Temperature of two stars are in the ratio 3 : 2 . If wavelenghth for the maximum intensity of the first body is `4000Å` , what is the corresponding wavelenghth of the second body ?

Options

A. `9000Å`
B. `6000Å`
C. `2000Å`
D. `8000Å`

Explanation

To solve the problem, we will use Wien's Displacement Law, which states that the product of the maximum wavelength of radiation emitted by a black body and its temperature is a constant. The formula can be expressed as:

\[ \lambda_{max} \cdot T = b \]

where \( \lambda_{max} \) is the maximum wavelength, \( T \) is the temperature in Kelvin, and \( b \) is Wien's displacement constant.

### Step-by-Step Solution:

1. **Identify the Given Ratios and Values**:
- The temperature ratio of the two stars is given as \( T_1 : T_2 = 3 : 2 \).
- The maximum wavelength for the first star (\( \lambda_{M1} \)) is given as \( 4000 \, \text{Å} \).

2. **Convert Wavelength to Meters**:
- Convert \( \lambda_{M1} \) from angstroms to meters:
\[ \lambda_{M1} = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \]

3. **Apply Wien's Displacement Law**:
- According to Wien's Law, we have:
\[ \frac{\lambda_{M1}}{\lambda_{M2}} = \frac{T_2}{T_1} \]

4. **Substituting the Temperature Ratio**:
- From the temperature ratio \( T_1 : T_2 = 3 : 2 \), we can express this as:
\[ \frac{T_2}{T_1} = \frac{2}{3} \]
- Therefore, we can rewrite the equation:
\[ \frac{\lambda_{M1}}{\lambda_{M2}} = \frac{2}{3} \]

5. **Rearranging to Find \( \lambda_{M2} \)**:
- Rearranging gives:
\[ \lambda_{M2} = \lambda_{M1} \cdot \frac{3}{2} \]

6. **Substituting the Value of \( \lambda_{M1} \)**:
- Now substituting \( \lambda_{M1} = 4000 \times 10^{-10} \, \text{m} \):
\[ \lambda_{M2} = 4000 \times 10^{-10} \cdot \frac{3}{2} \]

7. **Calculating \( \lambda_{M2} \)**:
- Performing the multiplication:
\[ \lambda_{M2} = 4000 \times 10^{-10} \cdot 1.5 = 6000 \times 10^{-10} \, \text{m} \]
- Converting back to angstroms:
\[ \lambda_{M2} = 6000 \, \text{Å} \]

### Final Answer:
The corresponding wavelength of the second star is \( 6000 \, \text{Å} \).

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