How to Calculate Remaining Radioactive Nuclei After Fractional Half-Lives

A radioactive sample starts with \(1.20\times10^{15}\) undecayed nuclei. Its half-life is \(8.0\ \text{h}\). After \(12\ \text{h}\), how many radioactive nuclei remain?

A tempting answer is \(3.00\times10^{14}\), because you might halve the sample twice. But \(12\ \text{h}\) represents only 1.5 half-lives, not two complete half-lives.

The correct answer is approximately:

\[ N=1.20\times10^{15}\times 2^{-1.5} \]

\[ \boxed{N=4.24\times10^{14}\text{ nuclei}} \]

The key is knowing what to do with the fractional half-life. You cannot simply round 1.5 down to 1 or up to 2. You must use the exponential decay relationship.

Before working through the method, make a prediction: after 1.5 half-lives, does 25%, 35%, or 50% of the original sample remain?

About 35.4% remains. The reason is that radioactive decay continues throughout the half-life, rather than occurring in sudden batches at the end of each interval.

01Calculating the remaining nuclei

For a radioactive isotope, the number of undecayed nuclei decreases exponentially.

The relationship is:

\[ N=N_0\left(\frac12\right)^n \]

where:

• \(N\) is the expected number of radioactive nuclei remaining.
• \(N_0\) is the initial number of radioactive nuclei.
• \(n\) is the number of half-lives elapsed.

To calculate \(n\), divide the elapsed time by the half-life:

\[ n=\frac{t}{T_{1/2}} \]

Here, \(t\) is the elapsed time and \(T_{1/2}\) is the half-life.

Combining these expressions gives the most useful equation for this type of question:

\[ \boxed{N=N_0\left(\frac12\right)^{t/T_{1/2}}} \]

This works whether the elapsed time represents one half-life, three half-lives, or any fraction in between.

Why does a fractional half-life work?

Imagine that a sample has a half-life of 8 hours.

After 8 hours, half the original radioactive nuclei are expected to remain.

What happens after another 4 hours?

It would be tempting to halve the remaining amount again, leaving 25% of the original sample. However, 4 hours is only half of a half-life.

To understand the correct result, consider two equal 4-hour intervals.

Suppose the fraction remaining after each 4-hour interval is \(r\).

After two such intervals, 8 hours have passed, so the remaining fraction must be \(1/2\).

Therefore:

\[ r\times r=\frac12 \]

\[ r=\sqrt{\frac12} \]

\[ r\approx0.7071 \]

So after half a half-life, approximately 70.7% of the current radioactive nuclei remain.

After 1.5 half-lives, the sample has undergone one complete halving followed by this additional reduction:

\[ \text{Fraction remaining} =\frac12\times0.7071 \]

\[ =0.3536 \]

Therefore, approximately 35.4% of the original radioactive nuclei remain.

Useful fractional half-life values

You can use the following table to check whether your calculator result makes physical sense.

Notice that the remaining percentage does not decrease in equal steps.

For example, the difference between 50% and 35.4% is 14.6 percentage points. The difference between 35.4% and 25% is 10.4 percentage points.

This is why linear interpolation is also incorrect. Radioactive decay follows an exponential relationship, not a straight line.

A calculator method that works every time

For an HSC calculation, follow these steps:

1. Identify the initial number of radioactive nuclei, \(N_0\).
2. Identify the elapsed time, \(t\), and half-life, \(T_{1/2}\).
3. Divide the elapsed time by the half-life.
4. Substitute the resulting decimal into the exponent.
5. Multiply by the initial number of nuclei.
6. Round the final answer to an appropriate number of significant figures.

For example, if \(t=15\ \text{h}\) and \(T_{1/2}=10\ \text{h}\):

\[ n=\frac{15}{10}=1.5 \]

Your calculator should evaluate:

\[ N=N_0\times2^{-1.5} \]

Keep the full calculator value until the final step.

A common mistake is to round \(1.5\) to \(2\) before calculating. That changes the physical situation and gives an incorrect result.

If a question gives you the decay constant, \(\lambda\), instead of the half-life, you can use the equivalent exponential decay law:

\[ N=N_0e^{-\lambda t} \]

The relationship between the decay constant and half-life is:

\[ \lambda=\frac{\ln 2}{T_{1/2}} \]

For a more detailed derivation of this relationship, see The Exponential Radioactive Decay Law: HSC Physics Guide.

02What does a fractional number of nuclei actually mean?

There is an important distinction between the mathematical prediction and what happens inside an individual sample.

You cannot have 35.4 nuclei. A nucleus either exists or it has decayed.

So why does the equation sometimes produce a decimal number?

Radioactive decay is random. We cannot predict exactly when an individual unstable nucleus will decay, but we can predict the statistical behaviour of a large population.

For a single nucleus, the probability of remaining undecayed after time \(t\) is:

\[ P(\text{survival})=2^{-t/T_{1/2}} \]

For a large sample, multiplying this probability by the initial number of nuclei gives the expected number remaining:

\[ N_{\text{expected}}=N_0P(\text{survival}) \]

Consider a sample containing 40 radioactive nuclei.

After half a half-life, the predicted number remaining is:

\[ N=40\times2^{-0.5} \]

\[ N\approx28.3 \]

An actual sample might contain 26, 29, or 31 undecayed nuclei. It cannot contain exactly 28.3.

