An electron of mass 'm', when accelerated through a potential 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. `lambdasqrt(M/m)`
B. `lambdasqrt(m/M)`
C. `lambda(M/m)`
D. `lambda(m/M)`

Explanation

To solve the problem, we need to find the de-Broglie wavelength associated with a proton when it is accelerated through the same potential difference as an electron. We know that the de-Broglie wavelength \(\lambda\) is given by:

\[ \lambda = \frac{h}{p} \]

where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

### Step 1: Determine the kinetic energy of the electron

When an electron is accelerated through a potential \(V\), its kinetic energy \(E\) can be expressed as:

\[ E = eV \]

where \(e\) is the charge of the electron.

### Step 2: Relate kinetic energy to momentum

The kinetic energy can also be expressed in terms of momentum \(p\):

\[ E = \frac{p^2}{2m} \]

where \(m\) is the mass of the electron.

### Step 3: Solve for momentum of the electron

Setting the two expressions for kinetic energy equal gives:

\[ eV = \frac{p^2}{2m} \]

Rearranging this to solve for momentum \(p\):

\[ p = \sqrt{2meV} \]

### Step 4: Calculate the de-Broglie wavelength of the electron

Substituting this expression for momentum into the de-Broglie wavelength formula gives:

\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} \]

### Step 5: Determine the kinetic energy of the proton

For a proton accelerated through the same potential \(V\), its kinetic energy is:

\[ E' = eV \]

where \(e\) is the charge of the proton (which is equal to the charge of the electron).

### Step 6: Relate kinetic energy to momentum for the proton

Using the same relationship for kinetic energy and momentum for the proton (mass \(M\)):

\[ E' = \frac{p'^2}{2M} \]

Setting this equal to the kinetic energy gives:

\[ eV = \frac{p'^2}{2M} \]

Rearranging to solve for momentum \(p'\):

\[ p' = \sqrt{2MeV} \]

### Step 7: Calculate the de-Broglie wavelength of the proton

Substituting this expression for momentum into the de-Broglie wavelength formula gives:

\[ \lambda' = \frac{h}{p'} = \frac{h}{\sqrt{2MeV}} \]

### Step 8: Find the ratio of the wavelengths

Now we can find the ratio of the wavelengths of the electron and the proton:

\[ \frac{\lambda}{\lambda'} = \frac{\frac{h}{\sqrt{2meV}}}{\frac{h}{\sqrt{2MeV}}} = \frac{\sqrt{2MeV}}{\sqrt{2meV}} = \sqrt{\frac{M}{m}} \]

### Step 9: Solve for \(\lambda'\)

Rearranging gives:

\[ \lambda' = \lambda \sqrt{\frac{m}{M}} \]

### Final Answer

Thus, the de-Broglie wavelength associated with the proton is:

\[ \lambda' = \lambda \sqrt{\frac{m}{M}} \]