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Question

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the de-Broglie wavelengths for a proton $(p)$ and an $\alpha$-particle $

Vedclass · Accepted Answer

$4 : 1$

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(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the de-Broglie wavelengths for a proton $(p)$ and an $\alpha$-particle $(\alpha)$ are equal, we have $\lambda_p = \lambda_{\alpha}$.Therefore, $\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$, which simplifies to $m_p v_p = m_{\alpha} v_{\alpha}$.The ratio of their velocities is $\frac{v_p}{v_{\alpha}} = \frac{m_{\alpha}}{m_p}$.We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton, so $m_{\alpha} = 4 m_p$.Substituting this, we get $\frac{v_p}{v_{\alpha}} = \frac{4 m_p}{m_p} = \frac{4}{1}$.Thus, the ratio of their velocities is $4 : 1$.

If the de-Broglie wavelengths for a proton and for an $\alpha$-particle are equal, then the ratio of their velocities will be

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