If lambda_0 is the de-Broglie wavelength for | vedclass

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(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda =

Vedclass · Accepted Answer

$\frac{\lambda_0}{2 \sqrt{2}}$

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(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton,$\lambda_0 = \frac{h}{\sqrt{2m_p q_p V}}$.For an $\alpha$-particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2q_p$.Substituting these values into the formula for the $\alpha$-particle:$\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(2q_p)V}} = \frac{h}{\sqrt{8(2m_p q_p V)}} = \frac{1}{\sqrt{8}} \left( \frac{h}{\sqrt{2m_p q_p V}} \right)$.Since $\sqrt{8} = 2\sqrt{2}$,we get $\lambda_{\alpha} = \frac{\lambda_0}{2\sqrt{2}}$.

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

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