A black body emits radiations of maximum intensity at a wavelength of Å 5000 , when the temperature of the body is `1227^(@)C`. If the temperature of the body is increased by `1000^(@)C` , the maximum intensity of emitted radiation would be observed at

Answer

To solve the problem, we will use Wien's Displacement Law, which states that the product of the wavelength at which the intensity of radiation is maximum (λ_max) and the absolute temperature (T) of the black body is a constant. This can be expressed mathematically as:

\[ \lambda_{1} \cdot T_{1} = \lambda_{2} \cdot T_{2} \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial wavelength (λ_1) = 5000 Å
- Initial temperature (T_1) = 1227 °C
- Increase in temperature = 1000 °C

2. **Convert the Initial Temperature to Kelvin**:
- To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T_1 = 1227 + 273 = 1500 \, K \]

3. **Calculate the New Temperature (T_2)**:
- The new temperature after the increase of 1000 °C:
\[ T_2 = 1227 + 1000 + 273 = 2500 \, K \]

4. **Set Up the Equation Using Wien's Displacement Law**:
- According to Wien's law:
\[ \lambda_{1} \cdot T_{1} = \lambda_{2} \cdot T_{2} \]
- Plugging in the known values:
\[ 5000 \, \text{Å} \cdot 1500 \, K = \lambda_{2} \cdot 2500 \, K \]

5. **Solve for λ_2**:
- Rearranging the equation to find λ_2:
\[ \lambda_{2} = \frac{5000 \, \text{Å} \cdot 1500 \, K}{2500 \, K} \]
- Simplifying:
\[ \lambda_{2} = \frac{7500000 \, \text{Å} \cdot K}{2500 \, K} \]
\[ \lambda_{2} = 3000 \, \text{Å} \]

6. **Conclusion**:
- The maximum intensity of emitted radiation would be observed at a wavelength of **3000 Å**.