Half-Life vs Decay Constant: What Is the Difference?

Half-life is a time, the time for half the nuclei in a sample to decay. The decay constant is a rate, an inverse time that sets how fast the exponential falls. They carry exactly the same information and convert into each other by lambda equals ln 2 divided by the half-life.

AP Physics: Unit 15 (topics 15.7 Fission, Fusion, and Nuclear Decay). AP Physics 2 Unit 15, Topic 15.7, learning objective 15.7.B, verified against the rendered CED pages: describe the radioactive decay of a given sample of material consisting of a finite number of nuclei. Essential knowledge 15.7.B.1 defines radioactive decay as the spontaneous transformation of a nucleus into one or more different nuclei, or to a lower energy level of the same nucleus. 15.7.B.1.i: the time at which an individual nucleus undergoes radioactive decay is indeterminable, but decay rates can be described using probability. 15.7.B.1.ii: the half-life, t one-half, of a radioactive material is the time it takes for half of the initial number of radioactive nuclei to have spontaneously decayed. 15.7.B.1.iii: the decay constant lambda can be related to the half-life with lambda equals ln 2 over t one-half. 15.7.B.2: a material's decay constant may be used to predict the number of nuclei remaining in a sample after a period of time, or the age of a material if the initial amount of material is known, with relevant equation N = N naught e to the minus lambda t and DERIVED equation ln of N over N naught equals minus lambda t. 15.7.B.3: different unstable elements and isotopes may have vastly different half-lives, ranging from fractions of a second to billions of years. So the exponential decay law is required content, not enrichment. Confirmed on the rendered Table of Information appendix: the Modern Physics group of the AP Physics 2 equation sheet prints N = N naught e to the minus lambda t and lambda = ln 2 over t one-half, but NOT the logarithmic derived form, and its symbol key defines lambda as wavelength or decay constant. Topic 15.7 carries NO boundary statement, checked on both of its printed pages; the relevant boundary is the Topic 15.8 one, which says AP Physics 2 does not expect students to memorize the processes by which specific isotopes decay or the half-lives of specific isotopes. Suggested skills for Topic 15.7: 1.B, 2.B, 2.C, 3.A, 3.B. SCOPE FINDINGS: no equation for activity or decay rate appears in the framework or on the sheet, and the units becquerel and curie appear nowhere in the CED; the phrases carbon dating and radiocarbon dating appear nowhere, though 15.7.B.2 does cover finding the age of a material; the mean lifetime is not in the framework or on the sheet. The CED's sample instructional activity 3 for Topic 15.7 has groups remove dice showing a 1 from a box of 200 each turn and states the resulting half-life is about 3.8 turns; recomputed here as ln 2 divided by ln of six fifths, which is 3.80 turns, confirming the CED's figure and showing that the effective decay constant is ln of six fifths, 0.182 per turn, rather than the naive one sixth. The CED's published sample multiple-choice set includes a question relating the number of nuclei at two later times using the decay constant, whose correct answer uses the elapsed interval t two minus t one and whose distractors add the times or introduce the initial count. Unit 15 is weighted at 12 to 15 percent of the multiple-choice section over a suggested 14 to 22 class periods, with a Progress Check of about 24 multiple-choice and 4 free-response questions.

The distinction, stated once

These are the same fact in two units, and the conversion between them is printed on the equation sheet.

[Half-life](/glossary/half-life) is a time. Essential knowledge 15.7.B.1.ii: the half-life, $t_{1/2}$ , of a radioactive material is the time it takes for half of the initial number of radioactive nuclei to have spontaneously decayed. Give a half-life in seconds and you have said something a person can picture.

The [decay constant](/glossary/decay-constant) is an inverse time. Essential knowledge 15.7.B.1.iii: the decay constant $\lambda$ can be related to the half-life of a radioactive material with the equation

\lambda = \frac{\ln 2}{t_{1/2}}

A larger $\lambda$ means a faster decay and therefore a shorter half-life. That inversion is the first thing to fix: the two quantities move in opposite directions.

