# Question #7fe9c

##### 1 Answer

The half-life of the element is

#### Explanation:

You can actually solve this problem without using the half-life equation.

When you're dealing with a sample of a radioactive element, you know that its initial mass will be **halved** with every half-life that passes.

In other words, you are left with **half** of what you started with after **one half-life**, you are left with **a quarter** of what you starter with after **two half-lives**, you are left with **one eighth** of what you started after **three half-lives**, and so on.

You can actually write this as

#A = A_0/2^(n)# , where

In your case, you know that you're left with *90 seconds*. This means that you can write

#underbrace(1/8 * color(red)(cancel(color(black)(A_0))))_(color(blue)(1/8"th of the original sample")) = color(red)(cancel(color(black)(A_0))) * 1/2^n#

This is equivalent to

#1/8 = 1/2^n#

#1/2^3 = 1/2^n implies n = 3#

So, **three half-lives** have passed in 90 seconds, which can only mean that

#3 * t_"1/2" = "90 s" implies t_"1/2" = "90 s"/3 = color(green)("30 s")#