Algebra II - Radioactive Decay Equations

The equation for radioactive decay is,

$A=A_{0}\cdot \left (\frac{1}{2} \right )^{t/h}$ .

Where $A_{0}$ is the original amount of a radioactive substance, is the final amount, is the half life of the substance, and is time.

The half life of Carbon-14 is about years. If a fossile contains grams of Carbon-14 at time , how much Carbon-14 remains at time years?

$8 \text{ grams}$

$96 \text{ grams}$

$16 \text{ grams}$

$4 \text{ grams}$

None of the other answers.

Explanation

Using the equation for radioactive decay, we get:

$A=64\cdot \left (\frac{1}{2} \right )^{17190/5730}$ .

$A = 64\cdot \left ( \frac{1}{2} \right )^{3} = 64 \cdot \frac{1}{8} = 8 \text{ grams}$

The number of butterflies in an exhibit is decreasing at an exponential rate of decay. The number of butterflies is decreasing by every year. There are butterflies in the exhibit right now. How many butterflies will be in the exhibit in years?

Explanation

Because the butterflies are decreasing exponentially, we can use this equation

$\small F = O\cdot(1-\text{decay})^{\text{time}}$

$\small F$ is the final value

$\small O$ is the original value

The decay for this problem is 5% or 0.05

The period of time is 7 years

Using this equation we can solve for $\small F$

$\small F = 500(1-0.05)^7 = 500\cdot0.95^7 \approx 349$

$\small F = 349$

The number of fish in an aquarium is decreasing with exponential decay. The population of fish is decreasing by each year. There are fish in the aquarium today. If the decay continues how many fish will be in the aquarium in years?

None of these answers are correct

Explanation

Every year the population of fish losses 7%. In other words, every year 93% of the fish remain from the previous year. Knowing this, we can use the original number of fish to find the number of fish for the next year. Since we want to know the number of fish 4 years from now, we multiply 1500 by 93% four times.

$\small \small 1500\cdot0.93\cdot0.93\cdot0.93\cdot0.93 = 1500\cdot(0.93)^4 \approx 1122$

The population of a city is decreasing. The city has a population of , people today, but the population decreases by every year. What will be the population of the city in years if this continues?

Explanation

Because the population of the city is decreasing every year at 10.5% we can find the population after each year by using

$\small 13,529\cdot(1-0.105) = 13,529\cdot(0.895)$

Because this decrease will continue every year for the 6 years, we can continue to multiply the population by the decay for every year.

$\small \small \small 13,529\cdot0.895\cdot0.895\cdot0.895\cdot0.895\cdot0.895\cdot0.895 \Rightarrow$

$\small \small \Rightarrow 13,529\cdot0.895^6 \approx 6953$

There is water leaking out of a cup. of the water is leaking out every minute. How many kilograms of water will be left in minutes and seconds, if there are kilograms of water , $\small W$ , in the cup right now?

kilograms

None of these answers are correct

Explanation

Because the water is leaking at a continuous rate, we can use the exponential decay equation.

$\small W_{new} = W\cdot(1-D)^t$

$\small D$ is the decay of the problem, 12% or 0.12. $\small t$ is equal to how many times the water will have a 12% decay. This can be calculated as

$\small t = \frac{4min 12secs}{1min}$

To calculate this we must first convert both time to seconds

$\small t = \frac{4\cdot60+12}{1\cdot60} = \frac{240+12}{60} = \frac{252}{60}$

Our equation is then

$\small \large \large W_{new} = 8\cdot(1-0.12)^{\frac{252}{60}} \approx 4.68$

$\small W_{new} = 4.68$

An animal population is dying out. There is a decrease in this number of animals by every year. In years, there will be of this animal left. What is the current population of this animal today?

Explanation

This is an exponential decay problem. Therefore, we can use this equation.

$\small P_{7} = P_{0}\cdot(1-d)^t$

$\small P_{7}$ is the animal population after the 7 years. $\small P_{0}$ is the animal population right now. $\small d$ is the decay of the animal population every year. $\small t$ is the time period of the animal populations decay.

From the problem we know after the 7 years the animal population will be 80, so

$\small P_{7} = 80$

The population is decreasing by 8% every year, therefore

$\small d$ = 8% = 0.08

$\small t$ is equal to the number of times a decay of 8% has occurred. Since an 8% decay happens every year, and the population is 7 years from now, the population has decayed 8%, 7 times. In other words,

$\small t = \frac{7 \text{ years}}{1 \text{ year}} = 7$

These values give us,

$\small 80 = P_{0}\cdot(1-0.08)^7$

Rearranging this equation we can solve for $\small P_{0}$

$\small P_{0} = \frac{80}{(1-0.08)^7} = \frac{80}{0.92^7} = 143$

A school is losing a certain number of students each year. This year, the school has students. Four years ago the school had students. The yearly rate of the school losing students has been the same for the last four years. What is the school's yearly rate of losing students?

Explanation

This is an exponential decay problem, meaning that the decay of the school's students can be found using

$\small S_{now} = S_{past}\cdot(1-d)^t$

$\small S_{now}$ is the number of students currently at the school, which is 242

$\small S_{past}$ is the number of students that were at the school 4 years ago, which is 591

$\small t$ is the number of times the decay has occurred. Since, we are trying to find the yearly decay, the decay that happened to the school from one year to the next, and we have the number of students from 4 years ago, $\small t =4$ .

$\small d$ is the decay rate of the school that we are trying to find. Because we have every number except for $\small d$ , we can plug the values into the equation to solve for $\small d$ .

$\small 242 = 591\cdot(1-d)^4$

$\small (1-d)^4 = \frac{242}{591} \approx 0.41$

$\small 1-d = \sqrt[4]{0.41} \approx 0.80$

$\small d = 0.20$ = 20%

Over the past few years, the number of students enrolled at a certain university has been decreasing. Each year there is a 12% decrease in student enrollement. Currently, 14,286 students are enrolled. If this trend continues, how many students will be enrolled in 5 years?