The half-life for radioactive decay of $^{14}C$ is 5730 years. An archaeological artifact containing wood had only 80% of the $^{14}C$ found in a living tree. Estimate the age of the sample.

The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only

Exercises · 3.14

^{14}C

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Concept understanding — Radioactive Decay Dating

Radioactive Decay Dating

Imagine you have a campfire that's been burning all night. When you wake up, you see the pile of ash and a few remaining embers. If you know how fast wood burns, you could look at the ratio of ash to unburnt wood and figure out roughly when the fire was lit. That's the core idea behind radioactive decay dating — except instead of wood turning to ash, we have unstable atoms turning into stable ones, and the "burn rate" is incredibly precise and constant.

The Intuition

Some atomic nuclei are unstable. They spontaneously transform into a different nucleus by emitting radiation — this is radioactive decay. The original unstable atom is called the parent, and the atom it becomes is the daughter.

Here's the key: every radioactive substance has a fixed half-life — the time it takes for exactly half of any sample to decay. If you start with 1000 parent atoms, after one half-life you'll have 500 parents and 500 daughters. After two half-lives, 250 parents and 750 daughters. After three, 125 parents and 875 daughters.

This is not a guess. It's a statistical certainty for large numbers of atoms. The half-life of carbon-14 is 5,730 years. The half-life of uranium-238 is 4.5 billion years. These numbers never change — not by heat, pressure, or chemical reactions.

So if you measure how much parent is left and how much daughter has formed, you can calculate how many half-lives have passed. Multiply by the half-life, and you get the age.

The Precise Statement

Radioactive decay follows first-order kinetics. The rate of decay at any instant is proportional to the number of parent atoms present:

-\frac{dN}{dt} = \lambda N

where $N$ is the number of parent atoms, $t$ is time, and $\lambda$ is the decay constant — a unique number for each radioactive isotope.

Solving this differential equation gives the exponential decay law:

N(t) = N_0 e^{-\lambda t}

where $N_0$ is the number of parent atoms at time $t = 0$ .

The half-life $t_{1/2}$ is related to $\lambda$ by:

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

t = \frac{1}{\lambda} \ln\left(1 + \frac{D}{P}\right)

This formula assumes no daughter atoms were present initially and none have been lost or added — a critical assumption we'll come back to.

How It's Actually Done

You can't count individual atoms in a rock. Instead, scientists measure the ratio of parent to daughter isotopes using a mass spectrometer. They also need to know the initial amount of daughter — often zero, but sometimes they use a trick called an isochron plot to figure it out.

Note

For carbon-14 dating, the "initial" amount of carbon-14 in a living organism is assumed constant because it's constantly replenished from the atmosphere. Once the organism dies, the carbon-14 decays with no new intake — that's when the clock starts.

The Three Big Assumptions

Every dating method rests on three assumptions. If any fails, the date is wrong.

The initial condition is known. You must know how much daughter was present at $t=0$ . For carbon-14, this means assuming the atmospheric ratio of $^{14}\text{C}$ to $^{12}\text{C}$ has been constant. (It hasn't been perfectly constant — that's why calibration curves exist.)
The system has been closed. No parent or daughter atoms have entered or left the sample since formation. If groundwater leached out uranium, or if heat drove off argon, the calculated age will be wrong.
The decay rate has been constant. This is on extremely solid ground — experiments and observations of supernova light curves confirm that half-lives haven't changed over billions of years.

Common Methods at a Glance

Method	Parent → Daughter	Half-Life	Dating Range	What It Dates
Carbon-14	$^{14}\text{C} \rightarrow ^{14}\text{N}$	5,730 years	Up to ~50,000 years	Organic remains
Potassium-argon	$^{40}\text{K} \rightarrow ^{40}\text{Ar}$	1.25 billion years	>100,000 years	Volcanic rocks

Why this formula?

Radioactive Decay Dating: Why the Formula Works

Radioactive decay dating (like carbon-14 dating) is based on a simple but profound observation: radioactive nuclei decay at a rate proportional to how many are left. This is not an assumption — it's a statistical law that emerges from quantum mechanics. Let's build the reasoning step by step.

