An electron is accelerated through a potential difference of V volt. It has a wavelength `lamda` associated must be accelerated so that its de Broglie wavelength is the same as the of a proton? Take mass of proton to be 1837 times larger than the mass of electron.

Options

A. V volt
B. 1837V volt
C. `(V)/(1837)` volt
D. `sqrt(1837) V` volt

Explanation

To solve the problem, we need to find the potential difference \( V_p \) through which a proton must be accelerated so that its de Broglie wavelength is the same as that of an electron accelerated through a potential difference \( V \).

### 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 Kinetic Energy**:
The momentum \( p \) can be expressed in terms of kinetic energy \( KE \):
\[ p = \sqrt{2m \cdot KE} \]
For a charged particle accelerated through a potential difference \( V \), the kinetic energy is given by:
\[ KE = qV \]
where \( q \) is the charge of the particle.

3. **Write the de Broglie Wavelength for Electron**:
For the electron, we have:
\[ \lambda_e = \frac{h}{\sqrt{2m_e \cdot eV}} \]
where \( m_e \) is the mass of the electron and \( e \) is the charge of the electron.

4. **Write the de Broglie Wavelength for Proton**:
For the proton, we have:
\[ \lambda_p = \frac{h}{\sqrt{2m_p \cdot eV_p}} \]
where \( m_p \) is the mass of the proton and \( V_p \) is the potential difference through which the proton is accelerated.

5. **Set the Wavelengths Equal**:
Since the de Broglie wavelengths are the same:
\[ \lambda_e = \lambda_p \]
Substituting the expressions for \( \lambda_e \) and \( \lambda_p \):
\[ \frac{h}{\sqrt{2m_e \cdot eV}} = \frac{h}{\sqrt{2m_p \cdot eV_p}} \]

6. **Cancel \( h \) and Rearrange**:
Canceling \( h \) from both sides and squaring gives:
\[ \frac{1}{2m_e \cdot eV} = \frac{1}{2m_p \cdot eV_p} \]
Rearranging this yields:
\[ m_p \cdot eV_p = m_e \cdot eV \]

7. **Substitute the Mass of Proton**:
Given that the mass of the proton \( m_p \) is 1837 times the mass of the electron \( m_e \):
\[ 1837m_e \cdot V_p = m_e \cdot V \]

8. **Solve for \( V_p \)**:
Dividing both sides by \( m_e \):
\[ 1837V_p = V \]
Thus,
\[ V_p = \frac{V}{1837} \]

### Final Answer:
The potential difference \( V_p \) through which the proton must be accelerated is:
\[ V_p = \frac{V}{1837} \]