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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Kinetic energy $K$ is related to momentum by $K = \frac{p^

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$4: 1$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Kinetic energy $K$ is related to momentum by $K = \frac{p^2}{2m}$.Substituting $p = \frac{h}{\lambda}$,we get $K = \frac{h^2}{2m\lambda^2}$.Since both particles have the same wavelength $\lambda$,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.Therefore,the ratio of kinetic energy of the proton $(K_p)$ to the kinetic energy of the alpha particle $(K_\alpha)$ is $\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p}$.Given that the mass of an alpha particle is approximately $4$ times the mass of a proton $(m_\alpha \approx 4m_p)$,we have $\frac{K_p}{K_\alpha} = \frac{4m_p}{m_p} = 4:1$.

$A$ proton moving with one-tenth of the velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having a certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of the kinetic energy of the proton to that of the alpha particle is:

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