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

Explanation:

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

$A=A_{o}.2^{\frac{-t}{h}}$   (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):

$A_{o}=\frac{A}{2^{\frac{-t}{h}}}$  

$A_{o}=A.2^{-\frac{-t}{h}}$  

This is the same as:

$A_{o}=A.2^{\frac{t}{h}}$   (2)

$A_{o}=(10g)(2)^{\frac{15.78y}{5.26y}}$  

$A_{o}=(10g)(2)^{3}$  

Finally:

$A_{o}=80g$>>> This is the amount of grams in the original sample