# Decay Energy/Gamma-ray production/Neutrino

## Homework Statement

Assume that 0.075 M of 56Co was produced by the decay of 56Ni following the explosion of
SN1987A.

a. Estimate the amount of energy released per second through the radioactive decay of 56Co
just after its formation and 1 yr after the explosion. Express your answers in both ergs s−1
and solar luminosities.

b. All of the 56Co decays produce a gamma-ray with an energy of of 847 keV and these were
observed coming from SN1987A 0.5 yr after the explosion. (A gamma-ray with an energy
of 1238 keV is also produced in 68% of the decays.) Estimate the rate of gamma-ray
production at that time and use this to calculate the maximum ﬂux of 847 keV gamma-
rays that could be observed at Earth. Assume a distance to SN1987A of 50 kpc and express
your ﬂux in units of photons cm−2 s−1.

c. The observed ﬂux in the 847 keV line at the above time was about 1.0 x 10−3
photons cm−2 s−1. Describe the most likely reason why your result from part b) is higher
than the observed ﬂux.

## Homework Equations

dN/dt = -λN = -$\frac{ln(2)}{τ(1/2)}$

N(1 year) = N0e-λt

Decay energy Q = KEf - KEi = mfc2 - mic2

## The Attempt at a Solution

a): Using the above equations I found:

56Co decays/s = 1.65 x 1047 decays/s
56Ni decays/s = 1.20 x 1045 decays/s

Q = c2(mf - mi)
Since,
mf = isotopic mass of 56Co = 55.940 u and
mi = 55.942 u
Q = 1.86 x 106

I then use the equation, Radioactive power = W = Q$\frac{A}{M}$
Molar mass 56Co = 56 g/mol
Molar mass 56Ni = 56 g/mol

This is where I am stuck for part a). How do I implement the 0.75 M$\odot$ given in the question? And how do these units combine to give me power? Do I multiply the results I get with my W equation by 0.75 M$\odot$ to get eV/s? If so, then the rest of the problem should be simply converting units, which I can do easily.

b): I'm having trouble determining the time they mean in this question. Do I use the .5 years, or do I have to calculate the rate at .5 years and use the resulting number? Additionally, as I am asking in another problem regaring neutrinos, how do I go about calculating the flux of the $\gamma$-rays which could be observed on the earth?

c): I believe this to be similar to how neutrinos are hard to observe. The energy of the $\gamma$-rays is so great that it is hard to "capture" them to detect them. Like neutrinos, they must react with out atmosphere so we may detect their child particles. We detect less than the theoretical flux because we only have a limited number of detectors as well as our detectors do not detect each $\gamma$-ray due to them not always reacting in our atmosphere (and above our detectors).