An electron of mass m has de Broglie wavelength `lambda` when accelerated through a potential difference V . When a proton of mass M is accelerated through a potential difference 9V, the de Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage ) .

Options

A. `(lambda)/(3)sqrt((M)/(m))`
B. `(lambda)/(3).(M)/(m)`
C. `(lambda)/(3)sqrt((m)/(M))`
D. `(lambda)/(3).(m)/(M)`

Answer

To solve the problem, we need to find the de Broglie wavelength of a proton when it is accelerated through a potential difference of 9V, given that an electron of mass \( m \) has a de Broglie wavelength \( \lambda \) when accelerated through a potential difference \( V \).

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle, and \( h \) is Planck's constant.

2. **Relate Momentum to Kinetic Energy**:
The momentum \( p \) can also be expressed in terms of kinetic energy \( K \):
\[ p = \sqrt{2mK} \]
where \( K \) is the kinetic energy of the particle.

3. **Kinetic Energy from Potential Difference**:
The kinetic energy \( K \) gained by a charged particle when accelerated through a potential difference \( V \) is given by:
\[ K = eV \]
where \( e \) is the charge of the particle.

4. **Substituting Kinetic Energy into the Wavelength Formula**:
For the electron, when it is accelerated through a potential difference \( V \):
\[ \lambda_e = \frac{h}{\sqrt{2m \cdot eV}} \]

5. **Finding the Wavelength for the Proton**:
For the proton, when it is accelerated through a potential difference of \( 9V \):
\[ \lambda_p = \frac{h}{\sqrt{2M \cdot e \cdot 9V}} \]

6. **Relating the Two Wavelengths**:
We can set up a ratio of the two wavelengths:
\[ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2m \cdot eV}}{\sqrt{2M \cdot e \cdot 9V}} \]
Simplifying this gives:
\[ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m}}{\sqrt{9M}} = \frac{1}{3} \cdot \frac{\sqrt{m}}{\sqrt{M}} \]

7. **Expressing the Wavelength of the Proton**:
Rearranging gives:
\[ \lambda_p = \lambda_e \cdot \frac{1}{3} \cdot \frac{\sqrt{m}}{\sqrt{M}} \]
Therefore, substituting \( \lambda_e \) with \( \lambda \):
\[ \lambda_p = \frac{\lambda}{3} \cdot \sqrt{\frac{m}{M}} \]

### Final Answer:
The de Broglie wavelength associated with the proton when accelerated through a potential difference of \( 9V \) is:
\[ \lambda_p = \frac{\lambda}{3} \sqrt{\frac{m}{M}} \]