A proton and an alpha-particle are accelerate | vedclass

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

Vedclass · Accepted Answer

$2\sqrt{2} : 1$

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(C) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mQV}}$.Since $h$ and $V$ are constant for both particles,we have $\lambda \propto \frac{1}{\sqrt{mQ}}$.Therefore,the ratio of the wavelength of the proton $(\lambda_p)$ to that of the $\alpha$-particle $(\lambda_\alpha)$ is $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha Q_\alpha}{m_p Q_p}}$.Given that the mass of an $\alpha$-particle $m_\alpha = 4m_p$ and its charge $Q_\alpha = 2Q_p$,we substitute these values:$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p 	imes 2Q_p}{m_p 	imes Q_p}} = \sqrt{8} = 2\sqrt{2}$.Thus,the ratio is $2\sqrt{2} : 1$.

$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100 \ V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is

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