Half-Life

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

9 formulas

10 exam-style examples

~60 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Half-Life within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what half-life describes physically and when it applies.
State every one of the 9 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Half-Life is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: half-life.
Capstone Studio instructional photograph

Environmental Engineering — Half-Life: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 9 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

N	Quantity produced by "N = N0e –0.693 t/τ" — read its definition and unit from the handbook line directly above the equation.
N0	Quantity produced by "N0 = original number of atoms" — read its definition and unit from the handbook line directly above the equation.
t	Quantity produced by "t = time" — read its definition and unit from the handbook line directly above the equation.
τ	Quantity produced by "τ = half-life" — read its definition and unit from the handbook line directly above the equation.
Flux at distance 2	Quantity produced by "Flux at distance 2 = (Flux at distance 1) (r1/r2)2" — read its definition and unit from the handbook line directly above the equation.
t1 2	Quantity produced by "t1 2 = 0.693" — read its definition and unit from the handbook line directly above the equation.
k	Quantity produced by "k = rate constant (time–1)" — read its definition and unit from the handbook line directly above the equation.
t1/2	Quantity produced by "t1/2 = half-life (time)" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where
where r1 and r2 are distances from source.
The half-life of a biologically degraded contaminant assuming a first-order rate constant is given by:
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Chronic daily intake, cancer risk and radioactive decay — Half-Life

Drinking water contains 0.0460 mg/L of a carcinogen. An adult of 67 kg drinks 2.3 L/day. With a cancer slope factor of 1.11 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 57-year half-life remaining after 101 years.

Given

C = 0.0460 mg/L
IR = 2.3 L/d
BW = 67 kg
CSF = 1.11 (mg/kg·d)⁻¹
t½ = 57 yr, t = 101 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0460 × 2.3)/67 = 1.579e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.579 e - 3 (1.11) = 1.75 e - 3$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 101/57) = 29.282$

Answer: CDI = 1.58e-3 mg/kg·d, risk = 1.75e-3, 29.28% of the radionuclide remains

Why the other options are there

Risk = 0.05106 (body weight and intake ignored)
11.4% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 2

Chronic daily intake, cancer risk and radioactive decay — Half-Life (2)

Drinking water contains 0.0620 mg/L of a carcinogen. An adult of 74 kg drinks 2.2 L/day. With a cancer slope factor of 1.96 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 42-year half-life remaining after 50 years.

Given

C = 0.0620 mg/L
IR = 2.2 L/d
BW = 74 kg
CSF = 1.96 (mg/kg·d)⁻¹
t½ = 42 yr, t = 50 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0620 × 2.2)/74 = 1.843e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.843 e - 3 (1.96) = 3.61 e - 3$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 50/42) = 43.816$

Answer: CDI = 1.84e-3 mg/kg·d, risk = 3.61e-3, 43.82% of the radionuclide remains

Why the other options are there

Risk = 0.12152 (body weight and intake ignored)
40.5% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 3

Chronic daily intake, cancer risk and radioactive decay — Half-Life (3)

Drinking water contains 0.0160 mg/L of a carcinogen. An adult of 80 kg drinks 2.1 L/day. With a cancer slope factor of 1.13 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 34-year half-life remaining after 44 years.

Given

C = 0.0160 mg/L
IR = 2.1 L/d
BW = 80 kg
CSF = 1.13 (mg/kg·d)⁻¹
t½ = 34 yr, t = 44 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0160 × 2.1)/80 = 4.200e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 4.200 e - 4 (1.13) = 4.75 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 44/34) = 40.779$

Answer: CDI = 4.20e-4 mg/kg·d, risk = 4.75e-4, 40.78% of the radionuclide remains

Why the other options are there

Risk = 0.01808 (body weight and intake ignored)
35.3% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 4

Chronic daily intake, cancer risk and radioactive decay — Half-Life (4)

Drinking water contains 0.0200 mg/L of a carcinogen. An adult of 68 kg drinks 1.5 L/day. With a cancer slope factor of 0.78 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 5-year half-life remaining after 114 years.

Given

C = 0.0200 mg/L
IR = 1.5 L/d
BW = 68 kg
CSF = 0.78 (mg/kg·d)⁻¹
t½ = 5 yr, t = 114 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0200 × 1.5)/68 = 4.412e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 4.412 e - 4 (0.78) = 3.44 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 114/5) = 0.000$

Answer: CDI = 4.41e-4 mg/kg·d, risk = 3.44e-4, 0.00% of the radionuclide remains

Why the other options are there

Risk = 0.01560 (body weight and intake ignored)
-1,040% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 5

Chronic daily intake, cancer risk and radioactive decay — Half-Life (5)

Drinking water contains 0.0680 mg/L of a carcinogen. An adult of 80 kg drinks 1.5 L/day. With a cancer slope factor of 0.83 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 9-year half-life remaining after 103 years.

