Electrons used in an electron microscope are accelerated by a voltage of `25 kV`. If the voltage is increased to `100 kV` then the de - Broglie wavelength associated with the electrons would

Options

A. increase by 2 times
B. decrease by 2 times
C. decrease by 4 times
D. increase by 4 times

Explanation

To solve the problem of how the de Broglie wavelength associated with electrons changes when the accelerating voltage is increased from 25 kV to 100 kV, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and voltage
The kinetic energy (K) of an electron accelerated through a potential difference (V) is given by:
\[ K = eV \]
where \( e \) is the charge of the electron and \( V \) is the accelerating voltage.

### Step 2: Calculate the initial and final kinetic energies
1. For the initial voltage \( V_1 = 25 \, \text{kV} \):
\[ K_1 = e \cdot V_1 = e \cdot 25 \times 10^3 \, \text{V} \]

2. For the final voltage \( V_2 = 100 \, \text{kV} \):
\[ K_2 = e \cdot V_2 = e \cdot 100 \times 10^3 \, \text{V} \]

### Step 3: Relate the de Broglie wavelength to kinetic energy
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy as:
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the electron.

### Step 4: Substitute momentum into the de Broglie wavelength formula
Substituting for momentum in the de Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
This shows that the wavelength is inversely proportional to the square root of the kinetic energy:
\[ \lambda \propto \frac{1}{\sqrt{K}} \]

### Step 5: Establish the ratio of wavelengths
From the relationship derived, we can write the ratio of the wavelengths at the two different kinetic energies:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} \]

### Step 6: Substitute the values of kinetic energies
Using \( K_1 = e \cdot 25 \times 10^3 \) and \( K_2 = e \cdot 100 \times 10^3 \):
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{100 \times 10^3}{25 \times 10^3}} = \sqrt{4} = 2 \]

### Step 7: Solve for the new wavelength
This implies:
\[ \lambda_2 = \frac{\lambda_1}{2} \]

### Conclusion
Thus, when the voltage is increased from 25 kV to 100 kV, the de Broglie wavelength associated with the electrons is halved.