Temperature of black body is `3000K`. When black body cools, then change in wevelength `Deltalambda = 9` micron corresponding to maximum energy density. Now temperature of black body is :-

Options

A. `300K`
B. `2700K`
C. `270K`
D. `1800K`

Explanation

To solve the problem, we will use Wien's displacement law, which states that the product of the wavelength of maximum emission (\( \lambda \)) and the absolute temperature (\( T \)) of a black body is a constant. The constant is given by:

\[ \lambda \cdot T = b \]

where \( b = 2.898 \times 10^{-3} \, \text{K m} \).

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial temperature \( T_1 = 3000 \, K \).
- Change in wavelength \( \Delta \lambda = 9 \, \mu m = 9 \times 10^{-6} \, m \).

2. **Calculate Initial Wavelength**:
- Using Wien's law, we can find the initial wavelength \( \lambda_1 \):
\[ \lambda_1 = \frac{b}{T_1} = \frac{2.898 \times 10^{-3}}{3000} \]
- Performing the calculation:
\[ \lambda_1 = \frac{2.898 \times 10^{-3}}{3 \times 10^{3}} = 0.966 \times 10^{-6} \, m \]

3. **Calculate New Wavelength**:
- The new wavelength \( \lambda_2 \) after cooling is:
\[ \lambda_2 = \lambda_1 + \Delta \lambda = 0.966 \times 10^{-6} + 9 \times 10^{-6} \]
- Performing the addition:
\[ \lambda_2 = 0.966 \times 10^{-6} + 9.000 \times 10^{-6} = 9.966 \times 10^{-6} \, m \]

4. **Calculate Final Temperature**:
- Using Wien's law again for the new wavelength:
\[ T_2 = \frac{b}{\lambda_2} = \frac{2.898 \times 10^{-3}}{9.966 \times 10^{-6}} \]
- Performing the calculation:
\[ T_2 \approx \frac{2.898 \times 10^{-3}}{9.966 \times 10^{-6}} \approx 290 \, K \]

5. **Conclusion**:
- The final temperature of the black body after cooling is approximately \( 290 \, K \).

### Final Answer:
The temperature of the black body after cooling is approximately **290 K**.