If a radioactive nuclide has a half-life of 1.3 billion years and it is determined that a rock contains 25% of its original amount of this nuclide, about how old is the rock?

Respuesta :

A=P(1/2)^(t/h)
A=final amount
P=initial amount
t=time
h=half life


so
h=1.3 billlion (just write 1.3 and remember to mulipy the end time by 1 billion)

and final amount is 25% of original or 0.25P
0.25P=P(1/2)^(t/1.3)
divide both sides by P
0.25=(1/2)^(t/1.3)
take ln of bot sides
ln(0.25)=ln((1/2)^(t/1.3))
property of logarithms
ln(0.25)=(t/1.3)ln(1/2)
divide both sides by ln(1/2)
(ln(0.25))/(ln(1/2))=t/1.3
times both sides by 1.3
(1.3ln(0.25))/(ln(1/2))=t
remember that this is billion so
(1,300,000,000ln(0.25))/(ln(1/2))=t
2600000000.0001=t
about 2.6 billion years