The proton and alpha-particle are accelerated | vedclass

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Question

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the mass, $q

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the mass, $q$ is the charge, and $V$ is the potential difference.Since $h$ and $V$ are the same for both particles, we have $\lambda \propto \frac{1}{\sqrt{mq}}$.Let $m_p$ and $q_p$ be the mass and charge of the proton, and $m_{\alpha}$ and $q_{\alpha}$ be the mass and charge of the $\alpha$-particle.Given: $m_{\alpha} = 4m_p$ and $q_{\alpha} = 2q_p$.The ratio of wavelengths is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$.Substituting the given values: $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p \cdot 2q_p}{m_p \cdot q_p}} = \sqrt{8} = 2\sqrt{2}$.

The proton and $\alpha$-particle are accelerated through the same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$-particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton).

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