Electrons used in an electron microscope are accelerated by a voltage of 25 k V. If the voltage is increased to 100 k V. then the be Broglie wavelength associated with the electrons would

Options

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

Explanation

To solve the problem of finding the de Broglie wavelength associated with electrons when the accelerating voltage is increased from 25 kV to 100 kV, 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: Relate Momentum to Kinetic Energy
The momentum (\( p \)) of an electron can be expressed in terms of its kinetic energy (\( KE \)):
\[ p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the electron.

### Step 3: Express Kinetic Energy in Terms of Voltage
For an electron accelerated through a potential difference \( V \), the kinetic energy can be expressed as:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron.

### Step 4: Substitute Kinetic Energy into the Momentum Equation
Substituting the expression for kinetic energy into the momentum equation gives:
\[ p = \sqrt{2m \cdot (e \cdot V)} \]

### Step 5: Substitute Momentum into the de Broglie Wavelength Formula
Now substituting the expression for momentum into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot (e \cdot V)}} \]

### Step 6: Analyze the Relationship with Voltage
From the equation, we can see that the de Broglie wavelength is inversely proportional to the square root of the voltage:
\[ \lambda \propto \frac{1}{\sqrt{V}} \]

### Step 7: Calculate the Change in Wavelength
Let \( \lambda_1 \) be the de Broglie wavelength at \( V_1 = 25 \, \text{kV} \) and \( \lambda_2 \) be the de Broglie wavelength at \( V_2 = 100 \, \text{kV} \):
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{100}{25}} = \sqrt{4} = 2 \]
Thus, we have:
\[ \lambda_2 = \frac{\lambda_1}{2} \]

### Conclusion
The de Broglie wavelength associated with the electrons when the voltage is increased to 100 kV will be half of the wavelength at 25 kV.