De Broglie wavelength `lambda` associated with neutrons is related with absolute temperature T as

Options

A. `lambda prop T`
B. `lambda prop (1)/(T)`
C. `lambda prop(1)/(sqrtT)`
D. `lambda prop T^(2)`

Explanation

To solve the problem regarding the relationship between the de Broglie wavelength \( \lambda \) associated with neutrons and absolute temperature \( T \), we can follow these steps:

### Step-by-Step Solution:

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 linear momentum of the particle.

2. **Express Linear Momentum**:
The linear momentum \( p \) can be expressed in terms of mass \( m \) and velocity \( v \):
\[ p = mv \]
For neutrons, we can also relate momentum to kinetic energy \( K \):
\[ K = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2K}{m}} \implies p = m \sqrt{\frac{2K}{m}} = \sqrt{2mK} \]

3. **Relate Kinetic Energy to Temperature**:
The kinetic energy of a particle is related to the absolute temperature \( T \) by the equation:
\[ K \propto T \]
This means that as temperature increases, the kinetic energy of the neutrons increases.

4. **Substituting Kinetic Energy into Momentum**:
Substituting \( K \) into the momentum expression gives:
\[ p = \sqrt{2mK} \propto \sqrt{2mT} \]
Therefore, we can express \( p \) as:
\[ p \propto \sqrt{T} \]

5. **Substituting Momentum Back into the Wavelength Formula**:
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \propto \frac{h}{\sqrt{T}} \]

6. **Conclusion**:
From the above relationship, we can conclude that:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
Thus, the de Broglie wavelength \( \lambda \) associated with neutrons is inversely proportional to the square root of the absolute temperature \( T \).

### Final Answer:
The correct relationship is:
\[ \lambda \text{ is inversely proportional to } \sqrt{T} \]