The half-life of C14 radioactive is 5760 yr. After how much time will 200 mg C14 sample be reduced to 25 mg?

Answer 1

k = 0.693/t_1/2

ln(No/N) = kt
No = 200
N = 25
k from above
Solve for t in years.

Answer 2

Ln(200/25)=Ln(2)/5760*t
Ln(8)=Ln(2)/5760*t
Ln(2^3)=Ln(2)/5760*t
3Ln(2)=Ln(2)/5760*t
3=t/5760
t=17280