac{\lambda_\text{p}}{\lambda_\text{e}}=\frac{\sqrt{\text{m}_\text{e}}}{\sqrt{\text{m}_\text{p}}}$
$\text{m}_\text{e}<\text{m}_\text{p}$
$\therefore\frac{\lambda_\text{p}}{\lambda_\text{e}}<1$
$\Rightarrow\lambda_ \text{p}<\lambda_\text{e}$
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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