The radioactive isotope 32P decays by first-order kinetics and has a half-life of 14.3 days.?

Question

How long does it take for 65.0% of a sample of 32P to decay?

Answer 1

If N0 is the intial amount of 32P, then the remaining amount N after time t is given by the equation for firts order:

N= No*e^-(kt)
For t= τ N=No/2, thus
No/2= No *e^-(kτ) =>
1/2 = e^-(kτ) =>
ln2 =kτ => k =ln2/τ

substitute k in the original equation and you get

N= No*e^-[(ln2/τ )*t]
When 65% decays, 100-65=35% remains thus

N=0.35No = No*e^-[(ln2/τ )*t] =>
0.35 = e^-[(ln2/τ )*t] =>
ln(0.35) = -[(ln2/τ )*t] =>
t= - τ *ln(0.35)/(ln(2) = -14.3 *ln(0.35)/(ln(2) = 21.7 days