Blackbody radiation frequency problem

In summary, the Cosmic microwave background radiation fits the Planck equations for a blackbody at 2.7 K. The wavelength at the maximum intensity of the spectrum is 1.0733*10^-3 m and the frequency is 2.795*10^11 /s. The total power incident on Earth from the background radiation is 3.013e-6 W/m^2. To calculate the total power incident upon the atmosphere of the Earth, the intensity of the background radiation should be multiplied by the area of the Earth.
  • #1
Benzoate
422
0

Homework Statement



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

Homework 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

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
 
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  • #2
Benzoate said:

The Attempt at a Solution



a) lambda(max)= 2.898*10^-3 m*K/(2.7 K) = 1.0733*10^-3 m

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

b)f=c/lambda=(3e8 m/s)/(1.0733*10^-3 m) = 2.795 *10^11 /s

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

c)R =sigma*T^4= (5.6703e-8 W/m^2*K^4)((2.7 K)^4) = 3.013e-6

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
dynamicsolo said:
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.

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
Benzoate said:
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?

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
dynamicsolo said:
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.)

Show I should multiply the intensity coming towards the atmosphere times the area of the Earth
 
  • #6
Benzoate said:
Show I should multiply the intensity coming towards the atmosphere times the area of the Earth

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.
 

1. What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by a perfect blackbody, which is an object that absorbs all radiation that falls on it and emits radiation at all wavelengths.

2. What is the blackbody radiation frequency problem?

The blackbody radiation frequency problem refers to the discrepancy between the predicted and observed frequencies of electromagnetic radiation emitted by a blackbody at different temperatures, as described by classical physics.

3. How was the blackbody radiation frequency problem solved?

The blackbody radiation frequency problem was solved by Max Planck in 1900 when he proposed the idea of quantization, which states that energy can only be emitted or absorbed in discrete packets, or quanta. This led to the development of the quantum theory of radiation.

4. What is Planck's law of blackbody radiation?

Planck's law of blackbody radiation is a mathematical equation that describes the spectral distribution of electromagnetic radiation emitted by a blackbody at a given temperature. It takes into account the relationship between the frequency of the radiation and the temperature of the blackbody.

5. How does the blackbody radiation frequency problem impact our understanding of the universe?

The blackbody radiation frequency problem was one of the key challenges that led to the development of quantum mechanics, which has greatly advanced our understanding of the fundamental workings of the universe. It also has practical applications in fields such as thermodynamics, astrophysics, and climate science.

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