The maximum spectral emissive power at black body temperature 5000 K is obtained at `lamda_m= 6000 `Å. If the temperature is increased by 10%, then decrease in `lamda_m` will be

Options

A. 0.15
B. 0.1
C. 0.05
D. 0.03

Explanation

To solve the problem step by step, we will use Wien's Displacement Law, which states that the product of the wavelength at which the spectral emissive power is maximum (λ_m) and the absolute temperature (T) of the black body is a constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature, \( T_1 = 5000 \, K \)
- Initial wavelength, \( \lambda_{m1} = 6000 \, \text{Å} \)
- Increase in temperature, \( \Delta T = 10\% \)

2. **Calculate the New Temperature:**
\[ T_2 = T_1 + 0.1 \cdot T_1 = 5000 + 0.1 \cdot 5000 = 5000 + 500 = 5500 \, K \]

3. **Apply Wien's Displacement Law:**
According to Wien's Displacement Law:
\[ \lambda_{m1} \cdot T_1 = \lambda_{m2} \cdot T_2 \]
Rearranging gives:
\[ \lambda_{m2} = \frac{\lambda_{m1} \cdot T_1}{T_2} \]

4. **Substitute the Values:**
\[ \lambda_{m2} = \frac{6000 \, \text{Å} \cdot 5000 \, K}{5500 \, K} \]

5. **Calculate \( \lambda_{m2} \):**
\[ \lambda_{m2} = \frac{6000 \cdot 5000}{5500} = \frac{30000000}{5500} \approx 5454.55 \, \text{Å} \]

6. **Calculate the Decrease in Wavelength:**
\[ \Delta \lambda = \lambda_{m1} - \lambda_{m2} = 6000 \, \text{Å} - 5454.55 \, \text{Å} \approx 545.45 \, \text{Å} \]

7. **Express the Decrease as a Percentage:**
\[ \text{Percentage Decrease} = \frac{\Delta \lambda}{\lambda_{m1}} \times 100 = \frac{545.45}{6000} \times 100 \approx 9.09\% \]

### Final Result:
The decrease in \( \lambda_m \) when the temperature is increased by 10% is approximately **545.45 Å**.