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What is the half life of an Egyptian papyrus with 63 percent of its original carbon-14 atoms?
The question is improperly asked. Papyrus does not have a half-life. Carbon-14 does. Half-life is constant, not variable, in this case, about 5730 years.
So, the question would be more properly asked as "what is the age of an Egyptian papyrus with 63 percent of its original carbon-14 atoms?"
The equation for radioactive decay is...
AT = A0 2(-T/H)
... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.
Substituting what we know, we get...
0.63 = (1) 2(-T/5730)
Solve for T...
log2(0.63) = -T/5730
T = -log2(0.63)(5730)
T = 3819 (conservatively rounded, T = 3800)
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