Question

15.57 The half-life for the radioactive decay of U−238 is 4.5 billion years and is independent of initial concentration. How long will it take for 20% of the U−238 atoms in a sample of U−238 to decay? Express your answer using two significant figures. If a sample of U−238 initially contained 1.1×1018 atoms when the universe was formed 13.8 billion years ago, how many U−238 atoms will it contain today? Express your answer using two significant figures.

Answer 1

For both parts, you will use the following expression:

N = No e^-kt   and k = ln2 / t_1/2

For the first part we'll use both equations, for the second wi will use only the first one. Here's the step to solve it:

a) Solving for t from the first expression: ln(No/N) = kt ----> t = ln(No/N) / k

k = ln2 / 4.5x10⁹ y

k = 1.54x10^-10 y^-1

Now, we'll consider No to be 100% and N (The final) to be 80 (cause it's decaying 20%) so:

t = ln (100/80) / 1.54x10^-10

t = 1.45x10⁹ years or simply 1.45 billions of years

b) In this part we have No = 1.1x10¹⁸ atoms, we have the time which is 13.8x10⁹ y and we will solve for N so, using the first expression (assuming the value of k the calculated in part a)

N = 1.1x10¹⁸ e(-1.54x10^-10 * 13.8x10⁹)

N = 1.31x10¹⁷ atoms.

Hope this helps.