If temperature becomes double, the emitted radiation will be :

Options

A. 16 times
B. 8 times
C. `2sqrt(2)` times
D. 32 times

Explanation

To solve the problem of how the emitted radiation changes when the temperature is doubled, we can use Stefan-Boltzmann Law, which states that the energy emitted by 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) by a black body at absolute temperature (T) is given by:
\[ E = \sigma T^4 \]
where \(\sigma\) is the Stefan-Boltzmann constant.

2. **Set Initial and Final Temperatures**:
Let the initial temperature be \(T_1 = T\) and the final temperature when it is doubled be \(T_2 = 2T\).

3. **Write the Emitted Radiation for Both Temperatures**:
- For the initial temperature \(T_1\):
\[ E_1 = \sigma T^4 \]
- For the final temperature \(T_2\):
\[ E_2 = \sigma (2T)^4 \]

4. **Calculate \(E_2\)**:
Expanding \(E_2\):
\[ E_2 = \sigma (2T)^4 = \sigma \cdot 16T^4 \]
Thus, we can express \(E_2\) in terms of \(E_1\):
\[ E_2 = 16 \sigma T^4 = 16E_1 \]

5. **Conclusion**:
Therefore, when the temperature is doubled, the emitted radiation becomes:
\[ E_2 = 16E_1 \]
This means the emitted radiation increases by a factor of 16.

### Final Answer:
If the temperature becomes double, the emitted radiation will be 16 times the initial radiation.

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