The practical difference is which one each tool wants. A question gives you the half-life, because that is the humane way to describe a material. The exponential equation needs the decay constant, because that is what sits in the exponent. Essential knowledge 15.7.B.2 gives that equation, $N = N_0 e^{-\lambda t}$ , and there is no version of it written in terms of $t_{1/2}$ anywhere in AP Physics 2. So the conversion is not a curiosity; it is the step between the data you are handed and the equation you are expected to use.

Which is why $\lambda = \ln 2 / t_{1/2}$ has its own essential knowledge statement and its own line on the sheet. The framework treats the translation as content in its own right.

Side by side

	Half-life $t_{1/2}$	Decay constant $\lambda$
CED statement	15.7.B.1.ii	15.7.B.1.iii
Kind of quantity	A time	An inverse time
Typical unit	$\mathrm{s}$ , or years for slow decays	$\mathrm{s^{-1}}$ , or per year
Definition	The time for half the initial nuclei to have decayed	The constant in the exponent of the decay law
Faster decay means	A smaller value	A larger value
Where it appears on the sheet	Only inside $\lambda = \ln 2 / t_{1/2}$	In $\lambda = \ln 2 / t_{1/2}$ and in $N = N_0 e^{-\lambda t}$
What you are usually given	This one	Rarely
What the exponential needs	Not this one	This one
Convert to the other by	$\lambda = \ln 2 / t_{1/2}$	$t_{1/2} = \ln 2 / \lambda$
Depends on the decay mode	No	No
Range the CED states	Fractions of a second to billions of years (15.7.B.3)	The CED states no range

Two rows deserve a second look.

The conversion row is symmetric, and that is worth noticing rather than memorising twice. Because $\ln 2$ appears in the numerator both ways, the same rearrangement runs in either direction: $\lambda t_{1/2} = \ln 2$ . If you remember the product form, you cannot get the fraction upside down.

The decay-mode row says something the rest of the unit backs up. Nothing about half-life or the decay constant refers to whether the nucleus is an alpha or a beta emitter. The whole of Topic 15.8, which distinguishes alpha from beta decay, has no bearing on either quantity. This is one number per material, whatever it is doing.

What the CED requires, including whether the exponential law is on the syllabus

Both quantities live under learning objective 15.7.B, in Topic 15.7, Fission, Fusion, and Nuclear Decay: describe the radioactive decay of a given sample of material consisting of a finite number of nuclei.

The answer to the obvious scope question is yes. The exponential decay law is required content, not an enrichment. Here is the objective in full:

15.7.B.1: radioactive decay is the spontaneous transformation of a nucleus into one or more different nuclei, or to a lower energy level of the same nucleus.
15.7.B.1.i: the time at which an individual nucleus undergoes radioactive decay is indeterminable, but decay rates can be described using probability.
15.7.B.1.ii: the half-life, $t_{1/2}$ , of a radioactive material is the time it takes for half of the initial number of radioactive nuclei to have spontaneously decayed.
15.7.B.1.iii: the decay constant $\lambda$ can be related to the half-life with $\lambda = \ln 2 / t_{1/2}$ .
15.7.B.2: a material's decay constant may be used to predict the number of nuclei remaining in a sample after a period of time, or the age of a material if the initial amount of material is known. Relevant equation: $N = N_0 e^{-\lambda t}$ . Derived equation: $\ln\!\left(\dfrac{N}{N_0}\right) = -\lambda t$ .
15.7.B.3: different unstable elements and isotopes may have vastly different half-lives, ranging from fractions of a second to billions of years.

So the framework asks for more than half-life reasoning. It names the exponential, it names the linearised form, and 15.7.B.2 explicitly extends the use to finding the age of a material when the initial amount is known.

