If the temperature of a black body becomes half of its original temperature, then the amount of radiation emitted by the body per second will be reduced to

Options

A. `1/2`
B. `1/4`
C. `1/16`
D. `1/8`

Explanation

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

\[ P = \sigma A T^4 \]

where:
- \( P \) is the power (amount of radiation emitted per second),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the body,
- \( T \) is the absolute temperature of the body.

### Step-by-Step Solution:

1. **Define the Initial Temperature**:
Let the initial temperature of the black body be \( T_0 \).

2. **Define the Final Temperature**:
According to the problem, the final temperature \( T_1 \) is half of the initial temperature:
\[ T_1 = \frac{T_0}{2} \]

3. **Calculate the Initial Power**:
Using the Stefan-Boltzmann Law, the initial power emitted by the black body is:
\[ P_0 = \sigma A T_0^4 \]

4. **Calculate the Final Power**:
Now, substituting \( T_1 \) into the Stefan-Boltzmann Law, the final power emitted is:
\[ P_1 = \sigma A T_1^4 = \sigma A \left(\frac{T_0}{2}\right)^4 \]
Simplifying this gives:
\[ P_1 = \sigma A \frac{T_0^4}{16} = \frac{1}{16} \sigma A T_0^4 \]

5. **Relate Final Power to Initial Power**:
Now we can relate the final power to the initial power:
\[ P_1 = \frac{1}{16} P_0 \]

6. **Conclusion**:
Thus, the amount of radiation emitted by the body per second when the temperature is halved is reduced to:
\[ P_1 = \frac{1}{16} P_0 \]

### Final Answer:
The amount of radiation emitted by the body per second will be reduced to \( \frac{1}{16} \) of its original value.

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