Blessed be the GOD and FATHER of our LORD JESUS CHRIST, who has blessed us with all spiritual blessings in the heavenly places in CHRIST. Amen.

- Ephesians 1:3

The Joy of a Teacher is the Success of his Students.

- Samuel Dominic Chukwuemeka

I greet you this day,

__First:__ read the notes.

__Second:__ view the videos.

__Third:__ solve the **solved examples** and **word problems.**

__Fourth:__ check your solutions with my **thoroughly-explained** solutions.

__Fifth:__ check your answers with the calculators as applicable.

I wrote the codes for these calculators using JavaScript, a client-side scripting language.

Please use the latest Internet browsers. The calculators should work.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.

If you are my student, please do not contact me here. Contact me via the school's system.

Thank you for visiting.

**Samuel Dominic Chukwuemeka** (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Students will:

(1.) Discuss exponential decay.

(2.) Discuss real-world applications of exponential decay.

(3.) Solve applied problems in exponential decay.

## Symbols and Meanings

- To solve for a specified variable for each formula, please review Solved Examples - Literal Equations
- $N_i$ = initial amount (amount at time = $0$)
- $N_r$ = remaining amount (amount remaining after some time, $t$)
- $N_d$ = decayed amount (amount that have decayed after some time, $t$)
- $N_{td}$ = total decayed amount (total amount that have decayed after some time, $t$)
- $\lambda$ = decay rate (decay constant)
- $t$ = time
- $T_{half}$ = half-life
- $p$ = mathematical constant

(1.) $ N_r = N_ie^{-\lambda t} $

(2.) $ N_r = \dfrac{N_i}{2^p} $

(3.) $ p = \dfrac{t}{T_{half}} $

(4.) $ N_i = N_r + N_{td} $

(5.) $ t = \dfrac{\ln{\left(\dfrac{N_r}{N_i}\right)}}{-\lambda} = \dfrac{-\ln{\left(\dfrac{N_r}{N_i}\right)}}{\lambda} $

(6.) $ \lambda = \dfrac{\ln{\left(\dfrac{N_r}{N_i}\right)}}{-t} = \dfrac{-\ln{\left(\dfrac{N_r}{N_i}\right)}}{t} $

(7.) $ T_{half} = \dfrac{t\ln 2}{-\ln{\left(\dfrac{N_r}{N_i}\right)}} = \dfrac{-t\ln 2}{\ln{\left(\dfrac{N_r}{N_i}\right)}} $

(8.) $ t =\dfrac{ T_{half}\ln{\left(\dfrac{N_r}{N_i}\right)}}{-\ln 2} = \dfrac{-T_{half}\ln{\left(\dfrac{N_r}{N_i}\right)}}{\ln 2} $

(9.) $ T_{half} = \dfrac{\ln 2}{\lambda} $

(10.) $ \lambda = \dfrac{\ln 2}{T_{half}} $

(11.) $ N_i = \dfrac{N_r}{e^{-\lambda t}} $

(12.) $ N_i = \dfrac{N_r}{e^{\dfrac{-t\ln 2}{T_{half}}}} $

(13.) $ N_i = N_r * 2^p $

(14.)
$N_i = N_r * 2$^{$\dfrac{t}{T_{half}}$}

**Half-life** is defined as the **time taken for half the initial amount of a radioactive substance to decay.**

**Decay constant** or **Disintegration constant** is defined as the **ratio of the rate of radioactive nuclear decay to the amount of radioactive nuclei remaining in the radioactive substance.**

**Pre-requisites:** Exponents and Logarithms

For ACT Students

The ACT is a timed exam...$60$ questions for $60$ minutes

This implies that you have to solve each question in one minute.

Some questions will typically take less than a minute a solve.

Some questions will typically take more than a minute to solve.

The goal is to maximize your time. You use the time saved on those questions you
solved in less than a minute, to solve the questions that will take more than a minute.

So, you should try to solve each question __correctly__ and __timely__.

So, it is not just solving a question correctly, but solving it __correctly on time__.

Please ensure you attempt __all ACT questions__.

There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students

Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students

Any question labeled WASCCE is a question for the WASCCE General Mathematics

Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students

All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.

Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students

__For the Questions:__

Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from
behind.

Any comma included in a number indicates a decimal point.

__For the Solutions:__

Decimals are used appropriately rather than commas

Commas are used to separate digits appropriately.

Solve all questions

Use ** at least two** methods whenever applicable.

Show all work

(1.) A radioactive element has a half-life of $20$ years.

If there are $2500mg$ of the element initially, how much of the element:

(a.) will remain after $40$ years?

(b.) will have decayed after $40$ years?

$ (a.) \\[3ex] N_r = N_ie^{-\lambda t} \\[3ex] T_{half} = 20\: years \\[3ex] t = 40\: years \\[3ex] N_i = 2500mg \\[3ex] \lambda = \dfrac{\ln 2}{T_{half}} \\[5ex] \lambda = \dfrac{\ln 2}{20} \\[5ex] -\lambda t = -\dfrac{\ln 2}{20} * 40 = -2\ln 2 \\[5ex] -2\ln 2 = \ln {2^{-2}} ...Law\: 5...Log \\[3ex] {2^{-2}} = \dfrac{1}{2^2} ...Law\: 6...Exp \\[5ex] {2^{-2}} = \dfrac{1}{4} \\[5ex] -2\ln 2 = \ln {\dfrac{1}{4}} \\[5ex] e^{-2\ln 2} = e^{\ln {\dfrac{1}{4}}} \\[5ex] e^{\ln \left({\dfrac{1}{4}}\right)} = \dfrac{1}{4} ...Law\: 7...Log \\[7ex] \Rightarrow N_r = 2500 * \dfrac{1}{4} \\[5ex] N_r = 625mg \\[3ex] (b.) \\[3ex] N_i = N_r + N_{td} \\[3ex] N_{td} = N_i - N_r \\[3ex] N_{td} = 2500 - 625 \\[3ex] N_{td} = 1875mg \\[3ex] $

$ (a.) \\[3ex] N_r = \dfrac{N_i}{2^p} \\[5ex] p = \dfrac{t}{T_{half}} \\[5ex] T_{half} = 20\: years \\[3ex] t = 40\: years \\[3ex] N_i = 2500mg \\[3ex] p = \dfrac{40}{20} \\[5ex] p = 2 \\[3ex] \Rightarrow N_r = \dfrac{2500}{2^2} \\[5ex] N_r = \dfrac{2500}{4} \\[5ex] N_r = 625mg \\[3ex] (b.) \\[3ex] N_i = N_r + N_{td} \\[3ex] N_{td} = N_i - N_r \\[3ex] N_{td} = 2500 - 625 \\[3ex] N_{td} = 1875mg $

(2.) A $1200 kg$ of a radioactive element kept decaying for $50$ years until $300 kg$ of the element remained.

What is the half-life of the element?

$ T_{half} = \dfrac{-t\ln 2}{\ln{\left(\dfrac{N_r}{N_i}\right)}} \\[7ex] N_i = 1200kg \\[3ex] N_r = 300 kg \\[3ex] t = 50\: years \\[3ex] T_{half} = \dfrac{-50 * \ln 2}{\ln{\left(\dfrac{300}{1200}\right)}} \\[7ex] T_{half} = \dfrac{-50 * 0.693147181}{\ln{(0.25)}} \\[5ex] T_{half} = \dfrac{-34.65735903}{-1.386294361} \\[5ex] T_{half} = 25\: years $

(3.) Sodium-24 is used to locate obstructions in blood flow.

It has a half-life of $15\: hours$.

If a procedure needs $0.75g$ and is scheduled to take place in $2\: days$, what is the minimum
amount of sodium-24 required now?

Round your answer to the nearest tenth.

$N_i = N_r * 2$

$ T_{half} = 15\: hours \\[3ex] t = 2\: days = 2 * 24 = 48\: hours \\[3ex] N_r = 0.75g $

$N_i = 0.75 * 2$

$ N_i = 0.75 * 2^{3.2} \\[3ex] N_i = 0.75 * 9.18958684 \\[3ex] N_i = 6.89219013 \\[3ex] N_{i} = 6.9g $

(4.) $175 g$ of a $200 g$ radioactive element decayed in $87$ hours.

What is the half-life of the element?