A detail about the sheet that the CED does not flag for you. Two of those three equations are printed in the Modern Physics group of the AP Physics 2 equation sheet, verified on the rendered Table of Information appendix:

N = N_0 e^{-\lambda t} \qquad \lambda = \frac{\ln 2}{t_{1/2}}

The derived form $\ln(N/N_0) = -\lambda t$ is not printed. It is required by 15.7.B.2 and you take a logarithm to get it, which is exactly why the CED labels it derived rather than relevant. That matters for a graphing question: taking logs is what turns an exponential decay into a straight line of slope $-\lambda$ , and Topic 15.7 lists skill 1.B, create quantitative graphs with appropriate scales and units including plotting data, among its five suggested skills.

The full skill list for Topic 15.7 is 1.B, 2.B, 2.C, 3.A and 3.B. Topic 15.7 carries no boundary statement, checked on both of its printed pages. The relevant boundary is next door, in Topic 15.8: AP Physics 2 does not expect students to memorize the processes by which specific isotopes decay or the half-lives of specific isotopes. So a half-life is data a question supplies, never something to recall.

And two things that are absent. There is no equation for activity or decay rate anywhere in the framework or on the sheet: nothing of the form $A = \lambda N$ is printed, and the becquerel and the curie never appear. Nor does the phrase radiocarbon dating, or carbon dating, though 15.7.B.2 does cover finding the age of a material. Use the mechanism the framework gives and skip the vocabulary it does not.

Why the relation carries a ln 2, and not a 1

The relation is not an arbitrary constant of convenience. It falls out of the definition in one line.

Start from the decay law, $N = N_0 e^{-\lambda t}$ . Ask what time makes $N$ equal to half of $N_0$ , which is the definition in 15.7.B.1.ii. Set $N/N_0 = 1/2$ and $t = t_{1/2}$ :

\frac{1}{2} = e^{-\lambda t_{1/2}}

Take logarithms of both sides:

\ln\frac{1}{2} = -\lambda t_{1/2} \quad \Rightarrow \quad -\ln 2 = -\lambda t_{1/2}

\lambda t_{1/2} = \ln 2 \approx 0.693

So the two forms in 15.7.B.1.iii and 15.7.B.2 are not independent facts. Either one implies the other, and $\ln 2$ is simply what you get when the threshold you chose to name was one half.

Why $\lambda$ is not the fraction lost per unit time. This is the misconception that survives longest, and it is easiest to kill with the CED's own dice activity. Sample instructional activity 3 for Topic 15.7 gives each group 200 dice, has them shake the box each turn and remove every die showing a 1, and asks them to determine the "half-life" of the dice, which the CED states is about 3.8 turns.

A sixth of the dice go per turn, so a naive rate would be $\lambda = 1/6 = 0.167$ per turn, giving a half-life of $0.693/0.167 = 4.16$ turns. That is not the CED's answer.

The correct treatment tracks the survivors: five sixths remain each turn, so after $n$ turns the fraction left is $(5/6)^n$ . Setting that to one half,

n = \frac{\ln 2}{\ln(6/5)} = \frac{0.693}{0.182} = 3.80 \ \text{turns}

which is the $3.8$ the CED states. The effective decay constant is $\ln(6/5) = 0.182$ per turn, not $0.167$ , and the difference is exactly the gap between "the fraction that goes in one step" and "the constant in the exponent".

The reason for the gap is that decay is continuous while the fraction is counted at intervals. During a turn the population is already shrinking, so a constant per-nucleus probability removes slightly fewer dice than a constant fraction of the starting number would. For small rates the two nearly agree, which is why the shortcut goes unpunished until it does not.

One further consequence of the same algebra. Because $\lambda$ and $t_{1/2}$ are inversely related, doubling the decay constant halves the half-life, and multiplying a half-life by ten divides the decay constant by ten. If a comparison question moves one, it moves the other the opposite way by the same factor.

Two routes to the same answer, and when each is faster

Every population question in this topic can be attacked two ways. Both are supported by the framework, and knowing which to reach for saves real time.

Route A, count the halvings. Divide the elapsed time by the half-life to get the number of half-lives $n = t/t_{1/2}$ , then the surviving fraction is $2^{-n}$ . This is direct arithmetic from 15.7.B.1.ii and needs no decay constant at all.

