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

Options

A. `2.5%`
B. `5.0%`
C. `7.5%`
D. `10%`

Answer

To solve the problem, 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. This constant is known as Wien's constant (B).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature, T = 6000 K
- Initial wavelength, λ_m = 5000 Å
- Temperature increase = 10%

2. **Calculate the New Temperature:**
- The new temperature (T') after a 10% increase can be calculated as:
\[ T' = T + 0.1T = 1.1T = 1.1 \times 6000 \text{ K} = 6600 \text{ K} \]

3. **Apply Wien's Displacement Law:**
- According to Wien's Displacement Law:
\[ \lambda_m \cdot T = B \]
- For the initial conditions:
\[ \lambda_m \cdot 6000 = B \]
- For the new conditions:
\[ \lambda_m' \cdot 6600 = B \]

4. **Relate the Two Conditions:**
- Since B is constant, we can set the two equations equal to each other:
\[ \lambda_m \cdot 6000 = \lambda_m' \cdot 6600 \]
- Rearranging gives:
\[ \lambda_m' = \frac{\lambda_m \cdot 6000}{6600} \]

5. **Substitute the Known Values:**
- Substitute λ_m = 5000 Å into the equation:
\[ \lambda_m' = \frac{5000 \cdot 6000}{6600} \]

6. **Calculate λ_m':**
- Performing the calculation:
\[ \lambda_m' = \frac{30000000}{6600} \approx 4545.45 \text{ Å} \]

7. **Determine the Change in Wavelength:**
- The change in wavelength (Δλ) is:
\[ \Delta \lambda = \lambda_m - \lambda_m' = 5000 - 4545.45 \approx 454.55 \text{ Å} \]

8. **Calculate the Percentage Decrease:**
- The percentage decrease in wavelength is given by:
\[ \text{Percentage Decrease} = \left(\frac{\Delta \lambda}{\lambda_m}\right) \times 100 = \left(\frac{454.55}{5000}\right) \times 100 \approx 9.1\% \]

### Final Answer:
The decrease in λ_m when the temperature is increased by 10% is approximately **9.1%**.

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