Blog of Veikko M.O.T. Nyfors, Hybrid Quantum ICT consultant

Quantum Mechanics demystified, a try

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Carbon14 Dating

$^{14}C$ is an isotope of Carbon with 6 protons and 8 neutrons in it’s nucleus. It is an unstable radioactive isotope. In radioctive decay $^{14}C$ turns to be $^{14}N$.
This is due to one of it’s neutrons turning into a proton. An electron as well as an anti-neutrino is emitted in this process. Electron flow is called Beta radiation and anti-neutrino turns mostly out as energy.
This is one form of Weak Nuclear Interaction

Amount of $^{14}C$ in atmosphere is one trillionth of that of other stable isotopes $^{12}C$ and $^{13}C$. Living organisms gather $^{14}C$ in the same proportion in their bodies when they are alive.
When they die, $^{12}C$ and $^{13}C$ will stay in dead bodies, but $^{14}C$ keeps on radioactive decaying. Thus it’s amount in the body will decrease over time.
Radioactive half-life of $^{14}C$ is 5700 years, i.e. in 5700 years it’s amount in a dead body will decrease to half of the original.
By analysing amount of $^{14}C$ in proportion to $^{12}C$ and $^{13}C$, one can deduce the age of the body by solving t from formula

$(\frac{1}{2})^{\frac{t}{T}} = \frac{A}{B} \iff t = \frac{T*ln(A/B)}{ln(1/2)}$,

where T is the half time, B is the original amount of C14 and A is the amount of it left.

So, if we have 20Kg of $^{12}C$ + $^{13}C$ in a fossil, there would had been $0.02mg$ of $^{14}C$ while it was alive. If we detect $0.000625mg$ of it left in fossil, the age of the fossil is

$t=\frac{5700*ln(0.000625/0.02)}{ln(1/2)} = 28500$