$ T_{half} = \dfrac{-t\ln 2}{\ln{\left(\dfrac{N_r}{N_i}\right)}} \\[7ex] N_i = 200 g \\[3ex] N_{td} = 175 g \\[3ex] N_i = N_r + N_{td} \\[3ex] N_r = N_i - N_{td} \\[3ex] N_r = 200 - 175 = 25 g \\[3ex] t = 87\: hours \\[3ex] T_{half} = \dfrac{-87 * \ln 2}{\ln{\left(\dfrac{25}{200}\right)}} \\[7ex] T_{half} = \dfrac{-87 * 0.693147181}{\ln{(0.125)}} \\[5ex] T_{half} = \dfrac{-60.30380471}{-2.079441542} \\[5ex] T_{half} = 29\: hours $

(5.) Americium-241 is a vital component of household smoke detectors.

It has a half-life of $432$ years.

How many years will it take a $10 g$ mass of Americium-241 to decay to $2.7 g$?

Round your answer to the nearest integer.

$ t = \dfrac{-T_{half}\ln{\left(\dfrac{N_r}{N_i}\right)}}{\ln 2} \\[7ex] T_{half} = 432\: years \\[3ex] N_i = 10 g \\[3ex] N_r = 2.7 g \\[3ex] t = \dfrac{-432 * \ln{\left(\dfrac{2.7}{10}\right)}}{\ln 2} \\[7ex] t = \dfrac{-432 * \ln{(0.27)}}{0.693147181} \\[7ex] t = \dfrac{-432 * -1.30933332}{0.693147181} \\[5ex] t = \dfrac{565.6319942}{0.693147181} \\[5ex] t = 816.0344725 \\[3ex] t = 816\: years $

(6.) Carbon-14 is a natural radioactive carbon isotope in living organisms.

It is constantly renewed as the organism lives, and begins to decay when the organism dies.

If the half-life of carbon-14 is $5730$ years, how old is a mommy having only $70\%$ of the
normal amount of carbon-14?

$ t = \dfrac{-T_{half}\ln{\left(\dfrac{N_r}{N_i}\right)}}{\ln 2} \\[7ex] T_{half} = 5730\: years \\[3ex] N_i = 100\% = 1 \\[3ex] N_r = 70\% = 0.7 \\[3ex] t = \dfrac{-5730 * \ln{\left(\dfrac{0.7}{1}\right)}}{\ln 2} \\[7ex] t = \dfrac{-5730 * \ln{(0.7)}}{0.693147181} \\[7ex] t = \dfrac{-5730 * -0.356674944}{0.693147181} \\[5ex] t = \dfrac{2043.747429}{0.693147181} \\[5ex] t = 2948.504278 \\[3ex] t = 2949\: years\: old $

(7.) Carbon-14 is a natural radioactive carbon isotope in living organisms.

It is constantly renewed as the organism lives, and begins to decay when the organism dies.

Assume the half-life of carbon-14 is $5750$ years.

Gregory, a paleontologist determined that the bones from a mastodon had lost $83.4\%$ of their carbon-14.

How old were the bones at the time they were discovered?

$ t = \dfrac{-T_{half}\ln{\left(\dfrac{N_r}{N_i}\right)}}{\ln 2} \\[7ex] T_{half} = 5750\: years \\[3ex] N_i = 100\% = 1 \\[3ex] N_{td} = 83.4\% = 0.834 \\[3ex] N_i = N_r + N_{td} \\[3ex] N_r = N_i - N_{td} \\[3ex] N_r = 1 - 0.834 = 0.166 \\[3ex] t = \dfrac{-5750 * \ln{\left(\dfrac{0.166}{1}\right)}}{\ln 2} \\[7ex] t = \dfrac{-5750 * \ln{(0.166)}}{0.693147181} \\[7ex] t = \dfrac{-5750 * -1.795767491}{0.693147181} \\[5ex] t = \dfrac{10325.66307}{0.693147181} \\[5ex] t = 14896.7829 \\[3ex] t = 14897\: years\: old $

