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Question

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

Vedclass · Accepted Answer

$2 \sqrt{2}: 1$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ of a charged 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 $\lambda \propto \frac{1}{\sqrt{mQ}}$.For a proton,$m_p = m$ and $Q_p = e$. For an $\alpha-$particle,$m_\alpha = 4m$ and $Q_\alpha = 2e$.The ratio of the wavelengths is $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha Q_\alpha}{m_p Q_p}}$.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 $2\sqrt{2} : 1$.

$A$ 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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