Find the ratio of the de Broglie wavelengths | vedclass

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Question

(A) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For

Vedclass · Accepted Answer

$2\sqrt{2} : 1$

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(A) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton $(p)$: $m_p = m$, $q_p = e$.For an $\alpha$-particle $(\alpha)$: $m_{\alpha} = 4m$, $q_{\alpha} = 2e$.The ratio of the wavelengths is:$\frac{\lambda_p}{\lambda_{\alpha}} = \frac{h / \sqrt{2m_p q_p V}}{h / \sqrt{2m_{\alpha} q_{\alpha} V}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$.Substituting the values:$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m 	imes 2e}{m 	imes e}} = \sqrt{8} = 2\sqrt{2}$.Thus, the ratio is $2\sqrt{2} : 1$.

Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.

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