1. The Core Idea: Exponential Decay

Imagine you have a large number $N$ of identical radioactive atoms. Each atom has a constant probability per unit time of decaying — call this decay constant $\lambda$ (units: $\text{time}^{-1}$ ).

If $\lambda = 0.1 \, \text{year}^{-1}$ , each atom has a 10% chance of decaying in any given year.
This probability does not change with the atom's age — no "memory" effect.

Why "proportional to $N$ "?

If you have $N$ atoms, the average number decaying in a small time $dt$ is:

\text{decays in } dt = \lambda N \, dt

This is a first-order rate law — the same form as in chemical kinetics for a unimolecular reaction. The minus sign appears because $N$ decreases:

\frac{dN}{dt} = -\lambda N

2. Solving the Differential Equation

This is a simple separable ODE:

\frac{dN}{N} = -\lambda \, dt

Integrate both sides:

\int_{N_0}^{N(t)} \frac{dN}{N} = -\lambda \int_0^t dt

\ln N(t) - \ln N_0 = -\lambda t

Exponentiate:

\boxed{N(t) = N_0 e^{-\lambda t}}

Key insight: The decay is exponential, not linear. After one half-life, half the atoms remain; after two half-lives, one-quarter remain — not zero.

3. Half-Life: A More Intuitive Constant

Define the half-life $t_{1/2}$ as the time when $N = N_0/2$ :

\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}}

Cancel $N_0$ :

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

Take natural log:

\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2} \quad \Rightarrow \quad -\ln 2 = -\lambda t_{1/2}

Thus:

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

Why this matters: Half-lives are measured experimentally (e.g., carbon-14: 5730 years). From $t_{1/2}$ , we get $\lambda$ , and then we can date samples.

4. The Dating Formula: From $N(t)$ to Age

In dating, we measure the current number of parent atoms $N$ and compare it to the initial number $N_0$ . But $N_0$ is often unknown — so we use the daughter product $D$ (the stable atom the parent decays into).

Conservation of atoms:

\text{Total atoms} = N(t) + D(t) = N_0

So:

N_0 = N(t) + D(t)

Plug into the decay law:

N(t) = (N(t) + D(t)) \, e^{-\lambda t}

Solve for $t$ :

\frac{N(t)}{N(t) + D(t)} = e^{-\lambda t}

Take natural log:

\ln\left( \frac{N(t)}{N(t) + D(t)} \right) = -\lambda t

Thus:

$\boxed{t = \frac{1}{\lambda} \ln\left( \frac{N(t) + D(t)}{N(t)} \right)}$ …

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Using the radioactive decay law and the known half-life of carbon-14 (5730 years), the age of the artifact that retains 80% of its original $^{14}C$ is approximately 1845 years.

Why Carbon Dating Works

Living trees constantly exchange carbon with the atmosphere, so the ratio of $^{14}C$ to stable carbon stays constant. Once the tree is cut and becomes an artifact, that exchange stops. The $^{14}C$ decays away at a fixed rate — its half-life is 5730 years. By measuring how much $^{14}C$ remains compared to a living tree, we can calculate how long ago the tree died.

The decay follows first-order kinetics: the number of radioactive nuclei decreases exponentially with time. The key relationship is:

N = N_0 e^{-\lambda t}

where $N_0$ is the initial number of $^{14}C$ atoms, $N$ is the number remaining after time $t$ , and $\lambda$ is the decay constant.

The half-life $t_{1/2}$ is related to $\lambda$ by:

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

Step-by-step solution

1. Identify what we know

Half-life: $t_{1/2} = 5730$ years
Fraction remaining: $\frac{N}{N_0} = 80\% = 0.80$
We need to find $t$ , the age of the sample.

2. Find the decay constant $\lambda$

From the half-life formula:

\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{5730 \text{ years}}

\lambda \approx 1.2097 \times 10^{-4} \text{ year}^{-1}

3. Apply the decay law

We have $N = N_0 e^{-\lambda t}$ , so:

\frac{N}{N_0} = e^{-\lambda t}

Substitute the fraction:

0.80 = e^{-\lambda t}

4. Solve for $t$

Take natural logarithms on both sides:

\ln(0.80) = -\lambda t

t = -\frac{\ln(0.80)}{\lambda}

Now $\ln(0.80) = \ln\left(\frac{4}{5}\right) = \ln 4 - \ln 5 \approx 1.3863 - 1.6094 = -0.2231$

So:

t = -\frac{-0.2231}{1.2097 \times 10^{-4}}

$t = \frac{0.2231}{1.2097 \times 10^{-4}}$ …

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Here are the most common mistakes students make when solving radioactive decay dating problems like this one, along with how to avoid each.

