The ratio of wavelengths of proton and deuteron accelerated by potential $$V_p$$ and $$V_d$$ is $$1:\sqrt{2}$$. Then, the ratio of $$V_p$$ to $$V_d$$ will be

We need to find the ratio $$\frac{V_p}{V_d}$$ given that the ratio of de Broglie wavelengths of proton and deuteron is $$1:\sqrt{2}$$.

Write the de Broglie wavelength for a charged particle accelerated through potential V.

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

where $$m$$ is the mass, $$q$$ is the charge, and $$V$$ is the accelerating potential.

Write the wavelengths for proton and deuteron.

For proton: $$\lambda_p = \frac{h}{\sqrt{2m_p e V_p}}$$

For deuteron: $$\lambda_d = \frac{h}{\sqrt{2m_d e V_d}}$$

Note: The deuteron has mass $$m_d = 2m_p$$ and charge $$e$$ (same as proton).

Find the ratio and solve.

$$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d V_d}{m_p V_p}} = \sqrt{\frac{2m_p V_d}{m_p V_p}} = \sqrt{\frac{2V_d}{V_p}}$$

Given $$\frac{\lambda_p}{\lambda_d} = \frac{1}{\sqrt{2}}$$:

$$\frac{1}{\sqrt{2}} = \sqrt{\frac{2V_d}{V_p}}$$

Squaring both sides:

$$\frac{1}{2} = \frac{2V_d}{V_p}$$

$$\frac{V_p}{V_d} = 4$$

The correct answer is Option D: $$4:1$$.