Question
An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 \mathrm{~V}$. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_{\mathrm{p}}$ respectively. The ratio $\frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}$, to the nearest integer, is

Answer

$ \frac{1}{2} \mathrm{mv}^2=\mathrm{qV} $

$ \lambda=\frac{\mathrm{h}}{\mathrm{mv}} $

$ \lambda=\sqrt{8} \simeq 3 .$

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