5. Gamma rays from the radioactive decay of nickel into iron drive most of the optical luminosity of a Type-Ia supernova. The process is given by \[^{56}N\longrightarrow ^{56}Co+\gamma\longrightarrow ^{56}Fe+\gamma\] where \(\gamma\) represents photons. The total atomic weights of \(^{56}N\) and \(^{56}Fe\) are 55.942135 and 55.934941, respectively. Let's calculate the total energy radiated in the optical wavelengths during the event, given that the characteristic times for the two decay processes are 8.8 days and 111 days, respectively.
a) Let's balance the decay process from \(^{56}N\) to \(^{56}Fe\) for a single atom, ignoring the intermediate step. According to the first law of thermodynamics, energy cannot be created or destroyed. Use the fact that \(E=mc^2\) to balance the equation.
The overall equation is \(^{56}N\longrightarrow^{56}Fe+\gamma\), but since the masses of \(^{56}N\) and \(^{56}Fe\) aren't the same, some mass must be converted to energy. We can figure out the amount of energy that is produced by balancing the equation and first finding the amount of "missing" mass.
\(55.942135g/mol=55.934941g/mol+m_{missing}\)
\(m_{missing}=0.007194g/mol\rightarrow7.194*10^{-6}kg\)
We can then use the missing mass to calculate the amount of energy released.
\(E=(7.194*10^{-6}kg)(3.0*10^8m/s)^2=6.47*10^{11}J/mol\)
We then divide by Avogadro's Number to calculate the amount of energy per atom.
\(E=\frac{6.47*10^{11}J/mol}{6.02*10^{23}atoms/mol}=1.08*10^{-12}J/atom\)
b) How many nickel atoms are there in the white dwarf? Use this number to estimate the total energy emitted in photons.
Since we know the mass of the white dwarf (\(1.4M_\odot\)) and the atomic mass of nickel-56, this is just a matter of dimensional analysis.
\(\frac{1.4M_\odot}{1}\frac{2*10^{33}g}{1M_\odot}\frac{1mol^{56}N}{55.942135g}\frac{6.02*10^{23}atoms}{1mol^{56}N}=3.0*10^{55}atoms\)
Since we already solved for the energy per atom, we can easily calculate the total energy.
\(\frac{3.0*10^{55}atoms}{1}\frac{1.08*10^{-12}J}{1atom}=3.25*10^{43}J\)
c) Now combine the characteristic times for the two processes to find a total characteristic time. Divide the energy found in part (b) by this time scale to find a characteristic luminosity.
Total time = 119.8 days = \(1.04*10^7s\).
\(L=\frac{3.25*10^{43}J}{1.04*10^7s}=3.14*10^{36}J/s\)
Nice! While this is perfectly correct, please try out cgs units next time! They take a while to get used to, but are really quite nice once you are used to them. They are the convention in astronomy.
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