Radioactive Decay Equations - Algebra II

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Question

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?

Answer

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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Question

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?

Answer

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

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:

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Question

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?

Answer

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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Question

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?

Answer

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 $\small F$

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

$\small F = 349$

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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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

Answer

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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Question

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?

Answer

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:

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Question

An element has a half life of 6 days. What is the approximate amount remaining for a 50 mg sample of this element after 5 days?

Answer

Write the formula for half life.

$y=a(\frac{1}{2})^x$

$a=\textup{ sample size} = 50$

$x=\textup{ time in days}$

Since the time requested is five out of the six day of the half life, the value of is:

$x=\frac{5}{6}$

Substitute all the known values into the equation.

$y=50(\frac{1}{2})^{\frac{5}{6}} = 28.06155$

The answer is:

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