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

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

Vedclass · Accepted Answer

$8$

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(B) The de-Broglie wavelength $\lambda$ of a particle of charge $q$ and mass $M$ accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2MqV}}$.For an $\alpha$-particle, $M_{\alpha} = 4M_p$ and $q_{\alpha} = 2e$, where $M_p$ is the mass of a proton and $e$ is the elementary charge.For a proton, $M_p = M_p$ and $q_p = e$.The ratio of wavelengths is $\frac{\lambda_{\alpha}}{\lambda_p} = \frac{\frac{h}{\sqrt{2(4M_p)(2e)V}}}{\frac{h}{\sqrt{2(M_p)(e)V}}} = \sqrt{\frac{2M_p e V}{16M_p e V}} = \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}}$.Comparing this with $\frac{1}{\sqrt{m}}$, we get $m = 8$.

The ratio of 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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