The temperature of a furnace is `2324^(@)C` and the intensity is maximum in its radiation spectrum nearly at `12,000Å`.If the intensity in the spectrum of star is maximum nearly at `4800Å` , then the surface temperature of star is nearly:

Options

A. `8400^(@)`
B. `7200^(@)C`
C. `6220^(@)C`
D. `5810^(@)C`

Explanation

To solve the problem, we will use Wien's Displacement Law, which states that the wavelength of maximum intensity (λ_max) of radiation emitted by a black body is inversely proportional to its absolute temperature (T). The formula can be expressed as:

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

where \( b \) is Wien's displacement constant, approximately equal to \( 2898 \, \mu m \cdot K \) or \( 2898000 \, Å \cdot K \).

### Step-by-Step Solution:

1. **Convert the temperature of the furnace to Kelvin**:
\[ T_1 = 2324 \, °C + 273.15 = 2597.15 \, K \]

2. **Identify the maximum wavelength for the furnace**:
\[ \lambda_{m1} = 12000 \, Å \]

3. **Apply Wien's Law for the furnace**:
\[ \lambda_{m1} \cdot T_1 = b \]
Substituting the known values:
\[ 12000 \, Å \cdot 2597.15 \, K = b \]

4. **Calculate the constant \( b \)**:
\[ b = 12000 \cdot 2597.15 = 31165800 \, Å \cdot K \]

5. **Identify the maximum wavelength for the star**:
\[ \lambda_{m2} = 4800 \, Å \]

6. **Use Wien's Law for the star to find its temperature \( T_2 \)**:
\[ \lambda_{m2} \cdot T_2 = b \]
Substituting the known values:
\[ 4800 \, Å \cdot T_2 = 31165800 \, Å \cdot K \]

7. **Solve for \( T_2 \)**:
\[ T_2 = \frac{31165800}{4800} \approx 6495.375 \, K \]

8. **Convert \( T_2 \) to Celsius** (if needed):
\[ T_2 \, (°C) = 6495.375 \, K - 273.15 \approx 6222.225 \, °C \]

### Final Answer:
The surface temperature of the star is approximately \( 6495.375 \, K \) or \( 6222.225 \, °C \).