Question

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?

Answer 1

1.45 K

I had the same question and i got it right.

Answer 2

T = 1.45 K

Step-by-step explanation:

As per Wein's law we know that

`\lambda = (2.9 * 10^6 nm K)/(T)`

here we know that

`\lambda` = wavelength of peak intensity

`T` = temperature of the object

so as per above formula we know that

`\lambda = 2 * 10^6 nm`

so we have

`2 * 10^6 = (2.9 * 10^6 nm K)/(T)`

`T = 1.45 K`