Question

A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope. How many half-lives have passed since the sample originally formed?

Answer 1

the answer to this question is a 3 half lives have passed

Answer 2

Answer: 3 half lives

Solution :

Formula used :

`a=(a_o)/(2^n)`

where,

a = amount of reactant left after n-half lives = initial amount - amount of daughter isotope= (1.28-1.12)= 0.16 g

`a_o` = Initial amount of the reactant = 1.28 g

n = number of half lives = ?

Putting values in above equation, we get:

`0.16=(1.28)/(2^n)`

`2^n=8`

`2^n=2^3`

`n=3`

Therefore, 3 half-lives have passed since the sample originally formed.