The de-Broglie wavelength of a particle is given by:

$$\lambda = \frac{h}{p}$$

where $$h$$ is the Planck's constant and $$p$$ is the momentum of the particle.

The momentum of a particle of mass $$m$$ and charge $$q$$ accelerated through a potential difference $$V$$ is given by:

$$p = \sqrt{2 m q V}$$

For a proton, $$m = 1.67 \times 10^{-27} \mathrm{~kg}$$ and $$q = 1.6 \times 10^{-19} \mathrm{~C}$$, and for an $$\alpha$$-particle, $$m = 6.64 \times 10^{-27} \mathrm{~kg}$$ and $$q = 2 \times 1.6 \times 10^{-19} \mathrm{~C}$$.

Therefore, the ratio of their de-Broglie wavelengths is:

$$\frac{\lambda_p}{\lambda_\alpha} = \frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha}{m_p} \cdot \frac{q_\alpha V_\alpha}{q_p V_p}} = \sqrt{\frac{6.64 \times 10^{-27}}{1.67 \times 10^{-27}} \cdot \frac{2 \times 1.6 \times 10^{-19} \times 4}{1.6 \times 10^{-19} \times 2}} = \sqrt{16} = 4$$

Therefore, the ratio of their de-Broglie wavelengths is 4 : 1.