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(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.For a particle of m

Vedclass · Accepted Answer

$1: 2 \sqrt{2}$

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(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.For a particle of mass $m$ and charge $q$ accelerated from rest through a potential $V$,the momentum is $p = \sqrt{2mqV}$.Thus,the wavelength is $\lambda = \frac{h}{\sqrt{2mqV}}$.For an $\alpha$-particle,$m_\alpha = 4m_p$ and $q_\alpha = 2e$. For a proton,$m_p = m_p$ and $q_p = e$.The ratio of their wavelengths is $\frac{\lambda_\alpha}{\lambda_p} = \frac{\sqrt{2m_p q_p V}}{\sqrt{2m_\alpha q_\alpha V}} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}}$.Substituting the values: $\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p 	imes e}{4m_p 	imes 2e}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.Therefore,the ratio is $1: 2\sqrt{2}$.

An $\alpha$-particle and a proton are accelerated from rest by the same potential. The ratio of their de-Broglie wavelengths is . . . . . .

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