The maximum wavelength of radiation emitted at 200 K is 4 μm. What will be the maximum wavelength of radiation emitted at 2400 K ?

Options

A. `3mum`
B. `0.3mum`
C. `2mum`
D. None of these

Explanation

To solve the problem of finding the maximum wavelength of radiation emitted at 2400 K given that the maximum wavelength at 200 K is 4 μm, we will use Wien's Displacement Law. This law states that the product of the maximum wavelength (λ_max) and the absolute temperature (T) is a constant.

### Step-by-Step Solution:

1. **Understand Wien's Displacement Law**:
According to Wien's Displacement Law, we have:
\[ \lambda_{max} \cdot T = b \]
where \( b \) is a constant.

2. **Set up the equation for two temperatures**:
For two different temperatures, we can express it as:
\[ \lambda_{m1} \cdot T_1 = \lambda_{m2} \cdot T_2 \]
where:
- \( \lambda_{m1} \) = maximum wavelength at temperature \( T_1 \)
- \( \lambda_{m2} \) = maximum wavelength at temperature \( T_2 \)

3. **Substitute the known values**:
Given:
- \( \lambda_{m1} = 4 \, \mu m \) (at \( T_1 = 200 \, K \))
- \( T_2 = 2400 \, K \)

We need to find \( \lambda_{m2} \):
\[ 4 \, \mu m \cdot 200 \, K = \lambda_{m2} \cdot 2400 \, K \]

4. **Rearrange the equation to solve for \( \lambda_{m2} \)**:
\[ \lambda_{m2} = \frac{4 \, \mu m \cdot 200 \, K}{2400 \, K} \]

5. **Calculate \( \lambda_{m2} \)**:
- First, simplify the fraction:
\[ \lambda_{m2} = \frac{800 \, \mu m \cdot K}{2400 \, K} \]
- Now, cancel out the common factors:
\[ \lambda_{m2} = \frac{800}{2400} \, \mu m = \frac{1}{3} \, \mu m \]

6. **Convert to micrometers**:
\[ \lambda_{m2} = 0.333... \, \mu m \approx 0.3 \, \mu m \]

### Final Answer:
The maximum wavelength of radiation emitted at 2400 K is approximately **0.3 μm**.