Route B, use the exponential. Convert with $\lambda = \ln 2 / t_{1/2}$ , then evaluate $N = N_0 e^{-\lambda t}$ from 15.7.B.2.

The two are algebraically identical, since $e^{-\lambda t} = e^{-(\ln 2)t/t_{1/2}} = 2^{-t/t_{1/2}}$ . They cannot disagree except by rounding.

Situation	Faster route	Why
$t$ is a whole number of half-lives	A	$2^{-3}$ is one eighth, done in your head
$t$ is an awkward multiple	B	The exponential does not care whether $n$ is an integer
You need the time to reach a given fraction	B	Rearrange $\ln(N/N_0) = -\lambda t$ for $t$
You are asked for a symbolic expression	B	$\lambda$ appears in the printed equation
You are asked to plot data as a straight line	B	$\ln(N/N_0)$ against $t$ has slope $-\lambda$
The question gives $\lambda$ and asks for a half-life	B, reversed	$t_{1/2} = \ln 2/\lambda$

Route A has one genuine limitation worth stating plainly: it works perfectly well for non-integer $n$ , since $2^{-2.4}$ is a legitimate number, but it is only quicker when $n$ is a small integer. Beyond that the exponential is less error-prone because you are pressing the same two keys either way.

The linearised form deserves its own mention because of skill 1.B. Plot $\ln(N/N_0)$ on the vertical axis against $t$ on the horizontal and 15.7.B.2's derived equation says the graph is a straight line through the origin with gradient $-\lambda$ . That is how a decay constant is measured rather than quoted, and it is why the derived form is in the framework at all. Then $t_{1/2}$ follows from $\ln 2$ divided by the gradient's magnitude.

A half-life can also be read straight off an untransformed graph of $N$ against $t$ : find the time at which the curve has fallen to $N_0/2$ . That reading is available without any logarithm, which is one more reason the framework treats half-life as the accessible quantity and the decay constant as the computational one.

The property that catches people out: the clock never restarts

Essential knowledge 15.7.B.1.i is the statement doing the work here: the time at which an individual nucleus undergoes radioactive decay is indeterminable, but decay rates can be described using probability.

An individual nucleus has no memory and no schedule. It does not age toward a decay, and a nucleus that has sat undecayed for three half-lives is exactly as likely to decay in the next second as a freshly made one. What decays predictably is the population, because a large number of independent random events averages out.

The consequence is worth stating as a rule: any moment can be treated as the start. If a sample has $N_1$ nuclei at time $t_1$ , then at a later time $t_2$ it has

N_2 = N_1 e^{-\lambda (t_2 - t_1)}

with the same $\lambda$ , and the original $N_0$ plays no part. Only the elapsed interval matters.

The CED tests exactly this. Its published sample multiple-choice set includes a question that begins with a sample of decay constant $\lambda$ having $N_0$ nuclei at $t = 0$ , then $N_1$ nuclei at $t = t_1$ and $N_2$ at a later $t = t_2$ , and asks which expression correctly relates $N_2$ to $N_1$ . The distractors are the ones you would expect: an option with $t_2 + t_1$ in the exponent instead of $t_2 - t_1$ , and two options that drag $N_0$ into a relation that does not need it. The correct relation is the one above.

In half-life language the same property reads: after one half-life from any starting moment, half of whatever was there then is left. Take a sample with a $5.0$ year half-life. If $660$ nuclei remain at some instant, then $5.0$ years later $330$ remain, whatever happened before that instant and however many there were originally.

Two related traps follow from the same idea.

Two half-lives do not remove everything. Half then half again leaves a quarter, not zero. The exponential never reaches zero, which is why questions ask for the fraction remaining rather than the time until nothing is left.

Half-life is not an average lifetime. The framework defines it only as the time for half the nuclei to have decayed, and offers no other timescale. The mean lifetime $1/\lambda$ is not in the AP Physics 2 framework or on the sheet, so do not import it.

