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Tue 5 Jun, 2007 10:42 am
I have this chem problem i cant figure out, can anyone help me?
Cobalt 60 is used in medicine, this isotope of cobalt has a half life of 5.26 years.
a)caculate K, the rate constant for this radioactive decay.
b)what fraction of a certin sample will remain after 12 years?
radioactive decay is a first order decay process
use the formulas:
t1/2 = ln(2)/K where t1/2 is the halflife
[A]t = [A]0*e^(-Kt) where [A]t is amount of cobalt-60 after time t, and [A]0 is the amount of cobalt-60 at the start
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