Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are $$\lambda_e$$ and $$\lambda_p$$ for electrons and protons respectively. The ratio of $$\frac{\lambda_e}{\lambda_p}$$ is given by: (given $$m_e$$ is mass of electron and $$m_p$$ is mass of proton).

The de-Broglie wavelength for a particle is given by $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle.

When a charged particle is accelerated through a potential difference, it gains kinetic energy. The kinetic energy $$K$$ acquired by a particle with charge $$q$$ accelerated through a potential difference $$U$$ is $$K = qU$$.

For electrons: charge $$q_e = e$$ (magnitude), accelerated through potential difference $$V$$, so kinetic energy $$K_e = e V$$.

For protons: charge $$q_p = e$$, accelerated through potential difference $$4V$$, so kinetic energy $$K_p = e \times 4V = 4eV$$.

The kinetic energy can also be expressed in terms of momentum and mass: $$K = \frac{p^2}{2m}$$. Therefore, we can write:

For electrons: $$K_e = \frac{p_e^2}{2m_e}$$, so $$p_e = \sqrt{2m_e K_e}$$.

For protons: $$K_p = \frac{p_p^2}{2m_p}$$, so $$p_p = \sqrt{2m_p K_p}$$.

Substituting the kinetic energies:

$$p_e = \sqrt{2m_e \times eV} = \sqrt{2m_e eV}$$

$$p_p = \sqrt{2m_p \times 4eV} = \sqrt{8m_p eV}$$

Now, the de-Broglie wavelengths are:

$$\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e eV}}$$

$$\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{8m_p eV}}$$

The ratio $$\frac{\lambda_e}{\lambda_p}$$ is:

$$\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e eV}}}{\frac{h}{\sqrt{8m_p eV}}} = \frac{h}{\sqrt{2m_e eV}} \times \frac{\sqrt{8m_p eV}}{h} = \frac{\sqrt{8m_p eV}}{\sqrt{2m_e eV}}$$

Simplify the expression:

$$\frac{\sqrt{8m_p eV}}{\sqrt{2m_e eV}} = \sqrt{\frac{8m_p eV}{2m_e eV}} = \sqrt{\frac{8m_p}{2m_e}} = \sqrt{4 \times \frac{m_p}{m_e}} = \sqrt{4} \times \sqrt{\frac{m_p}{m_e}} = 2 \sqrt{\frac{m_p}{m_e}}$$

Therefore, the ratio is $$\frac{\lambda_e}{\lambda_p} = 2 \sqrt{\frac{m_p}{m_e}}$$.

Comparing with the options:

A. $$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$$

B. $$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_e}{m_p}}$$

C. $$\frac{\lambda_e}{\lambda_p} = \frac{1}{2}\sqrt{\frac{m_e}{m_p}}$$

D. $$\frac{\lambda_e}{\lambda_p} = 2\sqrt{\frac{m_p}{m_e}}$$

Our result matches option D.

Hence, the correct answer is Option D.