Energy of radiation emitted by a black body at temperature 3000 K is `u_(1)` for wavelength between 8000 `Å` and `9000Å`, `u_(2)` for wavelength between 9000 `Å` and `10000Å` and `u_(3)` forwavelength between `10000Åand 11000Å`.Which of the following is true? [Wien’s constant `b=2.88xx10^(-3)mK]`

Options

A. (a)`u_(1)gtu_(2)gtu_(3)`
B. (b)`u_(3)gtu_(2)gtu_(1)`
C. (c)`u_(2)gtu_(1)andu_(2)gtu_(3)`
D. (d)`u_(2)gtu_(1)=u_(2)`

Explanation

To solve the problem, we will use Wien's Law, which relates the temperature of a black body to the wavelength at which it emits the maximum energy. The steps are as follows:

### Step 1: Understand Wien's Law
Wien's Law states that the wavelength at which the emission of a black body spectrum is maximized (λ_max) is inversely proportional to the temperature (T) of the black body. This can be expressed mathematically as:
\[ \lambda_{max} \cdot T = b \]
where \( b \) is Wien's constant.

### Step 2: Calculate λ_max for the given temperature
Given:
- Temperature \( T = 3000 \, K \)
- Wien's constant \( b = 2.88 \times 10^{-3} \, m \cdot K \)

We can rearrange the equation to find \( \lambda_{max} \):
\[ \lambda_{max} = \frac{b}{T} \]
Substituting the values:
\[ \lambda_{max} = \frac{2.88 \times 10^{-3} \, m \cdot K}{3000 \, K} \]

### Step 3: Convert the wavelength to Angstroms
First, calculate \( \lambda_{max} \):
\[ \lambda_{max} = \frac{2.88 \times 10^{-3}}{3000} = 9.6 \times 10^{-7} \, m \]
Now convert meters to Angstroms (1 m = \( 10^{10} \, Å \)):
\[ \lambda_{max} = 9.6 \times 10^{-7} \, m \times 10^{10} \, Å/m = 9600 \, Å \]

### Step 4: Determine the energy distribution
Now we compare \( \lambda_{max} \) with the given wavelength ranges:
- \( u_1 \): 8000 Å to 9000 Å
- \( u_2 \): 9000 Å to 10000 Å
- \( u_3 \): 10000 Å to 11000 Å

Since \( \lambda_{max} = 9600 \, Å \), it falls within the range of \( u_2 \).

### Step 5: Compare the energy emitted in each range
According to Wien's Law, since \( \lambda_{max} \) is in the range of \( u_2 \), we can conclude:
- The energy emitted in the range \( u_2 \) (which includes \( \lambda_{max} \)) is greater than that emitted in the ranges \( u_1 \) and \( u_3 \).

Thus, we can write:
\[ u_2 > u_1 \quad \text{and} \quad u_2 > u_3 \]

### Conclusion
The correct relationship is:
\[ u_2 > u_1 \quad \text{and} \quad u_2 > u_3 \]
Thus, the correct option is C.

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