Question

A lab technician had a sample of radioactive Americium. He knew it was one of the isotopes listed in the table below but did not know which one. The sample originally contained 0.008 g of the isotope, but 120 minutes later it contained 0.002 g.

a. How many half-lives have passed?
b. What is the half-life of the sample?
c. Which isotope did he have?

Answer 1

2 half lives. .008>.004>.002

the half life is 1 hour

he had Am-237

Answer 2

Answer: a. 2

b. 1 hour

c. Am-237

Explanation:

Half life is the time taken by a reactant to reduce to its original concentration. It is designated by the symbol
`t_(1)/(2)`.

Formula used :

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

where,

a = amount of reactant left after n-half lives = 0.002

`a_o` = Initial amount of the reactant = 0.008 g

n = number of half lives = ?

Putting values in above equation, we get:

`0.002=(0.008)/(2^n)`

`2^n=4\\\2^n=2^2`

`n=2`

Thus two half lives have passed.

2. If two half lives have passed in 120 minutes

one half life will be passed in =
`(120)/(2)* 1=60 minutes` or 1 hour

3. As half life is characteristic of a particular isotope.

Given : the half life of isotope Am-237 is 1 hour or 60 minutes. thus the isotope was Am-237.