When it costs a mark

Writing $\lambda = 1/t_{1/2}$ . The $\ln 2$ is not optional. Dropping it makes $\lambda$ too large by a factor of $1/0.693$ , about 44 percent, and therefore every surviving fraction too small. Keep the product form $\lambda t_{1/2} = \ln 2$ in mind and the error cannot survive a units check.

Inverting the relation. $\lambda = \ln 2/t_{1/2}$ , not $t_{1/2}/\ln 2$ . A quick sanity test: a long half-life must give a small decay constant.

Substituting a half-life directly into $N = N_0 e^{-\lambda t}$ . The exponent needs the decay constant. Putting $t_{1/2}$ where $\lambda$ belongs also breaks the units, since the exponent must be dimensionless and time multiplied by time is not.

Mixing time units. If $\lambda$ is per year, $t$ must be in years. Essential knowledge 15.7.B.3 warns that half-lives span fractions of a second to billions of years, so unit mismatches here can be enormous and still look plausible.

Using $\log_{10}$ instead of $\ln$ . The relation and the derived equation both use the natural logarithm. Using base ten introduces a factor of about $2.303$ .

Adding times instead of subtracting them. Between two later instants the exponent is $-\lambda(t_2 - t_1)$ . This is the distractor the CED's own sample question offers, and it is the most tempting wrong answer on the page.

Assuming a decayed sample eventually reaches zero. $e^{-\lambda t}$ is never zero. Answer with a fraction, not an endpoint.

Treating $\lambda$ as a wavelength. In the Modern Physics group of the equation sheet, $\lambda$ is a wavelength in five equations and a decay constant in two, and the sheet's own symbol key reads "wavelength or decay constant". The ambiguity is printed rather than invented, so read the equation you are in.

Reaching for the mean lifetime, activity, or the becquerel. None of them appears in the AP Physics 2 framework or on the sheet. There is no printed relation of the form activity equals $\lambda N$ .

Quoting a real isotope's half-life from memory. The Topic 15.8 boundary statement says students are not expected to memorize the half-lives of specific isotopes, so a question that needs one must give it.

Rounding $\ln 2$ too early. Carrying $0.693$ is fine for three significant figures; carrying $0.7$ is not, and it drifts by about one percent in the exponent, which compounds over several half-lives.

When they feel interchangeable, and why

Almost everything about these two is shared, which is precisely why the distinction slips.

They describe the same material property. One number characterises how fast a given isotope decays, and $t_{1/2}$ and $\lambda$ are two labels for it. Neither carries information the other lacks.

They live under one essential knowledge branch. 15.7.B.1.ii and 15.7.B.1.iii are consecutive statements under the same objective.

Neither depends on the decay mode, on the sample size, or on anything about the sample's history.

Both are useless for a single nucleus. Essential knowledge 15.7.B.1.i makes the time of an individual decay indeterminable, so both quantities are statements about a population.

Both are constants for a given material, which is what the word constant in decay constant is telling you.

So the two feel interchangeable because, as descriptions, they are. The place they stop being interchangeable is in an equation, and that is the whole practical content of the comparison: $N = N_0 e^{-\lambda t}$ accepts only $\lambda$ , and a stem quoting a time accepts only $t_{1/2}$ . Every mark lost on this pair is lost in the gap between those two sentences.

There is one situation where the difference genuinely disappears, and it is worth recognising so you do not waste time converting. If a question asks only for a ratio at a whole number of half-lives, the decay constant never needs to appear: three half-lives leaves one eighth, and the arithmetic is done. Reach for $\lambda$ when the elapsed time is not a tidy multiple, when you need to solve for the time, when you need a symbolic expression, or when you need a straight-line graph.

The three questions that decide it:

Is the given number a time or a rate? A time in seconds or years is a half-life. A number per second or per year is a decay constant.
Does the equation I am about to use contain $\lambda$ ? If so, convert first.
Is the elapsed time a whole number of half-lives? If so, count halvings and skip the conversion entirely.

