The de Broglie wavelength of a proton and alp | vedclass

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the

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$4:1$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,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 implies $m_p v_p = m_{\alpha} v_{\alpha}$.We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,so $m_{\alpha} = 4m_p$.Substituting this into the equation: $m_p v_p = (4m_p) v_{\alpha}$.Dividing both sides by $m_p$,we get $v_p = 4v_{\alpha}$.Thus,the ratio of their velocities is $\frac{v_p}{v_{\alpha}} = 4:1$.

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

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