For an enclosure maintained at 2000K, the maximum radiation occurs at wavelength `lambda_m`. If the temperature is raised to 3000K, the peak will shift to

Options

A. `0.5 lambda_m`
B. `lambda_m`
C. `2/3 lambda_m`
D. `3/2lambda_m`

Explanation

To solve the problem, we will use Wien's Displacement Law, which states that the wavelength corresponding to the maximum intensity of radiation emitted by a black body is inversely proportional to its absolute temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature \( T_1 = 2000 \, K \)
- Final temperature \( T_2 = 3000 \, K \)
- Wavelength at initial temperature \( \lambda_m \) (unknown value)

2. **Apply Wien's Displacement Law:**
According to Wien's Displacement Law:
\[ \lambda_m \cdot T = b \]
where \( b \) is a constant. Therefore, we can write:
\[ \lambda_{m1} \cdot T_1 = \lambda_{m2} \cdot T_2 \]
Here, \( \lambda_{m1} \) is the wavelength at temperature \( T_1 \) and \( \lambda_{m2} \) is the wavelength at temperature \( T_2 \).

3. **Set Up the Equation:**
Substituting the known values:
\[ \lambda_m \cdot 2000 = \lambda_{m2} \cdot 3000 \]

4. **Solve for \( \lambda_{m2} \):**
Rearranging the equation gives:
\[ \lambda_{m2} = \lambda_m \cdot \frac{2000}{3000} \]
Simplifying the fraction:
\[ \lambda_{m2} = \lambda_m \cdot \frac{2}{3} \]

5. **Conclusion:**
The peak wavelength at the new temperature \( T_2 = 3000 \, K \) will be:
\[ \lambda_{m2} = \frac{2}{3} \lambda_m \]

### Final Answer:
The peak will shift to \( \frac{2}{3} \lambda_m \).