An electron of mass `m` when accelerated through a potential difference `V` has de - Broglie wavelength `lambda`. The de - Broglie wavelength associated with a proton of mass `M` accelerated through the same potential difference will be

Options

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

Answer

To find the de Broglie wavelength associated with a proton of mass \( M \) when accelerated through the same potential difference \( V \) as an electron of mass \( m \) with a de Broglie wavelength \( \lambda \), we can follow these steps:

### Step 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.

### Step 2: Calculate the momentum of the electron
When an electron is accelerated through a potential difference \( V \), its kinetic energy \( KE \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron. The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
Equating the two expressions for kinetic energy:
\[ eV = \frac{p_e^2}{2m} \]
From this, we can solve for the momentum \( p_e \) of the electron:
\[ p_e = \sqrt{2meV} \]

### Step 3: Calculate the momentum of the proton
Similarly, for a proton accelerated through the same potential difference \( V \):
\[ KE = eV \]
The kinetic energy for the proton can also be expressed as:
\[ KE = \frac{p_p^2}{2M} \]
Equating the two expressions for kinetic energy:
\[ eV = \frac{p_p^2}{2M} \]
Solving for the momentum \( p_p \) of the proton:
\[ p_p = \sqrt{2MeV} \]

### Step 4: Relate the de Broglie wavelengths of the electron and proton
Using the de Broglie wavelength formula for both particles:
\[ \lambda_e = \frac{h}{p_e} \quad \text{and} \quad \lambda_p = \frac{h}{p_p} \]
We can write the ratio of the wavelengths:
\[ \frac{\lambda_e}{\lambda_p} = \frac{p_p}{p_e} \]
Substituting the expressions for momentum:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2MeV}}{\sqrt{2meV}} = \sqrt{\frac{M}{m}} \]

### Step 5: Solve for the wavelength of the proton
Rearranging the equation gives:
\[ \lambda_p = \lambda_e \cdot \sqrt{\frac{m}{M}} \]
Thus, the de Broglie wavelength associated with the proton is:
\[ \lambda_p = \lambda \cdot \sqrt{\frac{m}{M}} \]

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