A proton and alpha particle are accelerated through the same potential difference. The ratio of the…
A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha will be (mass of alpha particle is four times mass of proton.)
$1: 2$
$2 \sqrt{2}: 1$
$1: 1$
$2: 1$
Solution
De-Broglie wavelength is given by
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$
If $\lambda_1$ and $\lambda_2$ are de-Broglie wavelengths of proton and alpha particle then
$\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{m}_2 \mathrm{q}_2}{\mathrm{~m}_1 \mathrm{q}_1}}=\sqrt{4 \times 2}=\sqrt{8}=2 \sqrt{2}$
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