Question

The isotope samarium-151 decays into europium-151, with a half-life of around 96.6 years. A rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered.
The time since the rock reached its closure temperature is _____


years. When the rock was discovered, it had _____


grams of europium-151.

Answer 1

Part A:

The amount of substance left of a radioactive substance with a half life $t_{ \frac{1}{2} }$ and amount of initial substance $N_0$ is given by

$N(t)=N_0\left( \frac{1}{2} \right)^{ \frac{t}{t_{ \frac{1}{2} }} }$

Given that the isotope samarium-151 has a half-life of around 96.6 years and the rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered.

The time since the rock reached its closure temperature is obtained as follows:

$0.625=5\left( \frac{1}{2} \right)^{ \frac{t}{96.6} } \\ \\ \Rightarrow\left( \frac{1}{2} \right)^{ \frac{t}{96.6} }= \frac{0.625}{5} =0.125 \\ \\ \Rightarrow\frac{t}{96.6}\log{ \frac{1}{2} }=\log0.125 \\ \\ \Rightarrow\frac{t}{96.6}= \frac{\log0.125}{\log{ \frac{1}{2} }} =3 \\ \\ \Rightarrow t=3(96.6)=289.8$

Therefore, the time since the rock reached its closure temperature is 289.8 years.



Part B:

Given that the rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered. The number of grams of europium-151 when the rock was discovered is 5 - 0.625 = 4.375