A black body is at a temperature of `5760 K`. The energy of radiation emitted by the body at wavelength `250 nm` is `U_(1)` at wavelength `500 nm` is `U_(2)` and that at `1000 nm` is `U_(3)`. Wien's consant, `b = 2.88 xx 10^(6) nmK`. Which of the following is correct?

Options

A. `U_(1) = 0`
B. `U_(3) = 0`
C. `U_(1) gt U_(2)`
D. `U_(2) gt U_(1)`

Explanation

To solve the problem, we will use Wien's Displacement Law and analyze the emitted radiation at different wavelengths.

### Step-by-Step Solution:

1. **Understanding Wien's Displacement Law**:
Wien's Displacement Law states that the wavelength at which the emission of a black body spectrum is maximized (λm) is inversely proportional to the temperature (T) of the black body. The formula is given by:
\[ \lambda_m = \frac{b}{T} \]
where \( b \) is Wien's constant.

2. **Calculate λm**:
Given:
- Temperature \( T = 5760 \, K \)
- Wien's constant \( b = 2.88 \times 10^6 \, nm \cdot K \)

Substitute the values into the formula:
\[ \lambda_m = \frac{2.88 \times 10^6 \, nm \cdot K}{5760 \, K} \]

Performing the calculation:
\[ \lambda_m = 500 \, nm \]

3. **Identify the wavelengths**:
We have three wavelengths:
- \( \lambda_1 = 250 \, nm \) (U1)
- \( \lambda_2 = 500 \, nm \) (U2)
- \( \lambda_3 = 1000 \, nm \) (U3)

From the calculation, we see that \( \lambda_2 \) is the wavelength at which the maximum intensity occurs.

4. **Analyze the emitted energy**:
According to the black body radiation curve:
- At \( \lambda_2 = 500 \, nm \), the emitted energy \( U_2 \) is the maximum.
- At \( \lambda_1 = 250 \, nm \) and \( \lambda_3 = 1000 \, nm \), the emitted energies \( U_1 \) and \( U_3 \) will be less than \( U_2 \).

Therefore, we can conclude:
\[ U_2 > U_1 \quad \text{and} \quad U_2 > U_3 \]

5. **Conclusion**:
The correct statement based on the analysis is that \( U_2 \) is greater than both \( U_1 \) and \( U_3 \). Thus, the correct option is:
\[ U_2 > U_1 \quad \text{and} \quad U_2 > U_3 \]