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

Answer:

1.45 K

I had the same question and i got it right.

Answer 2

Answer:

T = 1.45 K

Explanation:

As per Wein's law we know that

$\lambda = \frac{2.9 \times 10^6 nm K}{T}$

here we know that

$\lambda$ = wavelength of peak intensity

$T$ = temperature of the object

so as per above formula we know that

$\lambda = 2 \times 10^6 nm$

so we have

$2 \times 10^6 = \frac{2.9 \times 10^6 nm K}{T}$

$T = 1.45 K$