The de Broglie wavelength of a proton and $$\alpha$$-particle are equal. The ratio of their velocities is

The de Broglie wavelength of a particle is given by $$\lambda = \frac{h}{mv}$$, where $$h$$ is Planck's constant, $$m$$ is the mass, and $$v$$ is the velocity of the particle.

For the proton, $$\lambda_p = \frac{h}{m_p v_p}$$, and for the alpha particle, $$\lambda_\alpha = \frac{h}{m_\alpha v_\alpha}$$.

Since the de Broglie wavelengths are equal, $$\lambda_p = \lambda_\alpha$$, we have $$\frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha}$$, which gives $$m_p v_p = m_\alpha v_\alpha$$.

An alpha particle has mass $$m_\alpha = 4m_p$$. Substituting, $$m_p v_p = 4m_p v_\alpha$$, so $$\frac{v_p}{v_\alpha} = 4$$.

Therefore, the ratio of their velocities is $$v_p : v_\alpha = 4 : 1$$.

The correct answer is 4 : 1.