Mistake 1: Confusing Half-Life with the Decay Constant ( $\lambda$ )

The Error: Students often plug the half-life ( $T_{1/2}$ ) directly into the decay equation as if it were the decay constant ( $\lambda$ ). For example, they might write $N = N_0 e^{-t / T_{1/2}}$ instead of $N = N_0 e^{-\lambda t}$ .
Why it happens: The formula $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$ is correct, but when switching to the exponential form $N = N_0 e^{-\lambda t}$ , students forget that $\lambda$ and $T_{1/2}$ are inversely related.
How to avoid: Always write the two key relationships side-by-side before solving:
- Relation 1: $\lambda = \frac{\ln 2}{T_{1/2}}$
- Relation 2: $N = N_0 e^{-\lambda t}$
- Step 1: Calculate $\lambda$ from the given half-life.
- Step 2: Use that $\lambda$ in the exponential decay equation.

Mistake 2: Misinterpreting "80% of the $^{14}C$ "

The Error: Students set up the ratio incorrectly. They might write $N = 0.8 \times N_0$ (which is correct) but then think the fraction remaining is 0.2 (20%) instead of 0.8 (80%).
Why it happens: The phrase "only 80% of the $^{14}C$ found in a living tree" means the sample has 80% left, not that 80% has decayed.
How to avoid: Read the problem statement carefully and write it as a mathematical statement:
- "Only 80% of the original" $\implies \frac{N}{N_0} = 0.8$
- Key check: If the problem said "lost 20%," then $\frac{N}{N_0} = 0.8$ as well. If it said "only 20% remains," then $\frac{N}{N_0} = 0.2$ . Always ask: What fraction is left?

Mistake 3: Forgetting to Take the Natural Log

The Error: After substituting into $N = N_0 e^{-\lambda t}$ , students try to solve for $t$ without using $\ln$ . They might write $0.8 = e^{-\lambda t}$ and then incorrectly cancel the $e$ or treat it as a simple algebraic term.
Why it happens: The exponential function is not linear; you cannot just "move" the $e$ to the other side.
How to avoid: Whenever you see $e^{\text{something}}$ $e^{something}$ , your next step is to apply the natural logarithm ( $\ln$ $ln$ ) to both sides of the equation.
- $0.8 = e^{-\lambda t}$
- $\ln(0.8) = -\lambda t$
- Then solve: $t = -\frac{\ln(0.8)}{\lambda}$

Mistake 4: Using the Wrong Base for the Logarithm

The Error: Students use $\log_{10}$ (common log) instead of $\ln$ (natural log) when solving $e^{-\lambda t}$ .
Why it happens: Calculators have both buttons, and students grab the first one they see.
How to avoid: Remember the rule: $e$ and $\ln$ are inverses. If the equation has $e$ , use $\ln$ . If the equation has $10^x$ , use $\log_{10}$ . For radioactive decay, it's almost always $\ln$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2024Set B-21 markMCQ
$3\, \text{A}$
›Reveal solutionSolution
Using Faraday’s laws of electrolysis, the charge passed is calculated from current and time, then converted to moles of electrons, and finally to mass of calcium. The mass deposited is 4.0 g.

The problem is a direct application of Faraday’s laws of electrolysis. When a current flows through a molten salt, the metal ions (here Ca²⁺) are reduced at the cathode. The key idea: the amount of substance deposited is proportional to the total charge passed, and the proportionality constant involves the molar mass and the number of electrons per ion.

Calcium in a molten salt exists as Ca²⁺ ions. Each Ca²⁺ ion requires 2 electrons to become neutral calcium metal:

$\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}$

So the number of moles of calcium deposited equals half the number of moles of electrons passed.

Let’s work through the calculation step by step.

Convert the total time into seconds.

