Question

The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial amount. The half-life of the radioactive gas radon is approximately 2.8 days. The initial amount of radon used in an experiment is 74 grams. If N represents the number of grams of radon remaining t days after the start of the experiment,
a. Write an equation that gives Nin terms of t.
b. how much gas radon approximately remains after 2.8 days?
c. approximately when will the amount of radon remaining be 10 grams?

Answer 1

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

$N = N_0 * (\frac{1}{2})^{t/p$

$N_0$ is the starting amount of parent element.

$N$ is the end amount of parent element

$t$ is the time elapsed

$p$ is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days$

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2} = 37g$

c.

Now we just gotta do some algebra. Use the original function but this time, replace $N(t)$ with 10g and solve algebraically.

$10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}$

Take the log of both sides.

$log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})$

Use the exponent rule for log laws that, $log(b^x) = x*log(b)$

$log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})$

$\frac{log(\frac{5}{37})}{log(\frac{1}{2})} = \frac{t}{2.8days}$

$2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})} = t$

slap that in your calculator and you get

$t = 8.1 days$