The graph showing the relation between de Broglie wavelength `lamda` of particle and its temperature T is

Options

A.
B.
C.
D.

Explanation

To solve the problem of finding the relationship between the de Broglie wavelength (λ) of a particle and its temperature (T), we will follow these steps:

### Step-by-Step Solution:

1. **Understand de Broglie Wavelength**:
The de Broglie wavelength (λ) 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.

2. **Relate Momentum to Kinetic Energy**:
The momentum \(p\) can be expressed in terms of kinetic energy (KE):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]

3. **Substitute Kinetic Energy**:
From the kinetic theory of gases, the average kinetic energy of a particle is given by:
\[ KE = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature. Substituting this into the momentum equation gives:
\[ p = \sqrt{2m \cdot \frac{3}{2} k T} = \sqrt{3mkT} \]

4. **Substitute Momentum Back into Wavelength**:
Now substitute \(p\) back into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

5. **Rearranging the Equation**:
This can be rearranged to show that:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
which implies:
\[ \lambda = \frac{h}{\sqrt{3mk}} \cdot \frac{1}{\sqrt{T}} \]
Here, \(\frac{h}{\sqrt{3mk}}\) is a constant.

6. **Taking Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln(\lambda) = -\frac{1}{2} \ln(T) + \ln\left(\frac{h}{\sqrt{3mk}}\right) \]
This can be rearranged to:
\[ \ln(\lambda) = -\frac{1}{2} \ln(T) + C \]
where \(C = \ln\left(\frac{h}{\sqrt{3mk}}\right)\).

7. **Identifying the Graph**:
The equation \(\ln(\lambda) = -\frac{1}{2} \ln(T) + C\) represents a straight line with:
- Slope \(m = -\frac{1}{2}\) (negative slope)
- Intercept \(C\) (positive value)

Therefore, when plotting \(\ln(\lambda)\) against \(\ln(T)\), we expect a straight line with a negative slope.

### Conclusion:
The graph showing the relation between the de Broglie wavelength \(\lambda\) of a particle and its temperature \(T\) will be a straight line with a negative slope.