Black holes in orbit around a normal star are detected from the earth due to the fricitional heating of infalling gas into the black hole, which can reach temperature greater than `10^(6)K`. Assuming that the infalling gas can be modelled as a blackbody radiator, then the wavelenght of maximum power lies

Options

A. in the visible region
B. in the X-ray region
C. in the microwave region
D. in the gamma-ray region of electromagnetic spectruem

Explanation

To solve the problem of determining the wavelength of maximum power emitted by the infalling gas into a black hole, we will use the principles of blackbody radiation and Wien's displacement law.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the infalling gas can be modeled as a blackbody radiator at a temperature \( T \) greater than \( 10^6 \) K. We need to find the wavelength at which the maximum power is emitted.

2. **Using Wien's Displacement Law**:
Wien's displacement law states that the wavelength \( \lambda_{\text{max}} \) at which the emission of a blackbody spectrum is maximized is inversely proportional to the temperature \( T \):
\[ \lambda_{\text{max}} = \frac{b}{T} \]
where \( b \) is Wien's displacement constant, approximately \( 2.898 \times 10^{-3} \, \text{m K} \).

3. **Substituting the Temperature**:
Given \( T > 10^6 \, \text{K} \), we can use this value to find \( \lambda_{\text{max}} \):
\[ \lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{10^6} \]

4. **Calculating \( \lambda_{\text{max}} \)**:
Performing the calculation:
\[ \lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{10^6} = 2.898 \times 10^{-9} \, \text{m} = 2.898 \, \text{nm} \]

5. **Identifying the Wavelength Range**:
The calculated wavelength \( 2.898 \, \text{nm} \) falls within the X-ray region of the electromagnetic spectrum (which ranges from approximately \( 0.01 \, \text{nm} \) to \( 10 \, \text{nm} \)).

6. **Conclusion**:
Therefore, the wavelength of maximum power emitted by the infalling gas around the black hole lies in the X-ray region.

### Final Answer:
The wavelength of maximum power lies in the X-ray region.

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