Solar radiation emitted by sun resembles that emitted by a body at a temperature of `6000 K` Maximum intensity is emitted at a wavelength of about `4800A^(@)` If the sun was cooled down from `6000K` to `3000K` then the peak intensity would occurs at a wavelength of .

Options

A. (a)`6000Å`
B. (b)`9600Å`
C. (c)`2400Å`
D. (d)`19200Å`

Answer

To solve the problem of finding the peak wavelength of solar radiation when the temperature of the sun is reduced from 6000 K to 3000 K, we can use Wien's Displacement Law. This law states that the product of the peak wavelength (λ_max) and the absolute temperature (T) of a black body is a constant.

### Step-by-Step Solution:

1. **Understand Wien's Displacement Law**:
\[ \lambda_{\text{max}} \cdot T = b \]
where \( b \) is Wien's displacement constant.

2. **Identify Given Values**:
- For the sun at 6000 K, the peak wavelength \( \lambda_1 = 4800 \) Å (angstrom).
- Temperature \( T_1 = 6000 \) K.
- New temperature \( T_2 = 3000 \) K.
- We need to find the new peak wavelength \( \lambda_2 \).

3. **Set Up the Equation Using Wien's Law**:
According to Wien's law:
\[ \lambda_1 \cdot T_1 = \lambda_2 \cdot T_2 \]

4. **Substitute the Known Values**:
\[ 4800 \, \text{Å} \cdot 6000 \, \text{K} = \lambda_2 \cdot 3000 \, \text{K} \]

5. **Solve for \( \lambda_2 \)**:
Rearranging the equation gives:
\[ \lambda_2 = \frac{4800 \, \text{Å} \cdot 6000 \, \text{K}}{3000 \, \text{K}} \]
\[ \lambda_2 = \frac{4800 \cdot 6000}{3000} \]
\[ \lambda_2 = \frac{28800000}{3000} \]
\[ \lambda_2 = 9600 \, \text{Å} \]

6. **Conclusion**:
The peak intensity would occur at a wavelength of \( 9600 \, \text{Å} \) when the sun is cooled down to \( 3000 \, \text{K} \).

### Final Answer:
The peak intensity occurs at a wavelength of \( 9600 \, \text{Å} \).