1. Oct 4, 2007

### Benzoate

1. The problem statement, all variables and given/known data

The Cosmic microwave background radiation fits the Planck equations for a blackbody at 2.7 K. a) What is the wavelength at the maximum intensity of the spectrum of the background radiation ? b) What is the frequency of the radiation at the maximum? c) what is the total power incident on Earth from the background radiation

2. Relevant equations

lambda(max)*T=2.898*10^-3 m*K
T=2.7 K
f=c/lambda(max)
sigma=5.6703e-8 W/m^2*K^4

R=sigma*T^4 or

3. The attempt at a solution

a) lambda(max)= 2.898*10^-3 m*K/(2.7 K)=1.0733*10^-3 m
b)f=c/lambda=(3e8 m/s)/(1.0733*10^-3 m)= 2.795 *10^11 /s
c)R =sigma*T^4= (5.6703e-8 W/m^2*K^4)((2.7 K)^4=3.013e-6

2. Oct 4, 2007

### dynamicsolo

Hence, the relatively new field of "mm-wave astronomy"...

I imagine they like to use "Hz" (Hertz), rather than sec^(-1), so this would be 0.280 THz.

So this gives the intensity of the "CMBR" in W/m^2. They're asking for the total power incident upon (the top of the atmosphere of) Earth, so you have one more step.

3. Oct 4, 2007

### Benzoate

So now I have to multiply the intensity of the "CMBR" times the area of the earth in order to calculate the power incident upon the atmosphere of the earth? Nah, that can't be right since the total power incident upon the atmospher is only the atmosphere of the earth ad ot the whole earth. Should I assume the area of the surface of the atmospher incident on is 1 square meter?

4. Oct 4, 2007

### dynamicsolo

The intensity you found is what is seen from everywhere on the sky, since the background radiation is (very nearly) uniform, so that value is what reaches each square meter of Earth. (We generally state that as "incident at the top of the atmosphere" since the microwave radiation largely is absorbed by the atmosphere in those wavelength bands and doesn't get to the Earth's surface.) I believe the question is asking for the total power over the entire Earth to stress the fact that this is really a small level. (However, radio telescopes are extremely sensitive and detectors responds to very tiny power levels.)

5. Oct 5, 2007

### Benzoate

Show I should multiply the intensity coming towards the atmosphere times the area of the Earth

6. Oct 5, 2007

### dynamicsolo

I believe that is what is being asked for. Upon reading the question again, I can see how it could seem unclear, but I'm pretty sure that's the value they're looking for.