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

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ for 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 $(p)$: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$For an $\alpha$-particle $(\alpha)$: $\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}}$Given that $m_{\alpha} = 4m_p$ and $q_{\alpha} = 2q_p$:$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{4m_p 	imes 2q_p}{m_p 	imes q_p}} = \sqrt{8} = 2\sqrt{2}$Thus, the ratio of their de-Broglie wavelengths is $2\sqrt{2}$.

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

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