Where this sits on the AP exam

Learning objective 15.7.B lives in Topic 15.7, inside Unit 15, Modern Physics, which the CED weights at 12 to 15 percent of the multiple-choice section over a suggested 14 to 22 class periods, with a Progress Check listed as about 24 multiple-choice questions and 4 free-response questions.

Topic 15.7's five suggested skills are 1.B, create quantitative graphs with appropriate scales and units including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.A, create experimental procedures appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

That list reads like a description of this comparison. Skill 2.C names "different times in a single scenario", which is what the memoryless property is about. Skill 1.B is the linearised plot. Skill 3.A is why the CED's dice activity exists: it is an experimental procedure for measuring a half-life without any radioactive material.

The unit's exam guidance is relevant too. The CED says the first free-response question on the AP Physics 2 exam is the Mathematical Routines question, which focuses on creating and using mathematical models, and that while Unit 15 offers content well suited to it, the question can draw on any of the seven units. An exponential decay with a conversion step and a graph is close to a model of that question type.

The CED's published sample multiple-choice set contains a question on exactly this material, asking how the number of nuclei at one late time relates to the number at an earlier late time given the decay constant. Its correct answer uses the elapsed interval $t_2 - t_1$ , and its distractors test whether you know the initial count is irrelevant.

For the neighbouring material in the same unit, alpha vs beta decay covers the modes, which neither of these quantities depends on, and fission vs fusion covers the reactions treated under the other objective of the same topic, 15.7.A.

From half-life to decay constant, then the same answer by two routes

A radioactive material has a half-life of $5.0$ years. (a) Find its decay constant. (b) Find the fraction of the original nuclei remaining after $12$ years, using the exponential law. (c) Confirm the answer by counting half-lives. (d) State how many nuclei remain from an initial sample of $8.0 \times 10^{6}$ .

(a) Use 15.7.B.1.iii, $\lambda = \ln 2 / t_{1/2}$ , with $\ln 2 = 0.693$ to three figures.
$\lambda = \dfrac{0.693}{5.0 \ \mathrm{yr}} = 0.139 \ \mathrm{yr^{-1}}$ to three significant figures. Note the unit: an inverse time, because $\ln 2$ is a pure number.
Sanity check the size before going on. A half-life of $5.0$ years should give a decay constant of roughly a seventh per year, and $0.139$ is close to $1/7.2$ , so nothing has been inverted.
(b) Apply 15.7.B.2, $N = N_0 e^{-\lambda t}$ , with $t = 12 \ \mathrm{yr}$ . The exponent is $\lambda t = (0.13863)(12) = 1.6636$ , keeping an extra figure through the working.
$\dfrac{N}{N_0} = e^{-1.6636} = 0.189$ to three significant figures.
(c) Count halvings instead. Twelve years is $12/5.0 = 2.4$ half-lives, so the surviving fraction is $2^{-2.4}$ .
$2^{-2.4} = 0.189$ , the same to three figures. That agreement is guaranteed algebraically, since $e^{-\lambda t} = 2^{-t/t_{1/2}}$ once $\lambda = \ln 2/t_{1/2}$ is substituted, so a disagreement would mean an arithmetic slip rather than a real difference.
Bracket the answer for reassurance: $2.4$ half-lives is between $2$ and $3$ , so the fraction must lie between $0.25$ and $0.125$ . It does.
(d) $N = (8.0 \times 10^{6})(0.189) = 1.5 \times 10^{6}$ nuclei, to two significant figures, matching the two figures in the given sample size.

(a) $\lambda = 0.139 \ \mathrm{yr^{-1}}$ . (b) $N/N_0 = 0.189$ . (c) $2^{-2.4} = 0.189$ , identical. (d) About $1.5 \times 10^{6}$ nuclei remain.

Running the decay law backwards to find a time

The same material, with $t_{1/2} = 5.0$ years and $\lambda = 0.1386 \ \mathrm{yr^{-1}}$ . (a) Find how long it takes for the sample to fall to $10.0$ percent of its original number. (b) Check the answer in half-lives. (c) A second sample of the same material is found to contain $660$ nuclei at one instant. Find how many remain $5.0$ years later, and explain what information was not needed.

