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))`

Explanation

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

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) 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: Relate momentum to kinetic energy
When a charged particle (like an electron or proton) is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = qV \]
where \( q \) is the charge of the particle. For an electron, \( q = e \) (the elementary charge), and for a proton, \( q = e \) as well.

### Step 3: Write the kinetic energy in terms of mass and velocity
The kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ qV = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity
From the equation \( qV = \frac{1}{2} mv^2 \), we can solve for \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 5: Find the momentum
The momentum \( p \) of the particle is given by:
\[ p = mv \]
Substituting the expression for \( v \):
\[ p = m \sqrt{\frac{2qV}{m}} = \sqrt{2mqV} \]

### Step 6: Substitute momentum into the de Broglie wavelength formula
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}} \]

### Step 7: Calculate the de Broglie wavelength for the proton
For a proton, the mass is \( M \) (instead of \( m \)), but the charge remains the same:
\[ \lambda_P = \frac{h}{\sqrt{2MQV}} \]

### Step 8: Compare the wavelengths
Now we can compare the de Broglie wavelengths of the electron and the proton. The de Broglie wavelength for the electron is:
\[ \lambda_e = \frac{h}{\sqrt{2mqV}} \]
And for the proton:
\[ \lambda_P = \frac{h}{\sqrt{2MQV}} \]

### Step 9: Express the proton's wavelength in terms of the electron's wavelength
To express \( \lambda_P \) in terms of \( \lambda_e \):
\[ \lambda_P = \lambda_e \cdot \sqrt{\frac{m}{M}} \]

### Final Result
Thus, the de Broglie wavelength associated with a proton of mass \( M \) accelerated through the same potential difference \( V \) is:
\[ \lambda_P = \lambda_e \cdot \sqrt{\frac{m}{M}} \]