Interstellar space is filled with radiation of frequency 160.23 GHz.
This radiation is considered to be a remnant of the "big bang." What is the corresponding blackbody temperature of this radiation?

Respuesta :

Answer:

1.548K

Explanation:

Given that f= 160.23GHz

c= 3E8m/s

b= 2.898*10^-3mk

So using

(Lambda)m x T= b

So T = b/ lambda

But wavelength ( lambda) = c/f

So T = bf/c

= 2.898E-3x 160.23E9/3E8

=1.548K