Question

Cobalt-56 has a decay constant of 8.77 × 10-3 (which is equivalent to a half life of 79 days). How many days will it take for a sample of cobalt-56 to decay to 62% of its original value?

Answer 1

As I read this one, is just a decay exponential equation... so

A = P(1 + r)ᵗ   where "t" is days passed. . hmm in this case is decay, so negative rate  A = P(1 - r)ᵗ, and the decimal amount would be 0.00877 for the rate

62% of P, the original value, is just 0.62P, now... if we hmm take P as just 1, it could be any amount, but 62% of 1,000,000 is just 62% of 1 times 1,000,000

so, for the sake of comparing it with a percentage, 1 will do

$\bf 0.62P=P(1-0.00877)^t\implies P=1\implies 0.62=(1-0.00877)^t \\\\\\ ln(0.62)=ln[(1-0.00877)^t]\implies ln(0.62)=t\cdot ln(1-0.00877) \\\\\\ \cfrac{ln(0.62)}{ln(0.99123)}=t$