An electron of mass $$m_e$$ and a proton of mass $$m_P$$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is:

We need to find the ratio of the de-Broglie wavelength associated with an electron to that with a proton when both are accelerated through the same potential difference.

1. Understand the de-Broglie Wavelength Formula

According to de-Broglie's hypothesis, a moving particle behaves as a wave, and its wavelength ($\lambda$) is given by:

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

Where:

• $$h$$ = Planck's constant
• $$p$$ = Momentum of the particle

2. Relate Momentum to Potential Difference

When a particle of charge $$q$$ and mass $$m$$ is accelerated from rest through a potential difference $$V$$, the kinetic energy ($$K$$) gained by the particle is:

$$K = qV$$

We also know that kinetic energy is related to momentum ($$p$$) by the formula:

$$K = \frac{p^2}{2m} \implies p = \sqrt{2mK}$$

Substituting the kinetic energy equation ($$K = qV$$) into the momentum formula gives:

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

3. Substitute Momentum into the Wavelength Equation

Now, substitute the expression for momentum ($$p$$) back into the de-Broglie wavelength formula:

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

4. Find the Ratio for Electron and Proton

Let's write out the wavelengths for both the electron and the proton:

• For the electron (mass = $$m_e$$, charge = $$e$$):

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

• For the proton (mass = $$m_p$$, charge = $$e$$):

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

(Note: Both the electron and proton have the same magnitude of charge $$q = e$$ and are accelerated through the same potential difference $$V$$.)

Taking the ratio of the de-Broglie wavelength of the electron to that of the proton:

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

Canceling out the common terms ($$h$$, $$2$$, $$e$$, and $$V$$), the equation simplifies inversely to the square root of their masses:

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

Final Answer: Option C ($$\sqrt{\frac{m_p}{m_e}}$$)