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The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is:
(1) 4 : 2
(2) 4 : 1
(3) 1 : 4
(4) 4 : 3
From de-Broglie wavelength: $\lambda = \frac{h}{mv}$. Given $\lambda_p = \lambda_\alpha$. $\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} = \frac{4m_p}{m_p} = \frac{4}{1}$.
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