Question

A 7.85x10^-5 mol sample of copper-61 emits 1.47x10^19 positrons in 90.0 minutes. What is the decay constant?

Answer 1

1mol = 6.022e23 molecules

so 7.85e-5 * 6.022e23 = 4.72727e19 positrons

The equation is Ln A = -kt + Ln Ao

lnA - lnAo = -kt

(lnA - lnA0) / t = -k

At A,T=90, it is4.72727e19 positrons - 1.47e19= 3.25727e19 positrons

At Ao, T=0 its 4.72727e19 positrons

so lets solve it

(ln (3.25727e19 )- ln(4.72727e19)) / 90 = -k

-0.0041384272 = -k

k = 0.0041384272 min^-1

-0.0041384272min^-1 * 60min/1hour = 0.248305632hour^-1

k = 0.24 hour^-1

Note check out some extra values which i used with yours if those are different then there may arise some variations in answer