Two stars A and B of surface area Sa and Sb & temperature `T_a and T_b` glow red and blue respectively. Choose the correct option.

Options

A. `T_4gtT_b`
B. `T_alt T_b`
C. `T_aS_a=T_bS_b`
D. `T_aS_b = T_bS_a`

Answer

To solve the problem, we will utilize Wien's Displacement Law and Stefan-Boltzmann Law to analyze the relationship between the temperatures and the colors of the stars A and B.

### Step-by-Step Solution:

1. **Understanding Wien's Displacement Law**:
- Wien's Displacement Law states that the wavelength (\( \lambda_m \)) at which the emission of a black body spectrum is maximized is inversely proportional to its absolute temperature (T). This can be expressed mathematically as:
\[ \lambda_m \cdot T = b \]
where \( b \) is a constant (approximately \( 2.898 \times 10^{-3} \, \text{m K} \)).

2. **Identifying the Colors**:
- In the question, star A glows red and star B glows blue. We know that red light has a longer wavelength than blue light. Therefore, we can denote:
\[ \lambda_A > \lambda_B \]
- Since \( \lambda \) is inversely proportional to \( T \), we can write:
\[ \lambda_A \cdot T_A = b \quad \text{and} \quad \lambda_B \cdot T_B = b \]

3. **Relating Temperatures**:
- From the above equations, we can express the temperatures in terms of wavelengths:
\[ T_A = \frac{b}{\lambda_A} \quad \text{and} \quad T_B = \frac{b}{\lambda_B} \]
- Since \( \lambda_A > \lambda_B \), it follows that:
\[ T_A < T_B \]
- This means that the temperature of star A (which glows red) is less than the temperature of star B (which glows blue).

4. **Analyzing Power Radiated**:
- According to Stefan-Boltzmann Law, the power radiated by a black body is given by:
\[ P = \sigma E A T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant, \( E \) is the emissivity, and \( A \) is the surface area.
- For stars A and B, the power radiated can be expressed as:
\[ P_A = \sigma E S_A T_A^4 \quad \text{and} \quad P_B = \sigma E S_B T_B^4 \]
- Without specific values for \( S_A \) and \( S_B \), we cannot determine which star radiates more power based solely on the information given.

5. **Conclusion**:
- From the analysis, we conclude that:
- The temperature of star A is less than that of star B.
- We cannot definitively compare the power radiated by the two stars without additional information about their surface areas.

### Final Answer:
The correct statement is that the temperature of star A (red) is less than the temperature of star B (blue).