Question

2. A 2.5 gram sample of an isotope of strontium-90 was formed in a 1960 explosion
of an atomic bomb at Johnson Island in the Pacific Test Site. The half-life of
strontium-90 is 28 years. In what year will only 0.625 grams of this strontium-90
remain?

Answer 1

Answer: 2016

Explanation:

In order to solve this problem, firstly we have to use the Radioactive Half Life Formula:  

$A=A_{o}.2^{\frac{-t}{h}}$ (1)  

Where:  

$A=0.625 g$ is the final amount of strontium-90

$A_{o}=2.5 g$ is the initial amount of strontium-90

$t$ is the time elapsed  

$h=28 years$ is the half life of strontium-90

Knowing this, let's substitute the values and find $t$ from (1):

$0.625 g=(2.5 g)2^{\frac{-t}{28 years}}$ (2)  

$\frac{0.625 g}{2.5 g}=2^{\frac{-t}{28 years}}$ (3)  

Applying natural logarithm in both sides:

$ln(\frac{0.625 g}{2.5 g})=ln(2^{\frac{-t}{28 years}})$ (4)  

$-1.386=-\frac{t}{28 years}ln(2)$ (5)  

Clearing $t$:

$t=\frac{-(1.386)(28 years)}{-ln (2)}$ (6)  

$t=56 years$ (7)  This is the time elapsed since the sample of strontium-90 was formed

If we add this time to 1960 (the year this isotope was formed) we finally find the year when only 0.625 grams of strontium-90 are left:

$1960 + 56=2016$

Therefore, the answer is 2016.