Four particles each of mass $$M$$, move along a circle of radius $$R$$ under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:

We need to determine the ratio of the de Broglie wavelengths of an electron and a proton when they are both accelerated from rest through the exact same potential difference $$V$$.

1. Establish the Relationship for de Broglie Wavelength

The de Broglie wavelength ($$\lambda$$) of a particle is related to its momentum ($$p$$) by the equation:

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

When a particle carrying a charge $$q$$ is accelerated from rest through a potential difference $$V$$:

• The kinetic energy ($$K$$) gained by the particle is given by:

  $$K = qV$$

• Since momentum is related to kinetic energy by $$p = \sqrt{2mK}$$, we can express momentum in terms of accelerating voltage:

  $$p = \sqrt{2mqV}$$

Substituting this momentum back into the wavelength formula gives:

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

2. Compare the Electron and the Proton

Let's look at the parameters for both particles:

• For the electron: mass is $$m_e$$ and charge magnitude is $$e$$.

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

• For the proton: mass is $$m_p$$ and charge magnitude is $$e$$.

  $$\lambda_p = \frac{h}{\sqrt{2m_p eV}}$$

Since Plank's constant ($$h$$), the fundamental electronic charge ($$e$$), and the accelerating potential ($$V$$) are identical for both particles, they cancel out entirely when taking a ratio.

3. Calculate the Ratio ($$\frac{\lambda_e}{\lambda_p}$$)

Taking the ratio of the electron's wavelength to the proton's wavelength:

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

Simplifying the radical expression yields:

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

Alternatively, this can be written as the fractional exponent power:

$$\frac{\lambda_e}{\lambda_p} = \left(\frac{m_p}{m_e}\right)^{\frac{1}{2}}$$

Conclusion

The ratio of their de Broglie wavelengths is $$\sqrt{\frac{m_p}{m_e}}$$ (or $$\left(\frac{m_p}{m_e}\right)^{\frac{1}{2}}$$), which corresponds to Option C.