Question

The half-life of cobalt is 5.26 years. If 10 grams are left after 15.78 years, how many grams were in the original sample?​

Answer 1

Answer: 80 grams

Step-by-step explanation:

This problem can be solved using the Radioactive Half Life Formula:

`A=A_(o).2^{(-t)/(h)}` (1)

Where:

`A=10g` is the final amount of the material

`A_(o)` is the initial amount of the material (the quantity we are asked to find)

`t=15.78y` is the time elapsed

`h=5.26y` is the half life of cobalt

Knowing this, let's find
`A_(o)` from (1):

`A_(o)=\frac{A}{2^{(-t)/(h)}}`

`A_(o)=A.2^{-(-t)/(h)}`

This is the same as:

`A_(o)=A.2^{(t)/(h)}` (2)

`A_(o)=(10g)(2)^{(15.78y)/(5.26y)}`

`A_(o)=(10g)(2)^(3)`

Finally:

`A_(o)=80g`>>> This is the amount of grams in the original sample