Given

C = 0.0680 mg/L
IR = 1.5 L/d
BW = 80 kg
CSF = 0.83 (mg/kg·d)⁻¹
t½ = 9 yr, t = 103 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0680 × 1.5)/80 = 1.275e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.275 e - 3 (0.83) = 1.06 e - 3$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 103/9) = 0.036$

Answer: CDI = 1.28e-3 mg/kg·d, risk = 1.06e-3, 0.04% of the radionuclide remains

Why the other options are there

Risk = 0.05644 (body weight and intake ignored)
-472.2% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 6

Chronic daily intake, cancer risk and radioactive decay — Half-Life (6)

Drinking water contains 0.0300 mg/L of a carcinogen. An adult of 63 kg drinks 1.7 L/day. With a cancer slope factor of 0.14 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 59-year half-life remaining after 43 years.

Given

C = 0.0300 mg/L
IR = 1.7 L/d
BW = 63 kg
CSF = 0.14 (mg/kg·d)⁻¹
t½ = 59 yr, t = 43 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0300 × 1.7)/63 = 8.095e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 8.095 e - 4 (0.14) = 1.13 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 43/59) = 60.340$

Answer: CDI = 8.10e-4 mg/kg·d, risk = 1.13e-4, 60.34% of the radionuclide remains

Why the other options are there

Risk = 0.00420 (body weight and intake ignored)
63.6% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 7

Chronic daily intake, cancer risk and radioactive decay — Half-Life (7)

Drinking water contains 0.0260 mg/L of a carcinogen. An adult of 65 kg drinks 2.0 L/day. With a cancer slope factor of 0.79 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 43-year half-life remaining after 101 years.

Given

C = 0.0260 mg/L
IR = 2.0 L/d
BW = 65 kg
CSF = 0.79 (mg/kg·d)⁻¹
t½ = 43 yr, t = 101 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0260 × 2.0)/65 = 8.000e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 8.000 e - 4 (0.79) = 6.32 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 101/43) = 19.630$

Answer: CDI = 8.00e-4 mg/kg·d, risk = 6.32e-4, 19.63% of the radionuclide remains

Why the other options are there

Risk = 0.02054 (body weight and intake ignored)
-17.4% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 8

Chronic daily intake, cancer risk and radioactive decay — Half-Life (8)

Drinking water contains 0.0400 mg/L of a carcinogen. An adult of 80 kg drinks 2.3 L/day. With a cancer slope factor of 0.69 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 25-year half-life remaining after 66 years.

Given

C = 0.0400 mg/L
IR = 2.3 L/d
BW = 80 kg
CSF = 0.69 (mg/kg·d)⁻¹
t½ = 25 yr, t = 66 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0400 × 2.3)/80 = 1.150e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.150 e - 3 (0.69) = 7.93 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 66/25) = 16.043$

Answer: CDI = 1.15e-3 mg/kg·d, risk = 7.93e-4, 16.04% of the radionuclide remains

Why the other options are there

Risk = 0.02760 (body weight and intake ignored)
-32.0% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 9

Chronic daily intake, cancer risk and radioactive decay — Half-Life (9)

Drinking water contains 0.0180 mg/L of a carcinogen. An adult of 63 kg drinks 1.9 L/day. With a cancer slope factor of 0.59 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 60-year half-life remaining after 104 years.

Given

C = 0.0180 mg/L
IR = 1.9 L/d
BW = 63 kg
CSF = 0.59 (mg/kg·d)⁻¹
t½ = 60 yr, t = 104 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0180 × 1.9)/63 = 5.429e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 5.429 e - 4 (0.59) = 3.20 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 104/60) = 30.076$

Answer: CDI = 5.43e-4 mg/kg·d, risk = 3.20e-4, 30.08% of the radionuclide remains

Why the other options are there

Risk = 0.01062 (body weight and intake ignored)
13.3% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Example 10

Chronic daily intake, cancer risk and radioactive decay — Half-Life (10)

Drinking water contains 0.0180 mg/L of a carcinogen. An adult of 64 kg drinks 2.2 L/day. With a cancer slope factor of 0.41 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 23-year half-life remaining after 18 years.

Given

C = 0.0180 mg/L
IR = 2.2 L/d
BW = 64 kg
CSF = 0.41 (mg/kg·d)⁻¹
t½ = 23 yr, t = 18 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0180 × 2.2)/64 = 6.188e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 6.188 e - 4 (0.41) = 2.54 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 18/23) = 58.131$

Answer: CDI = 6.19e-4 mg/kg·d, risk = 2.54e-4, 58.13% of the radionuclide remains

Why the other options are there

Risk = 0.00738 (body weight and intake ignored)
60.9% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Half-Life

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Half-Life contains 9 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Half-Life

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

N=N0​e−0.693t/τ

N0​=originalnumberofatoms

N=finalnumberofatoms

t=time

τ=half−life

Fluxatdistance2=(Fluxatdistance1)(r1​/r2​)2

t1​2=0.693

k=rateconstant(time−1)

t1​/2=half−life(time)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section