A beam of monoenergetic electrons, which have been accelerated from rest by a potential U, is used to form an interference pattern in a Young's Double slit experiment. The electrons are now accelerated by potential 4U. The fringe width -

Options

A. increae
B. decrease
C. remains same
D. no fringe pattern will be fromed

Explanation

To solve the problem, we need to analyze how the fringe width in a Young's Double Slit experiment changes when the potential applied to accelerate the electrons is increased from \( U \) to \( 4U \).

### Step-by-Step Solution:

1. **Understanding the Fringe Width Formula**:
The fringe width (\( \beta \)) in a Young's Double Slit experiment is given by the formula:
\[ \beta = \frac{n \lambda D}{d} \]
where:
- \( n \) is an integer (order of the fringe),
- \( \lambda \) is the wavelength of the electrons,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the two slits.

2. **Relating Wavelength to Potential**:
When electrons are accelerated through a potential \( U \), their kinetic energy is given by:
\[ KE = eU \]
This kinetic energy can also be expressed in terms of momentum (\( p \)):
\[ KE = \frac{p^2}{2m} \]
where \( e \) is the charge of the electron and \( m \) is its mass.

3. **Finding the Wavelength**:
The de Broglie wavelength (\( \lambda \)) of the electrons is given by:
\[ \lambda = \frac{h}{p} \]
Substituting \( p \) from the kinetic energy equation, we can relate \( \lambda \) to the potential:
\[ eU = \frac{h^2}{2m\lambda^2} \implies \lambda^2 = \frac{h^2}{2meU} \implies \lambda \propto \frac{1}{\sqrt{U}} \]

4. **Effect of Increasing Potential**:
When the potential is increased from \( U \) to \( 4U \):
\[ \lambda' \propto \frac{1}{\sqrt{4U}} = \frac{1}{2\sqrt{U}} = \frac{1}{2} \lambda \]
This means that the new wavelength (\( \lambda' \)) is half of the original wavelength (\( \lambda \)).

5. **Calculating the New Fringe Width**:
Since the fringe width is directly proportional to the wavelength:
\[ \beta' = \frac{n \lambda' D}{d} = \frac{n \left(\frac{1}{2} \lambda\right) D}{d} = \frac{1}{2} \beta \]
Therefore, the new fringe width (\( \beta' \)) is half of the original fringe width (\( \beta \)).

6. **Conclusion**:
Thus, when the potential is increased from \( U \) to \( 4U \), the fringe width decreases.

### Final Answer:
The fringe width will **decrease**.