Time given: 1 hour 47 minutes 13 seconds.

1 hour = $60 \times 60 = 3600$ s

47 minutes = $47 \times 60 = 2820$ s

Plus 13 seconds Total time $t = 3600 + 2820 + 13 = 6433$ s.

Calculate the total charge passed.

Current $I = 3$ A. Charge $Q = I \times t$

$Q = 3 \times 6433 = 19299$ C.

Find the number of moles of electrons.

Faraday constant $F = 96500$ C mol⁻¹ (the charge of one mole of electrons).

Moles of electrons $n_e = \dfrac{Q}{F} = \dfrac{19299}{96500}$

Compute: $19299 \div 96500 \approx 0.2$ (exactly $0.2$ ? Let’s check: $96500 \times 0.2 = 19300$ , very close to 19299 — the difference is just 1 C, negligible for our purpose).

So $n_e \approx 0.2$ mol. …
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KCET 2022Set B-31 markMCQ
$a^{n-1}$
›Reveal solutionSolution

Integrate the $n^{\text{th}}$ -order rate law to get $t_{1/2}\propto \dfrac{1}{a^{\,n-1}} = a^{\,1-n}$ .

Step 1 — Write the rate law and integrate it.

For a reaction of order $n$ ( $n\neq 1$ ) with initial concentration $a$ ,
$-\frac{d[A]}{dt}=k[A]^n\;\Longrightarrow\;\int_a^{[A]}\frac{d[A]}{[A]^n}=-k\int_0^t dt,$ $\frac{1}{n-1}\left[\frac{1}{[A]^{\,n-1}}-\frac{1}{a^{\,n-1}}\right]=kt .$
Step 2 — Put in the half-life condition.

At $t=t_{1/2}$ , $[A]=a/2$ :
$\frac{1}{n-1}\left[\frac{2^{\,n-1}}{a^{\,n-1}}-\frac{1}{a^{\,n-1}}\right]=k\,t_{1/2}$ $\boxed{\,t_{1/2}=\frac{2^{\,n-1}-1}{k\,(n-1)}\cdot\frac{1}{a^{\,n-1}}\,}$
Step 3 — Read off the dependence on $a$ .

Everything except $a$ is a constant for a given reaction, so

$t_{1/2}\;\propto\;\frac{1}{a^{\,n-1}}\;=\;a^{\,1-n}.$ …
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COMEDK 2022Set 20221 markMCQ
$^{-1}$
›Reveal solutionSolution

t(1/2) = 0.693 / 6.93 = 0.1 s

Concept: Radioactive decay is first order. Half-life and decay constant are related by

t(1/2) = ln 2 / k = 0.693 / k

Given k = 6.93 s^-1: …
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Q.The half-life for radioactive decay of $^{14}C$ is 5730 years. An archaeological artifact containing wood had only 80% of the $^{14}C$ found in a living tree. Estimate the age of the sample.

Concept understanding — Radioactive Decay Dating

Radioactive Decay Dating

The Intuition

The Precise Statement

How It's Actually Done

The Three Big Assumptions

Common Methods at a Glance

Why this formula?

Radioactive Decay Dating: Why the Formula Works

1. The Core Idea: Exponential Decay

Why "proportional to $N$ "?

2. Solving the Differential Equation

3. Half-Life: A More Intuitive Constant

4. The Dating Formula: From $N(t)$ to Age

Conservation of atoms:

Why Carbon Dating Works

Step-by-step solution

Method: First-Order Decay Law (Exponential Decay Method)

Steps

Mistake 1: Confusing Half-Life with the Decay Constant ( $\lambda$ )

Mistake 2: Misinterpreting "80% of the $^{14}C$ "

Mistake 3: Forgetting to Take the Natural Log

Mistake 4: Using the Wrong Base for the Logarithm

Q.The half-life for radioactive decay of 14C^{14}C14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C^{14}C14C found in a living tree. Estimate the age of the sample.

Why Carbon Dating Works

Step-by-step solution

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Q.The half-life for radioactive decay of $^{14}C$ is 5730 years. An archaeological artifact containing wood had only 80% of the $^{14}C$ found in a living tree. Estimate the age of the sample.

Unlock everything free for 14 days