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Question

(B) The de-Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqr

Vedclass · Accepted Answer

$2 \sqrt{2}: 1$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.Here,$h$ is Planck's constant,$m$ is the mass of the particle,and $q$ is the charge of the particle.For a proton,let mass be $m_p = m$ and charge be $q_p = e$.For an alpha particle,mass $m_{\alpha} = 4m$ and charge $q_{\alpha} = 2e$.Since both are accelerated through the same potential difference $V$,the ratio of their wavelengths is:$\frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_p q_p V}}}{\frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$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 $2\sqrt{2}: 1$.

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