During nuclear explosion, one of the products is $^{90}Sr$ with half-life of 28.1 years. If 1 μg of $^{90}Sr$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

> [!TLDR] > Radioactive decay follows first-order kinetics — the fraction remaining depends only on the half-life and elapsed time. After 10 years, about **0.78 μg** remains; after 60 years, about **0.23 μg** remains. ## Why this works: the idea behind radioactive dating Radioactive decay is a random, spontaneous process at the atomic level. For a large sample, the number of atoms that decay per unit time is proportional to the number present — that's first-order kinetics. The half-life ($t_{1/2}$) is the time after which exactly half the original atoms remain, regardless of the starting amount. This means we don't need to know the initial number of atoms — only the initial mass and the half-life. Since $^{90}\text{Sr}$ is chemically similar to calcium, it gets incorporated into bone. The problem says "if it is not lost metabolically", so the only removal mechanism is radioactive decay. > [!FORMULA] > $$N = N_0 \left(\frac{1}{2}\right)^{t / t_{1/2}} \quad \text{or} \quad m = m_0 \left(\frac{1}{2}\right)^{t / t_{1/2}}$$ where $m_0$ is the initial mass, $t_{1/2}$ is the half-life, and $t$ is the elapsed time. --- ## Step-by-step calculation **1. Identify the given data** - Initial mass: $m_0 = 1\ \mu\text{g}$ - Half-life: $t_{1/2} = 28.1\ \text{years}$ - Times of interest: $t_1 = 10\ \text{years}$, $t_2 = 60\ \text{years}$ **2. Find the decay constant (optional but clarifying)** The decay constant $\lambda$ relates to half-life by: $$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{28.1\ \text{yr}} \approx 0.02466\ \text{yr}^{-1}$$ We could use the exponential form $m = m_0 e^{-\lambda t}$, but the half-life formula is more direct here. **3. Apply the half-life formula for $t = 10$ years** The number of half-lives elapsed is: $$\frac{t}{t_{1/2}} = \frac{10}{28.1} \approx 0.3559$$ So the remaining fraction is: $$\left(\frac{1}{2}\right)^{0.3559} = 2^{-0.3559}$$ To evaluate this, take logs: $$\log_{10}(2^{-0.3559}) = -0.3559 \times \log_{10}2 \approx -0.35

During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 μg of ⁹⁰Sr was absorb

Exercises · 3.17

^{90}Sr

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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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Radioactive decay follows first-order kinetics — the fraction remaining depends only on the half-life and elapsed time. After 10 years, about 0.78 μg remains; after 60 years, about 0.23 μg remains.

Why this works: the idea behind radioactive dating

Radioactive decay is a random, spontaneous process at the atomic level. For a large sample, the number of atoms that decay per unit time is proportional to the number present — that's first-order kinetics. The half-life ( $t_{1/2}$ ) is the time after which exactly half the original atoms remain, regardless of the starting amount.

This means we don't need to know the initial number of atoms — only the initial mass and the half-life. Since $^{90}\text{Sr}$ is chemically similar to calcium, it gets incorporated into bone. The problem says "if it is not lost metabolically", so the only removal mechanism is radioactive decay.

N = N_0 \left(\frac{1}{2}\right)^{t / t_{1/2}} \quad \text{or} \quad m = m_0 \left(\frac{1}{2}\right)^{t / t_{1/2}}

where $m_0$ is the initial mass, $t_{1/2}$ is the half-life, and $t$ is the elapsed time.

Step-by-step calculation

1. Identify the given data

Initial mass: $m_0 = 1\ \mu\text{g}$
Half-life: $t_{1/2} = 28.1\ \text{years}$
Times of interest: $t_1 = 10\ \text{years}$ , $t_2 = 60\ \text{years}$

2. Find the decay constant (optional but clarifying)

The decay constant $\lambda$ relates to half-life by:

\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{28.1\ \text{yr}} \approx 0.02466\ \text{yr}^{-1}

We could use the exponential form $m = m_0 e^{-\lambda t}$ , but the half-life formula is more direct here.

