Electron are accelerated through a potential difference V and protons are accelerated through a potential difference of 4 V. The de-Broglie wavelength are `lamda_e and lamda_p` for electrons and protons, respectively The ratio of `lamda_e/lamda_p` is given by ( given , `m_e` is mass of electron and `m_p` is mass of proton )

Options

A. `lamda_e/lamda_p=sqrt((m_p)/m_e)`
B. `lamda_e/lamda_p=sqrt((m_e)/m_p)`
C. `lamda_e/lamda_p=1/2sqrt((m_e)/m_p)`
D. `lamda_e/lamda_p=2sqrt((m_p)/m_e)`

Answer

To solve the problem, we need to find the ratio of the de Broglie wavelengths of an electron and a proton when they are accelerated through different potential differences.

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can also be expressed in terms of mass (\( m \)) and velocity (\( v \)):
\[ p = mv \]
Thus, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{mv} \]

2. **Relating Velocity to Potential Difference**:
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done by the electric field:
\[ KE = qV \]
For an electron (charge \( e \)) accelerated through potential \( V \):
\[ \frac{1}{2} m_e v_e^2 = eV \]
Solving for \( v_e \):
\[ v_e = \sqrt{\frac{2eV}{m_e}} \]
For a proton (charge \( e \)) accelerated through potential \( 4V \):
\[ \frac{1}{2} m_p v_p^2 = e(4V) \]
Solving for \( v_p \):
\[ v_p = \sqrt{\frac{2e(4V)}{m_p}} = \sqrt{\frac{8eV}{m_p}} \]

3. **Finding the Wavelengths**:
Now substituting \( v_e \) and \( v_p \) back into the de Broglie wavelength equations:
- For the electron:
\[ \lambda_e = \frac{h}{m_e v_e} = \frac{h}{m_e \sqrt{\frac{2eV}{m_e}}} = \frac{h \sqrt{m_e}}{\sqrt{2eV}} \]
- For the proton:
\[ \lambda_p = \frac{h}{m_p v_p} = \frac{h}{m_p \sqrt{\frac{8eV}{m_p}}} = \frac{h \sqrt{m_p}}{\sqrt{8eV}} \]

4. **Finding the Ratio**:
Now we can find the ratio \( \frac{\lambda_e}{\lambda_p} \):
\[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h \sqrt{m_e}}{\sqrt{2eV}}}{\frac{h \sqrt{m_p}}{\sqrt{8eV}}} \]
Simplifying this gives:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_e}}{\sqrt{m_p}} \cdot \frac{\sqrt{8}}{\sqrt{2}} = \frac{\sqrt{m_e}}{\sqrt{m_p}} \cdot \sqrt{4} = 2 \frac{\sqrt{m_e}}{\sqrt{m_p}} \]

5. **Final Result**:
Thus, the final ratio of the de Broglie wavelengths is:
\[ \frac{\lambda_e}{\lambda_p} = 2 \sqrt{\frac{m_e}{m_p}} \]