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Physics · Ch 15 — Structure of Atoms and Nuclei

Half-life of Radioactive Material

15.10.1

Half-life of Radioactive Material

The half-life of a radioactive species, denoted T1/2T_{1/2}, is defined as the time required for the number of parent nuclei to fall to exactly HALF its initial value. Setting N(T1/2)=N0/2N(T_{1/2})=N_0/2 in the decay law N(t)=N0e−λtN(t)=N_0e^{-\lambda t} gives 12=e−λT1/2\frac{1}{2}=e^{-\lambda T_{1/2}}, which rearranges (taking the natural logarithm of both sides) to T1/2=ln⁡2λ=0.693λT_{1/2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} -- a species with a larger decay constant λ\lambda (decaying 'faster' in a statistical sense) therefore has a SHORTER half-life, and vice versa.

A subtle but important point about half-life is what happens after MORE than one half-life has elapsed. After one half-life, the population has fallen from N0N_0 to N0/2N_0/2 -- but it does NOT then fall all the way to zero over the next half-life; instead, exactly the same FRACTIONAL reduction applies again, so after a second half-life the population has fallen only to half of N0/2N_0/2, i.e. to N0/4N_0/4, not to zero. This same halving rule applies to any equal interval of length T1/2T_{1/2}, no matter how far along the decay process has already progressed -- af …

Misc Ex.15.10Half-life from a measured drop in activity over one hour

Worked out. An activity that decreases from 350 s−1^{-1} to 175 s−1^{-1} (exactly halving) over one hour (3600 s) is analysed with A(t)=A0e−λtA(t)=A_0e^{-\lambda t}: substituting 175=350 e−λ(3600)175=350\,e^{-\lambda(3600)} gives λ(3600)=ln⁡(350/175)=ln⁡2=0.6931\lambda(3600)=\ln(350/175)=\ln2=0.6931, so λ=1.925×10−4\lambda=1.925\times10^{-4} s−1^{-1}, and then T1/2=0.693/λ≈3600T_{1/2}=0.693/\lambda\approx3600 s (i.e. exactly one hour) -- which makes sense directly from the given data, since the activity was observed to exactly halve over that one-hour interval, so that interval must itself equal the half-life, providing a useful sanity check on the m …