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Temperature of Universe

  • #1

Homework Statement


Two scientists detected the cosmic microwave background radiation at a frequency of 160 GHz. What is the temperature of the universe?

Homework Equations


peak wavelength x temperature = 2.898 x 10^-3
c = f x wavelength

The Attempt at a Solution


I calculated the wavelength of the microwave radiation to be 1.875x10^-3m. So the temperature is 1.5456K.
But the answer is 3K... How come??
 

Answers and Replies

  • #2
963
213
Two scientists detected the cosmic microwave background radiation at a frequency of 160 GHz. What is the temperature of the universe?
The cosmic microwave background radiation has a thermal black body spectrum at a temperature of 2.72548±0.00057 K.
The spectral radiance dEν/dν peaks at 160.23 GHz.

see detail at
https://en.wikipedia.org/wiki/Cosmic_microwave_background
 
  • #3
phyzguy
Science Advisor
4,405
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The peak of the wavelength distribution is not at the same place as the peak of the frequency distribution, because the blackbody spectrum is not linear. So you cannot say: λpeak = c / νpeak. Try reading this link. Or try taking the blackbody distribution I(λ) and differentiate to find the peak, then compare it to the peak of I(ν). You will then see why they are different.
 
  • #4
The peak of the wavelength distribution is not at the same place as the peak of the frequency distribution, because the blackbody spectrum is not linear. So you cannot say: λpeak = c / νpeak. Try reading this link. Or try taking the blackbody distribution I(λ) and differentiate to find the peak, then compare it to the peak of I(ν). You will then see why they are different.
Thanks for replying!
From the link you gave me, "However the form of the law remains the same: the peak wavelength is inversely proportional to temperature (or the peak frequency is directly proportional to temperature)." Why couldn't I use the v=fλ to find peak frequency?
 
  • #5
phyzguy
Science Advisor
4,405
1,385
Thanks for replying!
From the link you gave me, "However the form of the law remains the same: the peak wavelength is inversely proportional to temperature (or the peak frequency is directly proportional to temperature)." Why couldn't I use the v=fλ to find peak frequency?
It's true that, "the peak wavelength is inversely proportional to temperature (or the peak frequency is directly proportional to temperature)". But the proportionality constants are different and the peaks occur in different places, so the peak frequency is not related to the peak wavelength by λpeak = c / νpeak. The reason is that the differential dν is not linearly related to the differential dλ. I'm going to repeat what I said before. Try finding the peaks of the two distributions and you will see why λpeak = c / νpeak does not work.
 

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