# Temperature of Universe

• mystreet123
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.

## 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??

mystreet123 said:
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

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.

phyzguy said:
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.