How many uranium-238 atoms are left after 1.338 x 10^10 years?

Question

The half-life of uranium-238 is about 4.46 x 10^9 years. How many will there be after 1.338 x 10^10 years? How can I figure this out? I know it's exponential, but how?

score 2 · Accepted Answer · answered Apr 14 '15 at 23:32

2

First you find the number of half lives in $1.338 \times 10^{10}$ years. That is 3. After every half life, the amount of uranium will halve. So there will be $\frac{1}{2^3}$ as much uranium-238 left.

answered Apr 14 '15 at 23:32

wlad

8,185

@ВалерийЗаподовников No it's right – wlad May 20 '23 at 08:33
It just so happened that in this case it is perfect x3. If it was not... – Валерий Заподовников May 20 '23 at 19:04

score 0 · Answer 2 · edited Apr 15 '15 at 00:08

Let X0 = the number of atoms at time t =o. Then dx/dt = -kt is the differential equation which governs the decay rate. When t = 4.46*10^9, the value of x0 will have decreased to 1/2 its value, or X0/2. The solution to the differential equation is: x(t) = X0*e^(-kt). Substitute t=4.46*10*9 and x(t) = X0/2 to evaluate k. Then, for t =1.338*10^10, just plug it into the solution of the differential equation. Edwin Gray

How many uranium-238 atoms are left after 1.338 x 10^10 years?

2 Answers2