The kinetic energy of an electron gets tripled, then the de-Broglie wavelength associated with it changes by a factor

Options

A. `1/3`
B. `sqrt(3)`
C. `(1)/(sqrt(3))`
D. 3

Explanation

To solve the problem, we need to determine how the de-Broglie wavelength of an electron changes when its kinetic energy is tripled.

### Step-by-Step Solution:

1. **Understand the de-Broglie Wavelength Formula**:
The de-Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

2. **Relate Momentum to Kinetic Energy**:
The momentum \( p \) of an electron can be expressed in terms of its mass \( m \) and velocity \( v \):
\[ p = mv \]
The kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

3. **Substituting Kinetic Energy into Momentum**:
We can substitute \( v \) into the momentum equation:
\[ p = m \cdot \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{2m \cdot KE} \]

4. **Substituting Momentum into the de-Broglie Wavelength Formula**:
Now substituting \( p \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

5. **Analyzing the Change in Kinetic Energy**:
If the initial kinetic energy is \( KE_1 \) and the final kinetic energy is \( KE_2 = 3 \cdot KE_1 \) (since it is tripled), we can express the final wavelength \( \lambda_2 \):
\[ \lambda_2 = \frac{h}{\sqrt{2m \cdot KE_2}} = \frac{h}{\sqrt{2m \cdot (3 \cdot KE_1)}} = \frac{h}{\sqrt{6m \cdot KE_1}} \]

6. **Finding the Ratio of Wavelengths**:
To find the ratio of the final wavelength to the initial wavelength:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\sqrt{6m \cdot KE_1}}{h} \cdot \frac{h}{\sqrt{2m \cdot KE_1}} = \sqrt{\frac{6}{2}} = \sqrt{3} \]
Therefore:
\[ \lambda_2 = \frac{\lambda_1}{\sqrt{3}} \]

7. **Conclusion**:
The de-Broglie wavelength changes by a factor of \( \frac{1}{\sqrt{3}} \).

### Final Answer:
The de-Broglie wavelength associated with the electron changes by a factor of \( \frac{1}{\sqrt{3}} \).