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

We have a proton and an alpha particle with the same de Broglie wavelength $$\lambda$$. The de Broglie wavelength is given by:

$$\lambda = \frac{h}{p}$$

Since both particles have the same wavelength $$\lambda$$, they have the same momentum: $$p_{proton} = p_{\alpha} = \frac{h}{\lambda}$$.

Now, the kinetic energy in terms of momentum is:

$$KE = \frac{p^2}{2m}$$

Since the momentum is the same for both particles:

$$\frac{KE_p}{KE_{\alpha}} = \frac{\frac{p^2}{2m_p}}{\frac{p^2}{2m_{\alpha}}} = \frac{m_{\alpha}}{m_p}$$

The mass of an alpha particle is approximately 4 times the mass of a proton ($$m_{\alpha} = 4m_p$$), so:

$$\frac{KE_p}{KE_{\alpha}} = \frac{4m_p}{m_p} = \frac{4}{1}$$

Hence, the ratio of kinetic energy of the proton to that of the alpha particle is $$4 : 1$$.