(8.) **JAMB** The particle emitted when $^{39}_{19}K$ decays to $^{39}_{19}K$ is

A. electron

B. beta

C. alpha

D. gamma

$ \gamma = ^{0}_{0}\gamma \\[3ex] 39 - 0 = 39 \\[3ex] 19 - 0 = 19 \\[3ex] ^{39}_{19}K - ^{0}_{0}\gamma = ^{39}_{19}K $

(9.) **JAMB** The percentage of the original nuclei of a sample of a radioactive substance left
after $5$ half-lives is

$
A.\:\: 8\% \\[3ex]
B.\:\: 5\% \\[3ex]
C.\:\: 3\% \\[3ex]
D.\:\: 1\%
$

$ N_i = 100\% \\[3ex] After\:\: 1 \:\:half-life,\:\: \dfrac{1}{2} * 100\% = 50\% \:\:decays,\:\: 100 - 50 = 50\% \:\:remain \\[5ex] After\:\: 2 \:\:half-lives,\:\: \dfrac{1}{2} * 50\% = 25\% \:\:decays,\:\: 50 - 25 = 25\% \:\:remain \\[5ex] After\:\: 3 \:\:half-lives,\:\: \dfrac{1}{2} * 25\% = 12.5\% \:\:decays,\:\: 25 - 12.5 = 12.5\% \:\:remain \\[5ex] After\:\: 4 \:\:half-lives,\:\: \dfrac{1}{2} * 12.5\% = 6.25\% \:\:decays,\:\: 12.5 - 6.25 = 6.25\% \:\:remain \\[5ex] After\:\: 5 \:\:half-lives,\:\: \dfrac{1}{2} * 6.25\% = 3.125\% \:\:decays,\:\: 6.25 - 3.125 = 3.125\% \:\:remain \\[5ex] 3.125\% \approx 3\% $

(10.) **JAMB** The process of energy production in the sun is

A. radioactive decay

B. electron collision

C. Nuclear fission

D. Nuclear fusion

The process of energy production in the sun is the nuclear fusion of the hydrogen atoms in the sun.

(11.) **JAMB** The particle that is responsible for nuclear fission in a nuclear reactor is

A. neutron

B. proton

C. electron

D. photon

The particle that is responsible for nuclear fission in a nuclear reactor is the neutron.

(12.) **JAMB** The count rate of a radioactive material is $800$ count/min.

If the half-life of the material is $4$ days, what would be the count rate $16$ days later?

$
A.\:\: 200 \:\:count/min \\[3ex]
B.\:\: 100 \:\:count/min \\[3ex]
C.\:\: 50 \:\:count/min \\[3ex]
D.\:\: 25 \:\:count/min
$

$ N_i = 800 \:\:count/min \\[3ex] T_{half} = 4 \:\:days \\[3ex] t = 16 \:\:days \\[3ex] N_r = ? \\[3ex] p = \dfrac{t}{T_{half}} = \dfrac{16}{4} = 4 \\[5ex] N_r = \dfrac{N_i}{2^p} \\[5ex] N_r = \dfrac{800}{2^4} = \dfrac{800}{16} = 50 \:\:count/min $

(13.) **JAMB** A radioisotope has a decay constant of $10^{-7}s^{-1}$.

The average life of the radioisotope is

$
A.\:\: 6.93 * 10^8s \\[3ex]
B.\:\: 1.00 * 10^{-7}s \\[3ex]
C.\:\: 1.00 * 10^7s \\[3ex]
D.\:\: 6.93 * 10^7s
$

The average life of the radioisotope is the half-life

It is important to know that $\ln 2 \approx 0.693$

$ \lambda = 10^{-7}s{-1} \\[3ex] T_{half} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{10^{-7}} = 0.693 * \dfrac{1}{10^{-7}} \\[5ex] T_{half} = 0.693 * 10^7 \\[3ex] T_{half} = 6.93 * 10^{-1} * 10^7 \\[3ex] T_{half} = 6.93 * 10^{-1 + 7} \\[3ex] T_{half} = 6.93 * 10^6s $

(14.) If the half-life of $Iodine-131$ is 8 days, how much of a 100 mg sample of $Iodine-131$ remains after 32 days?

