Q.Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of $100\ \text{V}$.

Equating de Broglie wavelengths and using $\lambda = h/\sqrt{2mqV}$ gives $V_p:V_\alpha = 8:1$.

For a charged particle of mass $m$ and charge $q$ accelerated from rest through a potential difference $V$, the kinetic energy gained is $qV = p^2/2m$, so the momentum is $p=\sqrt{2mqV}$ and the de Broglie wavelength is

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$

For the proton: mass $m_p$, charge $q_p = e$.

For the alpha particle: mass $m_\alpha = 4m_p$, charge $q_\alpha = 2e$.

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