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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.)
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The correct answer is:
$2 \sqrt{2}: 1$
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
$$
\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{m}_2 \mathrm{q}_2}{\mathrm{~m}_1 \mathrm{q}_1}}=\sqrt{4 \times 2}=\sqrt{8}=2 \sqrt{2}
$$
$$
\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}
$$
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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