Answer to Question #173517 in Physics for rebecca

Question

Uranium-237 is a radioactive isotope with a half-life of 7*10⁸ years. In a sample of initially 2*10¹⁵ atoms of ²³⁷U, how many are left after 2.1*10⁹ years?

(In the "Answer" field, write the number of atoms in standard computer notation for powers of 10. For example, for 7*10⁸ write "7E8".)

Answer 1

The number of atoms that are left after $t = 2.1\times 10^{9}years$ is given by the radioactive decay law:

$$N(t) = N_02^{-t/T}$$

where $N_0 = 2\times 10^{15}$ is the initial number of atoms, $T = 7\times 10^{8}years$ is the half-life. Thus, obtain:

$$N = 2\times 10^{15}\cdot 2^{-\left( \frac{2.1\times 10^{9}}{7\times 10^{8}} \right)} = 2.5\times 10^{14}$$

Answer. 2.5E14.