(a) Start from the derived equation named in 15.7.B.2, $\ln(N/N_0) = -\lambda t$ , and solve for $t$ : $t = -\ln(N/N_0)/\lambda$ .
With $N/N_0 = 0.100$ : $\ln(0.100) = -2.3026$ , so $t = 2.3026/0.1386$ .
$t = 16.6$ years to three significant figures.
(b) In half-lives that is $16.6/5.0 = 3.32$ half-lives. Check independently: falling to one tenth requires $2^{-n} = 0.100$ , so $n = \log_2 10 = 3.3219$ , and $3.3219 \times 5.0 = 16.6$ years. The two agree.
A rough bracket confirms it without a calculator: three half-lives leaves an eighth, which is $0.125$ , and four leaves a sixteenth, which is $0.0625$ . Ten percent sits between those, so the time must sit between $15$ and $20$ years.
(c) Essential knowledge 15.7.B.1.i makes each nucleus's decay independent of its history, so any instant may be treated as a fresh start and only the elapsed interval matters: $N_2 = N_1 e^{-\lambda (t_2 - t_1)}$ .
The interval here is exactly one half-life, so half remain: $660/2 = 330$ nuclei.
What was not needed: the original number of nuclei, and how long the sample had already been decaying. Neither appears in the relation, which is the point the CED's sample multiple-choice question on this material is built to test.

(a) About $16.6$ years. (b) $3.32$ half-lives, which agrees with $\log_2 10 = 3.32$ . (c) $330$ nuclei remain, and neither the original count nor the sample's prior history was required.

The CED's dice experiment, and why the decay constant is not one sixth

In sample instructional activity 3 for Topic 15.7, groups shake a box of $200$ dice each turn and remove every die showing a 1, repeating until all are removed and graphing dice against turns. The CED states the resulting "half-life" is about $3.8$ turns. (a) Derive that value. (b) Show that treating the decay constant as the fraction removed per turn gives the wrong answer. (c) Explain what the discrepancy illustrates about $\lambda$ .

(a) Track survivors rather than losses. Each die has a $1/6$ chance of being removed on a turn, so a $5/6$ chance of surviving it, and the turns are independent. After $n$ turns the expected surviving fraction is $(5/6)^n$ .
Set that to one half, matching the definition in 15.7.B.1.ii: $(5/6)^n = 1/2$ .
Take logarithms: $n \ln(5/6) = \ln(1/2)$ , so $n = \dfrac{\ln 2}{\ln(6/5)}$ .
Evaluate: $\ln 2 = 0.6931$ and $\ln(6/5) = 0.1823$ , so $n = 0.6931/0.1823 = 3.80$ turns. That is the CED's stated value of about $3.8$ .
Check it forwards: $(5/6)^{3.8} = 0.500$ , so $3.8$ turns does halve the population, and of $200$ dice about $100$ would remain.
(b) The naive route sets $\lambda = 1/6 = 0.1667$ per turn, the fraction removed. Then $t_{1/2} = \ln 2/\lambda = 0.6931/0.1667 = 4.16$ turns.
That is about $9$ percent longer than $3.80$ and does not match the CED's figure, so the naive identification of $\lambda$ with the per-turn fraction is wrong.
The correct effective decay constant is $\lambda = \ln(6/5) = 0.182$ per turn, which is larger than $1/6$ , and feeding it into $t_{1/2} = \ln 2/\lambda$ returns $3.80$ turns as it must.
(c) The decay constant is the constant in an exponent, not the fraction lost per interval. Real decay is continuous, so the population is already shrinking during an interval and a constant per-nucleus probability removes slightly fewer than a fixed share of the starting count. The two converge for small rates, which is why the shortcut usually goes unnoticed, and this activity is a case where it does not.

(a) $n = \ln 2 / \ln(6/5) = 3.80$ turns, matching the CED's stated $3.8$ . (b) Setting $\lambda = 1/6$ gives $4.16$ turns, which is wrong. (c) That $\lambda$ is the exponent constant, here $\ln(6/5) = 0.182$ per turn, and not the fraction removed per interval.

