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Algebra II : Radioactive Decay Equations

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Example Question #1 : Radioactive Decay Equations

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?

Possible Answers:

3,571

5,714

12,426

9,756

7,539

Correct answer:

7,539

Explanation:

This is an exponential decay problem. The formula for exponential decay is:

FV = PV(1- d)^n

Where

= future value

= present value

= rate of decay

= number of periods

This problem requests the number of students five years in the future. The rate of decay is twelve percent. Therefore:

FV = 14,286(1-.12)^5

FV = 7,539

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Example Question #1 : Radioactive Decay Equations

Sceintists recently discovered a new type of metal compound. They have roughly 15 grams of this compound, which has a half life of 16 hours. Approximately how much of this substance will the scientists have in 24 hours?

Possible Answers:

0.005 grams

0.50 grams

5.00 grams

0.06 grams

grams

Correct answer:

0.06 grams

Explanation:

Recall the radioactive decay formula:

$Q_{(t)}=Q_{(o)} (^{-r\cdot t})$

The half life formula is:

$r=\frac{ln2}{(h)}$ , where is the half life.

Plug in the given half life:

$r=\frac{ln2}{(16)}\approx 0.0433216$

Plug this value into the radioactive decay formula:

$Q_{(24)}=15 (^{-0.0433216\cdot 24})\approx 0.06$ grams

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Example Question #1 : 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 5730 years. If a fossile contains grams of Carbon-14 at time t = 0 , how much Carbon-14 remains at time t = 17190 years?

Possible Answers:

$16 \text{ grams}$

$4 \text{ grams}$

$8 \text{ grams}$

None of the other answers.

$96 \text{ grams}$

Correct answer:

$8 \text{ grams}$

Explanation:

Using the equation for radioactive decay, we get:

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

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

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Example Question #2 : Radioactive Decay Equations

The number of fish in an aquarium is decreasing with exponential decay. The population of fish is decreasing by $7\%$ each year. There are 1500 fish in the aquarium today. If the decay continues how many fish will be in the aquarium in years?

Possible Answers:

1200

1122

990

None of these answers are correct

Correct answer:

1122

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.

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

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Example Question #319 : Exponents

The population of a city is decreasing. The city has a population of , 529 people today, but the population decreases by $10.5\%$ every year. What will be the population of the city in years if this continues?

Possible Answers:

6565

7769

7312

6953

Correct answer:

6953

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.

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

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

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Example Question #320 : Exponents

There is water leaking out of a cup. $12\%$ 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?

Possible Answers:

5.40 kilograms

2.32 kilograms

4.68 kilograms

None of these answers are correct

Correct answer:

4.68 kilograms

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

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

$\small W_{new} = 4.68$

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Example Question #1 : Radioactive Decay Equations

An animal population is dying out. There is a decrease in this number of animals by $8\%$ every year. In years, there will be of this animal left. What is the current population of this animal today?

Possible Answers:

101

143

Correct answer:

143

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 $\small P_{0}$

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

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Example Question #2 : Radioactive Decay Equations

A school is losing a certain number of students each year. This year, the school has 242 students. Four years ago the school had 591 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?

Possible Answers:

$12\%$

$67\%$

$20\%$

$80\%$

Correct answer:

$20\%$

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%

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Example Question #3 : Radioactive Decay Equations

Cells in a dish have started to decay. The cells are decaying by $3\%$ every minutes. When you left the cells there were 1000 cells in the dish. Now there are cells in the dish. Approximately how long did you leave the cells for?

Possible Answers:

$78 \text{ hours and } 20 \text{ minutes}$

$23 \text{ hours and } 4 \text{ minutes}$

$90 \text{ hours and } 40 \text{ minutes}$

$26 \text{ hours and } 6 \text{ minutes}$

Correct answer:

$26 \text{ hours and } 6 \text{ minutes}$

Explanation:

This is an exponential decay problem. That means that after 20 minutes 3% of the cells decay and 97% of the cells in the dish are left. To find the new amount of cells in the dish, we multiple the original number by the 97%.

$\small 1000 \cdot(0.97)$

The number of cells after every 20 minute interval can be calculated this way. Therefore, to find how long the cell were decaying we use,

$\small \small \small 1000 \cdot(0.97) \cdot(0.97)\cdot... = 92$

Which can be rewritten as,

$\small \small 1000(0.97)^t = 92$

Now we can solve for $\small t$ , which is the amount of time that you were gone

$\small (1-0.03)^t = \frac{92}{1000}$

To solve for $\small t$ , we must take the log of both sides to base 10. This will give us,

$\small t\cdotlog(0.97) = log\bigg(\frac{92}{1000}\bigg)$

$\small \small t = \frac{log(\frac{92}{1000})}{log(0.97)} \approx 78.333$

Remember! $\small t$ is the number of decays that the cells went through. Each decay took 20 minutes to get through, however.

Therefore, the time that you were gone is

$\small \small t = 78.333\cdot\frac{20}{60} \approx 26 \text{ hours and} \ 6 \text{ minutes}$

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Example Question #4 : Radioactive Decay Equations

Suppose 5 milligrams of element X decays to 3.2 milligrams after 48 hours. What is the decay rate on a day to day basis?

Possible Answers:

$15\%$

$28\%$

$34\%$

$22\%$

$30\%$

Correct answer:

$22\%$

Explanation:

Write the formula for radioactive decay.

$X=X_0e^{-\lambda T }$

$X= \textup{final amount} = 3.2$

$X_0=\textup{ initial amount }= 5$

$T=\textup{time in days} = \frac{48 \textup{ hours}}{ \frac{24\textup{ hours}}{1\textup{ day}}} = 2$

Substitute the values in the equation and solve for lambda.

$5=3.2e^{-2\lambda }$

Divide by 3.2 on both sides.

$1.5625=e^{-2\lambda }$

Take the natural log of both sides to eliminate the exponential.

$ln(1.5625)=ln(e^{-2\lambda })$

$ln(1.5625)=-2\lambda$

Divide by negative two on both sides.

$\frac{ln(1.5625)}{-2}=\frac{-2\lambda}{-2}$

$\lambda=\frac{ln(1.5625)}{-2}=-0.223144$

Convert this to a percentage.

This element's decay rate is approximately: $22\%$

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Algebra II : Radioactive Decay Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Radioactive Decay Equations

Example Question #1 : Radioactive Decay Equations

Example Question #1 : Radioactive Decay Equations

Example Question #2 : Radioactive Decay Equations

Example Question #319 : Exponents

Example Question #320 : Exponents

Example Question #1 : Radioactive Decay Equations

Example Question #2 : Radioactive Decay Equations

Example Question #3 : Radioactive Decay Equations

Example Question #4 : Radioactive Decay Equations

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