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# Exponential Decay

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## Objectives

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
• $N_i$ = 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

## Formulas

(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}} ## Definitions 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. ## Solved Examples 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? First Method: First Formula (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] Second Method: Second Formula (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. First Method: First Formula N_i = \dfrac{N_r}{e^{\dfrac{-t\ln 2}{T_{half}}}} \\[7ex] T_{half} = 15\: hours \\[3ex] t = 2\: days = 2 * 24 = 48\: hours \\[3ex] N_r = 0.75g \\[3ex] N_i = \dfrac{0.75}{e^{\dfrac{-48 * \ln 2}{15}}} \\[7ex] N_i = \dfrac{0.75}{e^{\dfrac{-33.27106467}{15}}} \\[7ex] N_i = \dfrac{0.75}{e^{-2.218070978}} \\[7ex] N_i = \dfrac{0.75}{0.108818820} \\[7ex] N_i = 6.89219013 \\[3ex] N_{i} = 6.9g Second Method: Second Formula N_i = N_r * 2$$\dfrac{t}{T_{half}}$
$T_{half} = 15\: hours \\[3ex] t = 2\: days = 2 * 24 = 48\: hours \\[3ex] N_r = 0.75g$
$N_i = 0.75 * 2$$\dfrac{48}{15}$

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

(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 $2.7 g$?

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

(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
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
JAMB exam does not allow the use of a calculator.
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?

First Method: First Formula

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

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

## Exponential Decay / Half-Life Calculators

Make sure the applicable units are the same
Given: initial amount, half-life, time
To Find: other details

Given: initial amount, remaining amount, time
To Find: other details

Given: remaining amount, half-life, time
To Find: other details

Given: initial amount, remaining amount, half-life
To Find: other details