Exponential Decay

Pre-requisites: Exponents and Logarithms

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Solve all questions
Use at least two methods whenever applicable.
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(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 1200kg of a radioactive element kept decaying for 50 years until 300kg 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 \\[3ex] $ 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 $2.7 g$?
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
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

(16.) Assume that a dog takes 300mg of Rimadyl.
Each hour, the amount of Rimadyl in the dog's system decreases by about 19%.
How much Rimadyl is left in the system after 7 hours?

$ N(t) = a(1 - r)^t \\[3ex] a = 300\;mg \\[3ex] r = 19\% = \dfrac{19}{100} = 0.19/hour \\[5ex] t = 7\;hours \\[3ex] \implies \\[3ex] N(t) = 300(1 - 0.19)^7 \\[3ex] = 300(0.81)^7 \\[3ex] = 300(0.228767925) \\[3ex] = 68.6304 \\[3ex] \approx 69\;mg \\[3ex] $ At the rate of 19% per hour, the amount of Rimadyl left in the dog's system after 7 hours was approximately 69mg.

(17.) Radioactive isotope Carbon-14 decays at a rate proportional to the amount present.
If the decay rate is 12.10% per thousand years and the current mass is 135.2mg, what will the mass be 2.2 thousand years from now?

$ N(t) = a(1 - r)^t \\[3ex] a = 135.2\;mg \\[3ex] r = 12.1\% = \dfrac{12.1}{100} = 0.121/thousand\;\;years \\[5ex] t = 2.2\;thousand\;\;years \\[3ex] \implies \\[3ex] N(t) = 135.2(1 - 0.121)^{2.2} \\[3ex] = 135.2(0.879)^{2.2} \\[3ex] = 135.2(0.752966276) \\[3ex] = 101.80104 \\[3ex] \approx 101.8\;mg \\[3ex] $ At the rate of 12.10% per thousand years, the amount of Radioactive isotope Carbon-14 remaining after 2.2 thousand years from now is approximately 101.8mg.

Exponential Decay

Objectives

Definitions

Symbols and Meanings

Formulas

For Decay Factor and Decay Rate

For Exponential Decay Rate and Half-Life

Solved Examples

Exponential Decay / Half-Life Calculators

References