If the de Broglie wavelengths for a proton and an $\alpha$-particle are equal, then the ratio of their velocities respectively will be:

$4: 1$

The correct answer is Option (1) → $4: 1$

De Broglie wavelength, $λ=\frac{h}{P}=\frac{h}{mv}$

where,

P = Momentum of Particle

m = Mass of particle

v = velocity of particle

$λ_{proton}=\frac{h}{mv_1}$

$λ_{alpha}=\frac{h}{4mv_2}$

and,

$λ_{proton}=λ_{alpha}$ [given]

$\frac{h}{mv_1}=\frac{h}{4mv_2}$

$\frac{v_1}{v_2}=\frac{4}{1}$