The radiation emitted by the surface of the Sun emits maximum power at a wavelength of about 500 nm. Assuming the Sun to be a blackbody emitter. If its surface temperature (K) is given by `alpha . Beta xx10^(gamma)` then fill the value of `(alpha+beta+gamma)`. (Wien's constant is given by 2.898mm K)

Answer

To solve the problem, we will use Wien's Displacement Law, which states that the product of the wavelength at which the emission of a black body spectrum is maximized (λ_max) and the absolute temperature (T) of the black body is a constant (b). The formula is given by:

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

Where:
- \( \lambda_{\text{max}} \) is the wavelength in mm,
- \( T \) is the temperature in Kelvin,
- \( b \) is Wien's constant, approximately \( 2.898 \, \text{mm K} \).

### Step-by-Step Solution:

1. **Convert Wavelength to mm**:
The given wavelength \( \lambda_{\text{max}} \) is 500 nm. We need to convert this to mm:
\[ \lambda_{\text{max}} = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} = 500 \times 10^{-6} \, \text{mm} = 5 \times 10^{-4} \, \text{mm} \]

2. **Apply Wien's Law**:
Using Wien's Law:
\[ T = \frac{b}{\lambda_{\text{max}}} \]
Substitute the values:
\[ T = \frac{2.898 \, \text{mm K}}{5 \times 10^{-4} \, \text{mm}} \]

3. **Calculate Temperature**:
Performing the division:
\[ T = \frac{2.898}{5 \times 10^{-4}} = 5796 \, \text{K} \]

4. **Express Temperature in Required Format**:
The temperature can be expressed in the form \( \alpha \cdot \beta \times 10^{\gamma} \):
\[ T = 5.796 \times 10^{3} \, \text{K} \]
Here, we identify:
- \( \alpha = 5 \)
- \( \beta = 8 \)
- \( \gamma = 3 \)

5. **Calculate \( \alpha + \beta + \gamma \)**:
\[ \alpha + \beta + \gamma = 5 + 8 + 3 = 16 \]

### Final Answer:
The value of \( \alpha + \beta + \gamma \) is \( 16 \).

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