Question 7fe9c

Aug 12, 2015

The half-life of the element is ${t}_{\text{1/2" = "30 s}}$.

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 $\frac{1}{8} \text{th}$ of the original sample after 90 seconds. This means that you can write

${\underbrace{\frac{1}{8} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}}}}_{\textcolor{b l u e}{\frac{1}{8} \text{th of the original sample}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} \cdot \frac{1}{2} ^ n$

This is equivalent to

$\frac{1}{8} = \frac{1}{2} ^ n$

$\frac{1}{2} ^ 3 = \frac{1}{2} ^ n \implies n = 3$

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")#