$ N_r = N_ie^{-\lambda t} \\[3ex] T_{half} = 8\: days \\[3ex] t = 32\: days \\[3ex] N_i = 100mg \\[3ex] \lambda = \dfrac{\ln 2}{T_{half}} \\[5ex] \lambda = \dfrac{\ln 2}{8} \\[5ex] -\lambda t = -\dfrac{\ln 2}{8} * 32 = -4\ln 2 \\[5ex] -4\ln 2 = \ln {2^{-4}} ...Law\: 5...Log \\[3ex] {2^{-4}} = \dfrac{1}{2^4} ...Law\: 6...Exp \\[5ex] {2^{-4}} = \dfrac{1}{16} \\[5ex] -4\ln 2 = \ln {\dfrac{1}{16}} \\[5ex] e^{-4\ln 2} = e^{\ln {\dfrac{1}{16}}} \\[5ex] e^{\ln \left({\dfrac{1}{16}}\right)} = \dfrac{1}{16} ...Law\: 7...Log \\[7ex] \Rightarrow N_r = 100 * \dfrac{1}{16} \\[5ex] N_r = 6.25mg \\[3ex] $

$ N_r = \dfrac{N_i}{2^p} \\[5ex] p = \dfrac{t}{T_{half}} \\[5ex] T_{half} = 8\: days \\[3ex] t = 32\: days \\[3ex] N_i = 100mg \\[3ex] p = \dfrac{32}{8} \\[5ex] p = 4 \\[3ex] \Rightarrow N_r = \dfrac{100}{2^4} \\[5ex] N_r = \dfrac{100}{16} \\[5ex] N_r = 6.25mg \\[3ex] $ $6.25mg$ of $Iodine-131$ will remain after $32$ days

(15.) A tumor is injected with $0.62$ grams of Iodine-125

After 1 day, the amount of Iodine-125 has decreased by 1.15%

(a.) Write an exponential decay model with $A(t)$ representing the amount of Iodine-125 remaining in the tumor after $t$ days.

(b.) Use the formula for $A(t)$ to find the amount of Iodine-125 that would remain in the tumor after 8.5 days.

Round your answer to the nearest thousandth (3 decimal places) of a gram.

$ N_r = N_ie^{-\lambda t} \\[3ex] A_r = A_ie^{-\lambda t} \\[3ex] After\;\;1\;\;day \implies t = 1\;day \\[3ex] Decreased\;\;by\;\;1.15\% \;\; \implies \dfrac{1.15}{100} * A_i = 0.0115\;A_i \\[5ex] A_r = Remaining\;\;A_i = 1\;A_i - 0.0115\;A_i = 0.9885A_i \\[3ex] \implies \\[3ex] 0.9885A_i = A_ie^{-\lambda * 1} \\[3ex] A_ie^{-\lambda * 1} = 0.9885A_i \\[3ex] e^{-\lambda} = \dfrac{0.9885A_i}{A_i} \\[5ex] e^{-\lambda} = 0.9885 \\[3ex] Introduce\;\;natural\;\;logarithm\;\;to\;\;both\;\;sides \\[3ex] \ln e^{-\lambda} = \ln 0.9885 \\[3ex] -\lambda = \ln 0.9885 \\[3ex] \lambda = -\ln 0.9885 \\[3ex] (a.) \\[3ex] A_r = A_ie^{-\lambda t} \\[3ex] A_r = A(t) \\[3ex] A_i = 0.62\;grams \\[3ex] -\lambda = \ln 0.9885 \\[3ex] \implies \\[3ex] A(t) = 0.62e^{\ln 0.9885 t} \\[3ex] (b.) \\[3ex] t = 8.5\;days \\[3ex] A_r = ? \\[3ex] A_i = 0.62\;grams \\[3ex] -\lambda = \ln 0.9885 \\[3ex] A_r = A_ie^{-\lambda t} \\[3ex] A_r = 0.62e^{\ln 0.9885 * 8.5} \\[3ex] A_r = 0.62e^{-0.9831640916} \\[3ex] A_r = 0.62 * 0.9063620771 \\[3ex] A_r = 0.5619444878 \\[3ex] A_r \approx 0.562\;grams \;\;to\;\;the\;\;nearest\;\;thousandth \\[3ex] $ After 8.5 days, there will be approximately 0.562 grams of Iodine-125

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