Phys 132 Homework 23 Solutions
Professor Crystal Martin
TA: Ellie Hadjiyska
Problem 23: The number density of photons with energy Eabove E0is
n(E > E0) = Z∞
E0
n(E)dE =8π
h3c3Z∞
E0
E2dE
exp(E/kBT)−1
since photons are fermions. Our lower integration limit E0≈9.98MeV which is much larger than
the average thermal energy of particles in the star. In other words, EÀkBTfor all energies
within the integration range. Therefore exp(E /kBT)À1 and the denominator of the integrand is
approximately exp(E/kBT). The integral above becomes
Z∞
E0
E2dE
exp(E/kBT)−1≈Z∞
E0
e−E/kBTE2dE
We carry out the integral and simplify to find
n(E > E0)≈8πe−E0/kBT
h3c3(E2
0kBT+ 2E0k2
BT2+ 2k3
BT3)
If temperature rises from 109K(100 KeV) to 4 ×109K(400 KeV), the last term in the parenthesis
increases by a factor of 64 and we neglect change in the other two terms (which increase by factors
of 4 and 16, respectively). The exponential, however, increases by a whopping factor of 1032!
Therefore we increase the number of photons with enough energy to dissociate silicon by a factor
of approximately 1034 between the two temperatures in question.
