# Iodine-131 is a radioactive isotope with a half-life of 8 days. How many grams of a 64 g sample of iodine-131 will remain at the end of 24 days?

##### 2 Answers

#### Explanation:

The **nuclear half-life** of a radioactive isotope tells you how much time must pass in order for **half of the atoms** present in an initial sample to undergo *radioactive decay*.

In essence, the half-life tells you at what time intervals you can expect an initial sample of a radioactive isotope to be **halved**.

In your case, iodine-131 is said to have a half-life of **days**. This means that with **every** **days** that pass, your sample of iodine-131 will be **halved**.

If you take *be left with*

#1/2 * A_0 = A_0/2 -> # after the passing ofone half-life

#1/2 * A_0/2 = A_0/4 -># after the passing oftwo half-lives

#1/2 * A_0/4 = A_0/8 -># after the passing ofthree half-lives

#vdots#

and so on.

You can thus say that the amount of iodine-131 that **remains undecayed**,

#color(blue)(|bar(ul(color(white)(a/a)A = A_0 * 1/2^ncolor(white)(a/a)|)))#

Here **number of half-lives** that pass in

In your case, you have

#n = (24 color(red)(cancel(color(black)("days"))))/(8color(red)(cancel(color(black)("days")))) = 3#

This will get you

#A_"24 days" = "64 g" * 1/2^3#

#A_"24 days" = color(green)(|bar(ul(color(white)(a/a)"8 g"color(white)(a/a)|)))#

#### Explanation:

Half Life follows first Order kinetics

This graph shows what is going on during the decay

As you can see from the graph the half life is 8 days

Lets do the math and verify it with the graph

Recall

With this you can solve most problems regarding the above data

Now

**Greater the concentration of a sample greater the mass in in it**

So

Verifying with the graph

**Hence verified**