Two black bodies A and B emit radiations with peak intensities at wavelengths 4000 A and 8000 A respectively. Compare the total energy emitted per unit area per second by the two bodies.

Explanation

To solve the problem of comparing the total energy emitted per unit area per second by two black bodies A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
- For body A, the peak intensity wavelength (\( \lambda_A \)) is given as 4000 Å.
- For body B, the peak intensity wavelength (\( \lambda_B \)) is given as 8000 Å.

2. **Use Stefan-Boltzmann Law**:
- The energy radiated per unit area per unit time (E) by a black body is given by the Stefan-Boltzmann law:
\[ E = \sigma T^4 \]
- Here, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature of the body.

3. **Set Up the Ratio**:
- We need to find the ratio of the energy emitted by bodies A and B:
\[ \frac{E_A}{E_B} = \frac{\sigma T_A^4}{\sigma T_B^4} \]
- The \( \sigma \) cancels out:
\[ \frac{E_A}{E_B} = \left( \frac{T_A}{T_B} \right)^4 \]

4. **Apply Wien's Displacement Law**:
- Wien's Displacement Law states that:
\[ \lambda \cdot T = b \]
- Where \( b \) is Wien's constant. For bodies A and B, we can write:
\[ \lambda_A \cdot T_A = \lambda_B \cdot T_B \]
- Rearranging gives:
\[ \frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A} \]

5. **Substitute the Wavelengths**:
- Substitute the given wavelengths:
\[ \frac{T_A}{T_B} = \frac{8000 \, \text{Å}}{4000 \, \text{Å}} = 2 \]

6. **Calculate the Energy Ratio**:
- Now substitute \( \frac{T_A}{T_B} \) back into the energy ratio:
\[ \frac{E_A}{E_B} = \left( \frac{T_A}{T_B} \right)^4 = 2^4 = 16 \]

7. **Final Result**:
- Therefore, the ratio of the total energy emitted per unit area per second by bodies A and B is:
\[ \frac{E_A}{E_B} = 16 : 1 \]