Question

(30 Points) The peak intensity of radiation from a star named Sigma is 2 x 10^6 nm. What is the average surface temperature of Sigma rounded to the nearest whole number?
1.45 K
58 K
1,450 K
5.8 x 10^6 K

Answer 1

The surface temperature of the star is 1.45 K

Explanation:

The star can be thought as a black-body; the relationship between surface temperature and peak wavelength for black-body radiation is given by Wien's displacement law:

$\lambda T = b$

where

$\lambda$ is the peak wavelength

T is the surface temperature (in Kelvin)

$b=2.898\cdot 10^{-3} m\cdot K$ is Wien's displacement constant

For the star in this problem, we have

$\lambda=2\cdot 10^6 nm = 2\cdot 10^{-3} m$

Therefore, we can re-arrange the equation to find the surface temperature:

$T=\frac{b}{\lambda}=\frac{2.898\cdot 10^{-3}}{2\cdot 10^{-3}}=1.45 K$

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