A black body at high temperature `T` `K` radiates energy at the rate of `E W//m^(2)`. When the temperature falls to `(T//2) K`, the radiated energy will be

Options

A. `E//4`
B. `E//2`
C. `2E`
D. `E//16`

Explanation

To solve the problem, we will use the Stefan-Boltzmann law, which states that the energy radiated per unit area of a black body is proportional to the fourth power of its absolute temperature. The formula is given by:

\[ E = \sigma T^4 \]

where:
- \( E \) is the energy radiated per unit area (in watts per square meter),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-step Solution:

1. **Identify the initial conditions**:
- The initial temperature \( T_1 = T \) K.
- The initial energy radiated \( E_1 = E \) W/m².

2. **Apply the Stefan-Boltzmann law**:
- For the initial temperature \( T_1 \):
\[ E_1 = \sigma T_1^4 = \sigma T^4 \]

3. **Determine the new temperature**:
- The new temperature \( T_2 = \frac{T}{2} \) K.

4. **Calculate the energy radiated at the new temperature**:
- Using the Stefan-Boltzmann law for the new temperature \( T_2 \):
\[ E_2 = \sigma T_2^4 = \sigma \left(\frac{T}{2}\right)^4 \]

5. **Simplify the expression**:
- Calculate \( \left(\frac{T}{2}\right)^4 \):
\[ \left(\frac{T}{2}\right)^4 = \frac{T^4}{16} \]
- Therefore:
\[ E_2 = \sigma \frac{T^4}{16} \]

6. **Relate \( E_2 \) to \( E_1 \)**:
- Since \( E_1 = \sigma T^4 \), we can express \( E_2 \) in terms of \( E_1 \):
\[ E_2 = \frac{E_1}{16} \]

7. **Substituting \( E_1 \)**:
- Since \( E_1 = E \):
\[ E_2 = \frac{E}{16} \]

### Final Answer:
The radiated energy when the temperature falls to \( \frac{T}{2} \) K will be \( \frac{E}{16} \) W/m².

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