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(B) The kinetic energy $K$ of a charged particle accelerated through a potential difference $V$ is given by $K = qV$.The momentum $p$ of the particle

Vedclass · Accepted Answer

$\sqrt{8}$

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(B) The kinetic energy $K$ of a charged particle accelerated through a potential difference $V$ is given by $K = qV$.The momentum $p$ of the particle is related to kinetic energy by $p = \sqrt{2mK} = \sqrt{2mqV}$.The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.For a proton,$m_p = m$ and $q_p = e$. For an alpha particle,$m_{\alpha} = 4m$ and $q_{\alpha} = 2e$.Taking the ratio of the wavelengths:$\frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_p q_p V}}}{\frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{4m 	imes 2e}{m 	imes e}} = \sqrt{8}$.

$A$ proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is

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