A proton moving with one tenth of velocity of light has a certain...

As we know, de Broglie wavelength is given by,

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

Now kinetic energy can be written as,

$K = \dfrac{1}{2}m{v}^{2} = \dfrac{{p}^{2}}{2m} = \dfrac{{h}^{2}}{2m{\lambda}^{2}}$

Therefore,

$K ∝ \dfrac{1}{m}$

Hence, required ratio

$\dfrac{{K}_{p}}{{K}_{\alpha}} = \dfrac{{m}_{\alpha}}{{m}_{p}} = 4 ⇒ {m}_{\alpha}: {m}_{p} = 4: 1$