Question

Question #7fe9c

Answer 1

The half-life of the element is `t_"1/2" = "30 s"`.

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

`A` - the mass of the sample after `n` half-lives;
`A_0` - the initial mass of the sample;
`n` - the number of half-lives that have passed;

In your case, you know that you're left with `1/8"th"` of the original sample after 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")`