The maximum wavelength of radiation emitted at 2000 K is 4 `mu` m. The maximum wavelength of radiation emitted at 2400 K is

Options

A. 3.33`mu m`
B. 2.22`mu m`
C. 1.11nm
D. 0.111mum

Explanation

To solve the problem of finding the maximum wavelength of radiation emitted at 2400 K, we will use Wien's Displacement Law. This law states that the wavelength (λ) of the radiation emitted by a black body is inversely proportional to its absolute temperature (T).

### Step-by-Step Solution:

1. **Understand Wien's Displacement Law**:
According to Wien's Displacement Law, the relationship can be expressed as:
\[ \lambda \cdot T = b \]
where \( b \) is a constant.

2. **Identify the given values**:
- For the first case:
- \( T_1 = 2000 \, K \)
- \( \lambda_1 = 4 \, \mu m = 4 \times 10^{-6} \, m \)
- For the second case:
- \( T_2 = 2400 \, K \)
- We need to find \( \lambda_2 \).

3. **Set up the equation using Wien's Law**:
From Wien's Law, we can write:
\[ \lambda_1 \cdot T_1 = \lambda_2 \cdot T_2 \]

4. **Rearranging the equation to find \( \lambda_2 \)**:
\[ \lambda_2 = \frac{\lambda_1 \cdot T_1}{T_2} \]

5. **Substituting the known values**:
\[ \lambda_2 = \frac{(4 \times 10^{-6} \, m) \cdot (2000 \, K)}{2400 \, K} \]

6. **Calculating \( \lambda_2 \)**:
\[ \lambda_2 = \frac{(4 \times 10^{-6}) \cdot 2000}{2400} \]
\[ \lambda_2 = \frac{8 \times 10^{-3}}{2400} \]
\[ \lambda_2 = 3.33 \times 10^{-6} \, m = 3.33 \, \mu m \]

### Final Answer:
The maximum wavelength of radiation emitted at 2400 K is \( 3.33 \, \mu m \).

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