# Average photon energy from a galaxy cluster

1. Mar 12, 2016

### throneoo

1. The problem statement, all variables and given/known data
I want to calculate/estimate the average photon energy from a galaxy cluster incident on the mirror of a X ray telescope (Chandra to be specific) . The cluster has redshift z and at constant uniform temperature T

2. Relevant equations
Specific Intensity of thermal bremsstrahlung : I∝Λ(ν)*T-1/2*exp(-hν/kT)
z=(νemittedobserved)/νobserved
3. The attempt at a solution
At high temperatures the X ray emission is dominated by thermal bremsstrahlung.
The average photon energy should be found via the integral ( ∫ I dν ) / ( ∫ [I/hν] dν ) from ν=0 to infinity, then that energy should be divided by (1+z) due to the redshift, but since the full expression is not given in the problem, it's not really a viable approach.

Therefore I'm left with the approximation approach. I think <E>=hv≈kT should be good enough since in the x-ray region, photons' momentum and energy is comparable to that of electrons, and that it seem to be the value at which I is equal to its average value (referring to the curve below).

The classical spectrum (valid for moderate temperature): https://upload.wikimedia.org/wikipe...ower2.svg/731px-Bremsstrahlung_power2.svg.png

2. Mar 13, 2016

### Staff: Mentor

Galaxies radiate a lot more visible light than x-rays, and those two cannot be described by a single temperature. I don't think starting the integration at 0 will lead to a useful result.

You'll need a temperature value if you want to follow the approach of thermal emission. And probably the sensitivity range of Chandra, unless you really care about all the low-energetic photons.

3. Mar 14, 2016

### throneoo

Okay so I can effectively put the sensitivity range of chandra in my integration, then I~exp(-x) with the range. However exp(x)/x is still a nasty integral so I'll have to look up values of exponential integral.

And by thermal emission do you mean black-body emission?

4. Mar 14, 2016

### Staff: Mentor

If it is not black-body-like, you need to know more about the radiation.

5. Mar 14, 2016

### throneoo

I think that would be the braking radiation as given in OP