The ratio of the de-Broglie wavelength of an | vedclass

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(C) The de-Broglie wavelength $\lambda$ of a particle of mass $M$ and charge $q$ accelerated through a potential $V$ is given by $\lambda = \frac{h}{\

Vedclass · Accepted Answer

$8$

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(C) The de-Broglie wavelength $\lambda$ of a particle of mass $M$ and charge $q$ accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2MqV}}$.For an $\alpha$-particle,$M_{\alpha} = 4M_p$ and $q_{\alpha} = 2e$. For a proton,$M_p = M_p$ and $q_p = e$.The ratio is $\frac{\lambda_{\alpha}}{\lambda_p} = \sqrt{\frac{M_p q_p}{M_{\alpha} q_{\alpha}}} = \sqrt{\frac{M_p \cdot e}{4M_p \cdot 2e}} = \sqrt{\frac{1}{8}}$.Comparing this with $\frac{1}{\sqrt{m}}$,we get $m = 8$.

The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is $........$

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