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Question

(A) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda

Vedclass · Accepted Answer

$\frac{V}{8}$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For the proton $(p)$: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$.For the $\alpha$-particle $(\alpha)$: $\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}}$.Given $\lambda_p = \lambda_\alpha$, we have $\sqrt{2m_p q_p V} = \sqrt{2m_\alpha q_\alpha V_\alpha}$.Squaring both sides: $m_p q_p V = m_\alpha q_\alpha V_\alpha$.We know that $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$.Substituting these values: $m_p q_p V = (4m_p)(2q_p) V_\alpha$.$m_p q_p V = 8 m_p q_p V_\alpha$.Therefore, $V_\alpha = \frac{V}{8}$.

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

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