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 at which the intensity of radiation is maximum (λ_max) is inversely proportional to the absolute temperature (T) of the black body. The formula can be expressed as:

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

where \( b \) is Wien's displacement constant, approximately equal to \( 2898 \, \text{μm K} \).

### Step-by-Step Solution:

1. **Convert the temperature of the furnace to Kelvin:**
The temperature of the furnace is given as \( 2324^\circ C \). To convert this to Kelvin, we use the formula:
\[ T_1 = 2324 + 273 = 2597 \, K \]

2. **Identify the maximum wavelength for the furnace:**
The maximum wavelength for the furnace is given as \( \lambda_1 = 12000 \, \text{Å} \) (which is \( 12000 \times 10^{-10} \, m \) or \( 1.2 \times 10^{-6} \, m \)).

3. **Use Wien's Law to find the temperature of the star:**
According to Wien's Law, we can set up the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{T_2}{T_1} \]
where \( \lambda_2 = 4800 \, \text{Å} \) (which is \( 4800 \times 10^{-10} \, m \) or \( 4.8 \times 10^{-7} \, m \)).

4. **Substituting the values into the equation:**
\[ \frac{12000}{4800} = \frac{T_2}{2597} \]
Simplifying the left side:
\[ \frac{12000}{4800} = 2.5 \]
Therefore, we have:
\[ 2.5 = \frac{T_2}{2597} \]

5. **Solving for \( T_2 \):**
Rearranging the equation gives:
\[ T_2 = 2.5 \times 2597 \]
Calculating \( T_2 \):
\[ T_2 = 6492.5 \, K \]

6. **Convert \( T_2 \) back to Celsius:**
To convert the temperature back to Celsius, we subtract 273:
\[ T_2 = 6492.5 - 273 = 6219.5^\circ C \]

### Final Answer:
The surface temperature of the star is approximately \( 6219^\circ C \).