An electron accelerated under a potential difference of V volt has a certain wavelength `lamda` It is known that the mass of a proton is about 1800 times the mass of an electron . If a proton has to have the same wavelength `lamda`, then it will have to be accelerated under the potential difference of

Options

A. (a)V volt
B. (b)1800 volt
C. (c)`V/1800 `volt
D. `sqrt1800` volt

Answer

To solve the problem, we need to find the potential difference \( V' \) that a proton must be accelerated through to have the same de Broglie wavelength \( \lambda \) as an electron accelerated through a potential difference \( V \).

### Step-by-step Solution:

1. **Understanding de Broglie's 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.

2. **Relating Momentum to Kinetic Energy**:
The momentum \( p \) can be related to the kinetic energy \( K \) of the particle:
\[ K = \frac{p^2}{2m} \]
Rearranging gives:
\[ p = \sqrt{2mK} \]

3. **Kinetic Energy from Potential Difference**:
When a charged particle is accelerated through a potential difference \( V \), its kinetic energy \( K \) is given by:
\[ K = qV \]
where \( q \) is the charge of the particle.

4. **For the Electron**:
For the electron, we have:
\[ K_e = eV \]
where \( e \) is the charge of the electron. Substituting this into the momentum equation gives:
\[ p_e = \sqrt{2m_e eV} \]
Therefore, the wavelength for the electron becomes:
\[ \lambda = \frac{h}{\sqrt{2m_e eV}} \]

5. **For the Proton**:
For the proton, we want to find \( V' \) such that:
\[ \lambda = \frac{h}{\sqrt{2m_p eV'}} \]
where \( m_p \) is the mass of the proton.

6. **Setting the Wavelengths Equal**:
Since both wavelengths are equal, we can set the two equations equal to each other:
\[ \frac{h}{\sqrt{2m_e eV}} = \frac{h}{\sqrt{2m_p eV'}} \]
Canceling \( h \) and \( e \) from both sides, we get:
\[ \sqrt{2m_e V} = \sqrt{2m_p V'} \]

7. **Squaring Both Sides**:
Squaring both sides results in:
\[ 2m_e V = 2m_p V' \]
Simplifying gives:
\[ m_e V = m_p V' \]

8. **Substituting the Mass of Proton**:
We know that \( m_p = 1800 m_e \). Substituting this into the equation gives:
\[ m_e V = 1800 m_e V' \]
Dividing both sides by \( m_e \) (assuming \( m_e \neq 0 \)):
\[ V = 1800 V' \]

9. **Finding \( V' \)**:
Rearranging gives:
\[ V' = \frac{V}{1800} \]

### Final Answer:
The potential difference \( V' \) that the proton must be accelerated through to have the same wavelength \( \lambda \) as the electron is:
\[ V' = \frac{V}{1800} \]