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

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$2.8$

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(D) 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,$\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$.For an $\alpha$ particle,$\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}}$.The ratio is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$.We know that the mass of an $\alpha$ particle is $m_{\alpha} = 4m_p$ and its charge is $q_{\alpha} = 2e$,while for a proton $m_p = m_p$ and $q_p = e$.Substituting these values: $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p 	imes 2e}{m_p 	imes e}} = \sqrt{8} = 2\sqrt{2}$.Using $\sqrt{2} \approx 1.414$,we get $\frac{\lambda_p}{\lambda_{\alpha}} = 2 	imes 1.414 = 2.828 \approx 2.8$.

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \ V$. After this,their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$ is

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