3. Apply the half-life formula for $t = 10$ years

The number of half-lives elapsed is:

\frac{t}{t_{1/2}} = \frac{10}{28.1} \approx 0.3559

So the remaining fraction is:

\left(\frac{1}{2}\right)^{0.3559} = 2^{-0.3559}

To evaluate this, take logs:

\log_{10}(2^{-0.3559}) = -0.3559 \times \log_{10}2 \approx -0.3559 \times 0.3010 = -0.1071

Thus $2^{-0.3559} = 10^{-0.1071} \approx 0.781$

Therefore:

m_{10} = 1\ \mu\text{g} \times 0.781 \approx 0.78\ \mu\text{g}

Tip

You can also compute $2^{-0.3559}$ directly on a calculator as $e^{-0.3559 \ln 2} \approx e^{-0.2467} \approx 0.781$ . Either way, the result is the same. …

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Here are the common mistakes students make on this exact radioactive decay problem, along with how to avoid each one.

Mistake 1: Using the wrong formula (Linear instead of Exponential)

The error:

Students treat decay as a linear process, e.g., "half-life is 28.1 years, so after 28.1 years half remains, after 56.2 years one-quarter remains." Then they try to interpolate linearly for 10 years and 60 years.

This gives a wrong answer because decay is exponential, not linear.

How to avoid:

Always use the exponential decay formula:

N = N_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}

where:

$N_0$ = initial amount
$t$ = time elapsed
$T_{1/2}$ = half-life

Key insight: The fraction remaining after time $t$ is $2^{-t/T_{1/2}}$ , not $1 - \frac{t}{T_{1/2}}$ .

Mistake 2: Forgetting to convert units (μg to g or vice versa)

The error:

The problem gives $1\ \mu\text{g}$ but students sometimes convert to grams incorrectly (e.g., $1\ \mu\text{g} = 10^{-6}\ \text{g}$ is correct, but they might write $10^{-9}\ \text{g}$ or forget to convert at all).

How to avoid:

Remember: $1\ \mu\text{g} = 10^{-6}\ \text{g}$
Keep the unit consistent throughout. Since the answer is asked in μg, stay in μg — no conversion needed.
Only convert if the final answer must be in grams.

Correct approach:

Initial amount $N_0 = 1\ \mu\text{g}$ . Final answer will also be in μg.

Mistake 3: Misinterpreting "not lost metabolically"

The error:

Some students think "not lost metabolically" means the decay rate changes or that the substance is removed by the body. They then try to add an extra decay constant or subtract some amount.

How to avoid:

The phrase "if it is not lost metabolically" simply means:

The only way the amount decreases is through radioactive decay — no biological excretion.

So you only use the radioactive half-life. No extra terms.

Mistake 4: Rounding off too early in intermediate steps

The error:

Students compute $2^{-10/28.1}$ and round to 0.78, then multiply by 1 μg to get 0.78 μg. But the exact value is needed for the 60-year case, and early rounding compounds errors.

How to avoid:

Keep at least 4 significant figures in intermediate calculations.
Use the exact exponent: $t/T_{1/2}$ as a fraction, not a rounded decimal.
For 60 years: $60/28.1 \approx 2.1359$ , not 2.1.

Example of correct calculation:

For $t = 10$ years:

\frac{t}{T_{1/2}} = \frac{10}{28.1} \approx 0.3559

N = 1 \times 2^{-0.3559} \approx 0.781\ \mu\text{g}

For $t = 60$ years:

\frac{t}{T_{1/2}} = \frac{60}{28.1} \approx 2.1359

N = 1 \times 2^{-2.1359} = 0.2276 \approx 0.228\ \mu\text{g}

Mistake 5: Using natural log formula incorrectly

The error: …

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Q.During nuclear explosion, one of the products is $^{90}Sr$ with half-life of 28.1 years. If 1 μg of $^{90}Sr$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

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 this works: the idea behind radioactive dating

Step-by-step calculation

Method: Exponential Decay Law (First-Order Kinetics)

Steps

Mistake 1: Using the wrong formula (Linear instead of Exponential)

Mistake 2: Forgetting to convert units (μg to g or vice versa)

Mistake 3: Misinterpreting "not lost metabolically"

Mistake 4: Rounding off too early in intermediate steps

Mistake 5: Using natural log formula incorrectly

Q.During nuclear explosion, one of the products is 90Sr^{90}Sr90Sr with half-life of 28.1 years. If 1 μg of 90Sr^{90}Sr90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Why this works: the idea behind radioactive dating

Step-by-step calculation

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Q.During nuclear explosion, one of the products is $^{90}Sr$ with half-life of 28.1 years. If 1 μg of $^{90}Sr$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Unlock everything free for 14 days