An α particle and a carbon 12 atom has same kinetic energy K . The...

De-Broglie wavelength is stated as $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}}$, where, $K$ is kinetic energy and $m$ is mass.

Here, ${\left(K\right)}_{\alpha} = {\left(K\right)}_{C ‐ 12}$.

Now, the ratio $\dfrac{{\lambda}_{\alpha}}{{\lambda}_{C12}} = \dfrac{\sqrt{12m}}{\sqrt{4m}} = \sqrt{3}: 1$