A proton and an electron are accelerated by the same potential difference, let `lambda_(e)` and `lambda_(p)` denote the de-Broglie wavelengths of the electron and the proton respectively

Options

A. `lambda_e = lambda_p`
B. `lambda_e lt lambda_p`
C. `lambda_egt lambda_p`
D. The relation between ` lambda_e and lambda_p` depends on the accelerating potential difference

Answer

To solve the problem, we need to analyze the de Broglie wavelengths of a proton and an electron when both are accelerated by the same potential difference.

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
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.

2. **Relate Momentum to Velocity**:
The momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

3. **Determine Velocity from Potential Difference**:
When a charged particle is accelerated through a potential difference \( V \), the kinetic energy gained by the particle is equal to the work done on it:
\[ KE = qV \]
where \( q \) is the charge of the particle. For an electron, \( q = e \) (the elementary charge), and for a proton, \( q = e \) as well.

The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]

Setting these two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = qV \]

4. **Solve for Velocity**:
Rearranging the equation to solve for \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

5. **Substituting Velocity into the de Broglie Wavelength Formula**:
Now substituting \( v \) into the momentum formula:
\[ p = mv = m\sqrt{\frac{2qV}{m}} = \sqrt{2mqV} \]

Then substituting this into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}} = \frac{h}{\sqrt{2m \cdot e \cdot V}} \]

6. **Comparing Wavelengths of Electron and Proton**:
Now we can express the de Broglie wavelengths for the electron (\( \lambda_e \)) and the proton (\( \lambda_p \)):
\[ \lambda_e = \frac{h}{\sqrt{2m_e e V}} \quad \text{and} \quad \lambda_p = \frac{h}{\sqrt{2m_p e V}} \]

7. **Finding the Relation**:
To find the relation between \( \lambda_e \) and \( \lambda_p \), we can take the ratio:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_p e V}}{\sqrt{2m_e e V}} = \sqrt{\frac{m_p}{m_e}} \]
Since the mass of the proton (\( m_p \)) is much greater than the mass of the electron (\( m_e \)), we have:
\[ \frac{\lambda_e}{\lambda_p} > 1 \quad \Rightarrow \quad \lambda_e > \lambda_p \]

### Conclusion:
Thus, the relation between the de Broglie wavelengths is:
\[ \lambda_e > \lambda_p \]

### Final Answer:
The correct option is \( \lambda_e \) is greater than \( \lambda_p \).