The peak intensity of radiation from a star named sigma is 2 x 10^6 mmkay. What is the average surface temperature of Sigma rounded to the nearest whole number?
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Answer:
T = 1.45 K
Explanation:
As per Wein's law we know that
[tex]\lambda = \frac{2.9 \times 10^6 nm K}{T}[/tex]
here we know that
[tex]\lambda[/tex] = wavelength of peak intensity
[tex]T [/tex] = temperature of the object
so as per above formula we know that
[tex]\lambda = 2 \times 10^6 nm[/tex]
so we have
[tex]2 \times 10^6 = \frac{2.9 \times 10^6 nm K}{T}[/tex]
[tex]T = 1.45 K[/tex]