A proton and an alpha-particle are accelerate | vedclass

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Question

(A) The de-Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $\Delta V$ is given by $\lambda = \frac{h}{\s

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $\Delta V$ is given by $\lambda = \frac{h}{\sqrt{2mq\Delta V}}$.For a proton $(p)$ and an $\alpha$-particle $(\alpha)$:Mass of proton $m_p = m$, charge $q_p = e$.Mass of $\alpha$-particle $m_\alpha = 4m$, charge $q_\alpha = 2e$.Given potentials: $V_p = 2\,V$ and $V_\alpha = 4\,V$.The ratio of wavelengths is $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha V_\alpha}{m_p q_p V_p}}$.Substituting the values:$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m 	imes 2e 	imes 4V}{m 	imes e 	imes 2V}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.Thus, the ratio $\lambda_p : \lambda_\alpha = 4:1$.

$A$ proton and an $\alpha$-particle are accelerated from rest by $2\,V$ and $4\,V$ potentials, respectively. The ratio of their de-Broglie wavelength is: (in $:1$)

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