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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton,$\lambda_p = \frac{h}{\sqrt{2 m_p q_p V}}$.For an $\alpha$-pa

Vedclass · Accepted Answer

$\frac{\lambda}{2 \sqrt{2}}$

Vedclass · Answer

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton,$\lambda_p = \frac{h}{\sqrt{2 m_p q_p V}}$.For an $\alpha$-particle,the mass $m_{\alpha} = 4 m_p$ and the charge $q_{\alpha} = 2 q_p$.Substituting these values,$\lambda_{\alpha} = \frac{h}{\sqrt{2 (4 m_p) (2 q_p) V}} = \frac{h}{\sqrt{16 m_p q_p V}} = \frac{1}{4} \frac{h}{\sqrt{m_p q_p V}}$.Comparing this with $\lambda_p$,we get $\lambda_{\alpha} = \frac{\lambda}{2 \sqrt{2}}$.

$A$ proton accelerated through a potential $V$ has de-Broglie wavelength $\lambda$. Then the de-Broglie wavelength of an $\alpha$-particle,when accelerated through the same potential $V$ is :

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