Question

The half-life for the process 238U 206Pb is 4.5 109 yr. A mineral sample contains 45.9 mg of 238U and 16.8 mg of 206Pb. What is the age of the mineral?

i used the equation: n(T)=n(O)/2^(t/halflife) and i got 6.53e9 but it is wrong. what am i doing wrong?

Answer 1

I suspect you didn't change the Pb to U.
16.8 mg Pb x (238/206) = 19.4 mg U
19.4 + original U = 19.4 + 45.9 = 65.3; therefore, the original amount of U238 = 65.3.
Now k = 0.693/t_1/2 = 0.693/4l.5E9.
Then ln(No/N) = kt
No = 65.3
N = 45.9
k from above.
Solve for t in years.

Answer 2

thanks! so k=1.669e-11
so ln(65.3/45.9)=1.669e-11(t)
so does t=2.112e10 years?

Answer 3

I entered the asnwer I got and it is wrong.