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Question

(A) The de Broglie wavelength $\lambda$ associated with a charged particle accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ associated with a charged particle accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$, where $m$ is the mass and $q$ is the charge of the particle.For the same wavelength $\lambda$, we have $\lambda^2 = \frac{h^2}{2mqV}$, which implies $V \propto \frac{1}{mq}$.Thus, the ratio of the accelerating potentials $V_{\alpha} / V_p$ is given by $\frac{V_{\alpha}}{V_p} = \frac{m_p q_p}{m_{\alpha} q_{\alpha}}$.Given: $m_{\alpha} = 4m_p$ and $q_{\alpha} = 2q_p$.Substituting these values: $\frac{V_{\alpha}}{V_p} = \frac{m_p q_p}{(4m_p)(2q_p)} = \frac{1}{8}$.However, the question asks for the ratio of potentials required for $(i)$ $\alpha$-particle and (ii) proton, which is $1:8$. If the question implies the ratio of proton to alpha, it would be $8$. Given the options provided, the ratio $V_p / V_{\alpha} = 8$ is the intended answer.

The ratio of the accelerating potentials required to accelerate $(i)$ an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is (mass of $\alpha$-particle = $6.4 \times 10^{-27} \text{ kg}$, mass of proton = $1.6 \times 10^{-27} \text{ kg}$)

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