An alpha-particle and a proton are accelerate | vedclass

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(B) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m q V}}$.Since both particle

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$1: 2 \sqrt{2}$

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(B) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m q V}}$.Since both particles are accelerated from rest by the same potential $V$,the ratio of their wavelengths is:$\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{h / \sqrt{2 m_{\alpha} q_{\alpha} V}}{h / \sqrt{2 m_{p} q_{p} V}} = \sqrt{\frac{m_{p} q_{p}}{m_{\alpha} q_{\alpha}}}$.We know that the mass of an $\alpha$-particle is $4$ times the mass of a proton $(m_{\alpha} = 4 m_{p})$ and the charge of an $\alpha$-particle is $2$ times the charge of a proton $(q_{\alpha} = 2 q_{p})$.Substituting these values:$\frac{\lambda_{\alpha}}{\lambda_{p}} = \sqrt{\frac{m_{p} q_{p}}{(4 m_{p})(2 q_{p})}} = \sqrt{\frac{1}{8}} = \frac{1}{2 \sqrt{2}}$.Thus,the ratio is $1 : 2 \sqrt{2}$.

An $\alpha$-particle and a proton are accelerated from rest by the same potential,then the ratio of their de-Broglie wavelength is

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