Frequently asked questions

What is the difference between half-life and the decay constant?

Half-life is a time and the decay constant is an inverse time, and they carry the same information about a material. AP Physics 2 essential knowledge 15.7.B.1.ii defines the half-life as the time it takes for half of the initial number of radioactive nuclei to have spontaneously decayed. Essential knowledge 15.7.B.1.iii gives the link: the decay constant equals the natural logarithm of 2 divided by the half-life. Because the relation is inverse, a long half-life means a small decay constant. In practice a question gives you the half-life, while the exponential equation for the number of nuclei remaining needs the decay constant, so converting between them is a routine step.

What is the formula relating half-life and the decay constant?

The decay constant equals the natural logarithm of 2 divided by the half-life. AP Physics 2 states it as essential knowledge 15.7.B.1.iii and prints it in the Modern Physics group of the equation sheet. Since the natural logarithm of 2 is about 0.693, a half-life of 5.0 years gives a decay constant of 0.139 per year. The relation is easiest to remember as a product: the decay constant multiplied by the half-life always equals the natural logarithm of 2, which makes it impossible to write the fraction upside down.

Does AP Physics 2 require the exponential decay equation?

Yes. Essential knowledge 15.7.B.2 states that a material's decay constant may be used to predict the number of nuclei remaining in a sample after a period of time, or the age of a material if the initial amount is known, and gives N equals N naught times e to the minus lambda t as the relevant equation. It also lists the natural logarithm of N over N naught equals minus lambda t as a derived equation. The exponential form and the half-life relation are both printed on the AP Physics 2 equation sheet; the logarithmic form is not printed, which is why the CED labels it derived.

Why is the decay constant not just one over the half-life?

Because the half-life is defined by the population falling to one half, and solving the exponential decay law for that condition introduces a natural logarithm of 2. Setting N over N naught equal to one half in N equals N naught e to the minus lambda t and taking logarithms gives lambda times the half-life equals the natural logarithm of 2, which is about 0.693 rather than 1. Dropping that factor makes every decay constant too large by a factor of 1 over 0.693, which is about 44 percent too big, and every predicted surviving fraction correspondingly too small.

Do you have to know the half-lives of specific isotopes for AP Physics 2?

No. The Topic 15.8 boundary statement says AP Physics 2 does not expect students to memorize the processes by which specific isotopes decay or the half-lives of specific isotopes, so any half-life a question needs will be supplied. What the framework does state, at essential knowledge 15.7.B.3, is that different unstable elements and isotopes may have vastly different half-lives, ranging from fractions of a second to billions of years. That range is worth carrying because it is a warning about time units: a mismatch between seconds and years in an exponent produces an answer that is wrong by an enormous factor while still looking reasonable.

If a sample is halfway through a half-life, does the clock restart when you start measuring?

There is no clock to restart. AP Physics 2 essential knowledge 15.7.B.1.i says the time at which an individual nucleus undergoes radioactive decay is indeterminable, but decay rates can be described using probability, so a nucleus does not age toward its decay. Any instant can therefore be treated as a starting point: if a sample has N one nuclei at one time, then at a later time it has N one multiplied by e to the minus lambda times the elapsed interval. The original number and the sample's prior history play no part. The CED's own published sample multiple-choice set tests exactly this, with distractors that add the two times instead of subtracting them.

Is there an equation for activity or decay rate on the AP Physics 2 equation sheet?

No. The Modern Physics group of the AP Physics 2 equation sheet prints the exponential decay law and the half-life relation, and nothing of the form activity equals the decay constant multiplied by the number of nuclei. The units becquerel and curie do not appear anywhere in the course and exam description either. What the framework does give you, at essential knowledge 15.7.B.2, is the number of nuclei remaining and the age of a material if the initial amount is known. Work with numbers of nuclei rather than with activity, since numbers of nuclei are what the printed equation returns.