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(C) According to the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mas

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$4 : 1$

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(C) According to the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.Given that the de Broglie wavelengths are equal,$\lambda_p = \lambda_{\alpha}$.Therefore,$\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$.This simplifies to $m_p v_p = m_{\alpha} v_{\alpha}$,or $\frac{v_p}{v_{\alpha}} = \frac{m_{\alpha}}{m_p}$.Since the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton $(m_{\alpha} = 4m_p)$,we substitute this into the ratio:$\frac{v_p}{v_{\alpha}} = \frac{4m_p}{m_p} = \frac{4}{1}$.Thus,the ratio of the velocities of the proton and the $\alpha$-particle is $4:1$.

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

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