If the de Broglie wavelengths associated with a proton and an $α$-particle are equal, then the ratio of velocities of the proton and the a-particle will be:

$4:1$

$\text{Debroglie wavelength }\lambda = \frac{h}{p}$

$ \text{If Debroglie wavelength of two particles are equal then momentum of the two particles are also equal.}$

$\Rightarrow P_\alpha = P_p$

$\Rightarrow m_\alpha v_\alpha = m_p v_p$

$\Rightarrow \frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} = 4$