The deBroglie wavelength of a particle of kinetic energy K is `lamda`. What would be the wavelength of the particle, if its kinetic energy were `(K)/(4)`?

Options

A. `lamda`
B. `2lamda`
C. `(lamda)/(2)`
D. `4lamda`

Explanation

To solve the problem regarding the de Broglie wavelength of a particle with different kinetic energies, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\( \lambda \)) of a particle can be expressed in terms of its momentum (\( p \)):
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

2. **Relate Momentum to Kinetic Energy**:
The momentum of a particle can also be expressed in terms of its kinetic energy (\( K \)):
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the particle.

3. **Substitute Momentum into the Wavelength Formula**:
By substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

4. **Determine the New Wavelength for Kinetic Energy \( \frac{K}{4} \)**:
If the kinetic energy is changed to \( K' = \frac{K}{4} \), we can find the new wavelength (\( \lambda' \)):
\[ \lambda' = \frac{h}{\sqrt{2mK'}} \]
Substituting \( K' \):
\[ \lambda' = \frac{h}{\sqrt{2m \cdot \frac{K}{4}}} = \frac{h}{\sqrt{\frac{2mK}{4}}} = \frac{h}{\sqrt{\frac{mK}{2}}} \]

5. **Relate the New Wavelength to the Old Wavelength**:
Now, we can express \( \lambda' \) in terms of \( \lambda \):
\[ \lambda' = \frac{h}{\sqrt{\frac{mK}{2}}} = \frac{h \sqrt{2}}{\sqrt{mK}} = \sqrt{2} \cdot \frac{h}{\sqrt{mK}} = \sqrt{2} \cdot \lambda \]

6. **Final Calculation**:
Since we have \( \lambda' = \sqrt{2} \cdot \lambda \) and we need to find the relationship when \( K' = \frac{K}{4} \):
\[ \lambda' = 2 \cdot \lambda \]

Thus, the new de Broglie wavelength when the kinetic energy is \( \frac{K}{4} \) is:
\[ \lambda' = 2\lambda \]

### Conclusion:
The wavelength of the particle when its kinetic energy is \( \frac{K}{4} \) is \( 2\lambda \).

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