The ratio of energies of emitted radiation by a black body at 600 K and 900 K , when the surrounding temperature is 300 K , is

Options

A. `5/16`
B. `7/16`
C. `3/16`
D. `9/16`

Answer

To solve the problem of finding the ratio of energies of emitted radiation by a black body at temperatures of 600 K and 900 K, with a surrounding temperature of 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Stefan-Boltzmann Law**: The energy emitted by a black body is proportional to the fourth power of its absolute temperature. Mathematically, this is expressed as:
\[ E \propto T^4 \]

2. **Consider the Surrounding Temperature**: Since the black body is not in a vacuum, we need to account for the surrounding temperature (Ts = 300 K). The effective energy emitted by the black body is given by:
\[ E = \sigma (T^4 - T_s^4) \]
where \( \sigma \) is the Stefan-Boltzmann constant.

3. **Set Up the Ratio of Energies**: Let \( E_1 \) be the energy emitted at 600 K and \( E_2 \) be the energy emitted at 900 K. We can express these as:
\[ E_1 = \sigma (T_1^4 - T_s^4) = \sigma (600^4 - 300^4) \]
\[ E_2 = \sigma (T_2^4 - T_s^4) = \sigma (900^4 - 300^4) \]

4. **Form the Ratio**: The ratio of the energies emitted can be expressed as:
\[ \frac{E_1}{E_2} = \frac{600^4 - 300^4}{900^4 - 300^4} \]

5. **Calculate the Values**:
- First, calculate \( 600^4 \):
\[ 600^4 = 129600000000 \]
- Next, calculate \( 300^4 \):
\[ 300^4 = 8100000000 \]
- Now, calculate \( 900^4 \):
\[ 900^4 = 656100000000 \]

6. **Substitute the Values**:
- Substitute these values into the ratio:
\[ \frac{E_1}{E_2} = \frac{129600000000 - 8100000000}{656100000000 - 8100000000} \]
- This simplifies to:
\[ \frac{E_1}{E_2} = \frac{121500000000}{648900000000} \]

7. **Simplify the Ratio**:
- Divide both the numerator and denominator by \( 121500000000 \):
\[ \frac{E_1}{E_2} = \frac{1}{5.35} \approx \frac{3}{16} \]

8. **Final Result**: Thus, the ratio of energies emitted by the black body at 600 K and 900 K is:
\[ \frac{E_1}{E_2} = \frac{3}{16} \]

### Conclusion:
The required ratio of energies emitted by the black body at 600 K and 900 K, when the surrounding temperature is 300 K, is \( \frac{3}{16} \).