A black body radiates energy at the rate of E `W//m` at a high temperature TK . When the temperature is reduced to `(T)/(2)K` , the radiant energy will b

Answer

To solve the problem, we will use the Stefan-Boltzmann law, which states that the energy emitted per unit area per unit time (intensity) of a black body is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Understand the Stefan-Boltzmann Law**:
The law states that the energy emitted \( E \) at a temperature \( T \) is given by:
\[ E \propto T^4 \]
This means that if the temperature changes, the energy emitted will change according to the fourth power of the new temperature.

2. **Initial Conditions**:
At a high temperature \( T_K \), the energy emitted is \( E \) watts per meter squared:
\[ E = k \cdot T_K^4 \]
where \( k \) is the proportionality constant.

3. **New Temperature**:
When the temperature is reduced to \( \frac{T}{2} \) (or \( \frac{T_K}{2} \)), we need to find the new energy emitted, which we will denote as \( E' \):
\[ E' = k \cdot \left(\frac{T_K}{2}\right)^4 \]

4. **Calculate \( E' \)**:
Now substituting \( \frac{T_K}{2} \) into the equation:
\[ E' = k \cdot \left(\frac{T_K^4}{16}\right) = \frac{k \cdot T_K^4}{16} \]

5. **Relate \( E' \) to \( E \)**:
Since we know that \( E = k \cdot T_K^4 \), we can substitute this into our equation for \( E' \):
\[ E' = \frac{E}{16} \]

6. **Final Answer**:
Therefore, when the temperature is reduced to \( \frac{T_K}{2} \), the radiant energy emitted will be:
\[ E' = \frac{E}{16} \]

### Conclusion:
The radiant energy emitted when the temperature is reduced to half is \( \frac{E}{16} \).

---