Step-by-Step Solution

1. The de-Broglie wavelength is given by the formula: \( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \), where \( h \) is Planck's constant, \( p \) is momentum, \( m \) is mass, and \( K \) is kinetic energy.

2. The kinetic energy gained by a charged particle accelerated through a potential \( V \) is given by \( K = qV \), where \( q \) is the charge of the particle.

3. Therefore, the de-Broglie wavelength can be written as: \( \lambda = \frac{h}{\sqrt{2mqV}} \).

4. Let \( \lambda_p \) and \( \lambda_\alpha \) be the de-Broglie wavelengths of the proton and alpha particle, respectively. Let \( m_p \) and \( q_p \) be the mass and charge of the proton, and \( m_\alpha \) and \( q_\alpha \) be the mass and charge of the alpha particle.

5. We know that \( m_\alpha = 4m_p \) and \( q_\alpha = 2q_p \).

6. The ratio of the de-Broglie wavelengths is:

   \( \frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2m_p q_p V}}}{\frac{h}{\sqrt{2m_\alpha q_\alpha V}}} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{(4m_p)(2q_p)}{m_p q_p}} = \sqrt{8} = 2\sqrt{2} \)

7. Therefore, the ratio is \( 2\sqrt{2}:1 \).