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 wavelength at which the maximum power is emitted by a blackbody is inversely proportional to its temperature. The formula is given by:

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

where:
- \(\lambda_{\text{max}}\) is the wavelength at which maximum power is emitted,
- \(T\) is the absolute temperature in Kelvin,
- \(b\) is Wien's constant, approximately \(2.898 \, \text{mm K}\).

Given that the maximum wavelength \(\lambda_{\text{max}} = 500 \, \text{nm}\), we first convert this into millimeters:

\[ 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} = 500 \times 10^{-6} \, \text{mm} = 5 \times 10^{-7} \, \text{mm} \]

Now we can substitute the values into Wien's law:

\[ (5 \times 10^{-7} \, \text{mm}) \cdot T = 2.898 \, \text{mm K} \]

To find the temperature \(T\), we rearrange the equation:

\[ T = \frac{2.898 \, \text{mm K}}{5 \times 10^{-7} \, \text{mm}} \]

Calculating this gives:

\[ T = \frac{2.898}{5} \times 10^{7} \, \text{K} = 0.5796 \times 10^{7} \, \text{K} \approx 5.796 \times 10^{3} \, \text{K} \]

Now, we can express the temperature in the form \( \alpha \cdot \beta \times 10^{\gamma} \):

Here, we have:
- \(\alpha = 5\)
- \(\beta = 8\)
- \(\gamma = 3\)

Now we can calculate \( \alpha + \beta + \gamma \):

\[ \alpha + \beta + \gamma = 5 + 8 + 3 = 16 \]

Thus, the final answer is:

\[ \boxed{16} \]