The de Broglie wavelengths of a proton and an $$\alpha$$ particle are $$\lambda$$ and $$2\lambda$$ respectively. The ratio of the velocities of proton and $$\alpha$$ particle will be :

We are asked to find the ratio of velocities of a proton and an alpha particle given their de Broglie wavelengths as $$\lambda$$ and $$2\lambda$$, respectively. The de Broglie relation is $$\lambda = \frac{h}{mv}$$, which implies $$v = \frac{h}{m\lambda}$$.

Thus, for a proton the velocity is $$v_p = \frac{h}{m_p\lambda}$$, while for an alpha particle with mass $$m_\alpha = 4m_p$$ and wavelength $$2\lambda$$, the velocity becomes $$v_\alpha = \frac{h}{m_\alpha \times 2\lambda} = \frac{h}{4m_p \times 2\lambda} = \frac{h}{8m_p\lambda}$$.

Taking the ratio gives $$\frac{v_p}{v_\alpha} = \frac{h/(m_p\lambda)}{h/(8m_p\lambda)} = \frac{8m_p\lambda}{m_p\lambda} = 8$$ so that $$v_p : v_\alpha = 8 : 1$$. The correct answer is Option D: 8 : 1.