If at temperature `T_(1)=1000K`, the wavelength is `2.2xx10^(-6)` m, then at what temperature the wavelength will be `4.4xx10^(-5)m`?

Options

A. 52 K
B. 47 K
C. 49 K
D. 48 K

Explanation

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

\[ \lambda_1 T_1 = \lambda_2 T_2 \]

Where:
- \( \lambda_1 \) is the initial wavelength,
- \( T_1 \) is the initial temperature,
- \( \lambda_2 \) is the final wavelength,
- \( T_2 \) is the final temperature.

Given:
- \( T_1 = 1000 \, K \)
- \( \lambda_1 = 2.2 \times 10^{-6} \, m \)
- \( \lambda_2 = 4.4 \times 10^{-5} \, m \)

We need to find \( T_2 \).

### Step-by-Step Solution:

1. **Write down the formula from Wien's Displacement Law**:
\[ \lambda_1 T_1 = \lambda_2 T_2 \]

2. **Rearrange the formula to solve for \( T_2 \)**:
\[ T_2 = \frac{\lambda_1 T_1}{\lambda_2} \]

3. **Substitute the known values into the equation**:
\[ T_2 = \frac{(2.2 \times 10^{-6} \, m)(1000 \, K)}{4.4 \times 10^{-5} \, m} \]

4. **Calculate the numerator**:
\[ 2.2 \times 10^{-6} \times 1000 = 2.2 \times 10^{-3} \]

5. **Now, divide by \( \lambda_2 \)**:
\[ T_2 = \frac{2.2 \times 10^{-3}}{4.4 \times 10^{-5}} \]

6. **Perform the division**:
\[ T_2 = \frac{2.2}{4.4} \times 10^{(-3 + 5)} = 0.5 \times 10^{2} = 50 \, K \]

7. **Final answer**:
\[ T_2 = 50 \, K \]

### Conclusion:
The temperature at which the wavelength will be \( 4.4 \times 10^{-5} \, m \) is \( 50 \, K \).