However, if you repeated the experiment with many identical samples, the average number remaining would approach 28.3.

For typical HSC questions, the number of nuclei is usually enormous, so the statistical prediction is an excellent approximation.

The important distinction is:

• The equation predicts the expected number of undecayed parent nuclei.
• An individual nucleus decays randomly.
• A fractional answer represents an average prediction, not a literal fraction of a nucleus.

03How do scientists measure half-life experimentally?

The half-life equation is a mathematical model, but how can we determine whether a radioactive sample actually follows it?

The measurement problem is straightforward: we cannot directly watch individual nuclei inside a sample. Instead, we measure the radiation emitted as those nuclei decay.

A standard experimental arrangement uses a Geiger-Müller tube.

What should the experiment show?

The prediction is that after one half-life, the number of undecayed parent nuclei will halve.

The activity of a radioactive sample is related to the number of undecayed nuclei by:

\[ A=\lambda N \]

Here, \(A\) is activity, measured in becquerels (Bq), and \(\lambda\) is the decay constant.

Since \(\lambda\) remains constant for a particular isotope, activity is proportional to the number of undecayed nuclei.

A detector with constant efficiency will therefore record a count rate proportional to the sample’s activity.

This means the corrected count rate should also halve after one half-life.

The background radiation problem

There is one complication.

The detector does not only record radiation from the radioactive sample. It also detects background radiation from the surroundings.

Its reading can be represented as:

\[ C_{\text{measured}}=C_{\text{sample}}+C_{\text{background}} \]

Suppose an illustrative experiment produces the following readings.

The corrected sample count rate halves every 6 minutes.

Therefore:

\[ T_{1/2}=6\ \text{min} \]

The raw detector readings do not halve exactly because the background contribution stays constant.

The revised measurement procedure is to record the background with the source removed, subtract it from each measured reading, and keep the source-detector distance unchanged.

Longer counting intervals and repeated measurements also help reduce the effect of random fluctuations.

Now suppose we want to predict the sample’s corrected count rate after 9 minutes.

\[ n=\frac{9}{6}=1.5 \]

\[ C_{\text{sample}}=800\times2^{-1.5} \]

\[ C_{\text{sample}}\approx283\ \text{counts/min} \]

The detector would therefore read approximately:

\[ C_{\text{measured}}=283+20 \]

\[ \boxed{303\ \text{counts/min}} \]

This is a useful reminder that the radioactive sample follows the decay law, while a constant background does not.

04Worked examples

05What happens to the nuclei that decay?

When a radioactive nucleus decays, it does not simply disappear.

It transforms into another nucleus and may emit a particle or electromagnetic radiation.

For example, in alpha decay, a nucleus emits an alpha particle containing two protons and two neutrons.

Consider the decay of polonium-210:

\[ {}^{210}_{84}\mathrm{Po} \rightarrow {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\mathrm{He} \]

The original polonium nucleus becomes a lead nucleus and an alpha particle.

Two conservation laws must be satisfied in this nuclear equation.

Conservation of nucleon number

The nucleon number, \(A\), is the total number of protons and neutrons.

For polonium-210:

\[ 210=206+4 \]

The total number of nucleons is conserved.

Conservation of nuclear charge

The atomic number, \(Z\), represents the number of protons.

For the same decay:

\[ 84=82+2 \]

The total nuclear charge is also conserved.

These checks are essential when writing nuclear equations.

Beta-minus decay provides a useful contrast.

In beta-minus decay, a neutron transforms into a proton, emitting an electron and an electron antineutrino:

\[ {}^{14}_{6}\mathrm{C} \rightarrow {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \]

Notice what changes:

• The nucleon number stays at 14.
• The atomic number increases from 6 to 7.
• The emitted electron carries a charge of \(-1\), balancing the increase in nuclear charge.
• The antineutrino carries neither electric charge nor nucleon number.

The parent nucleus has changed identity, but the conservation laws still hold.

Parent nuclei are not the same as all nuclei

This is an important distinction when interpreting half-life calculations.

Suppose you start with \(N_0\) radioactive parent nuclei, and every parent undergoes alpha decay to a stable daughter nucleus.

The number of parent nuclei remaining is:

\[ N_{\text{parent}}=N_0 2^{-t/T_{1/2}} \]

The number of parent nuclei that have decayed is:

\[ N_{\text{decayed}}=N_0-N_{\text{parent}} \]

If there were initially no daughter nuclei, and each parent decay produces one stable daughter, then:

\[ N_{\text{daughter}}=N_0-N_{\text{parent}} \]

However, if the daughter is also radioactive, some of those daughter nuclei may decay again.

In that situation, the number of daughter nuclei produced is not necessarily the number of daughter nuclei still present at a later time.

06Nuclear mass, atomic mass, binding energy, and released energy

A half-life calculation tells you how many parent nuclei remain. A mass-energy calculation answers a different question: how much energy is released when those nuclei decay?

To handle these questions correctly, you need to distinguish four related quantities.

Nuclear mass versus atomic mass

A nucleus contains protons and neutrons. A neutral atom contains the same nucleus plus electrons.

Therefore:

\[ m_{\text{atom}}=m_{\text{nucleus}}+Zm_e \]

where:

• \(m_{\text{atom}}\) is the atomic mass.
• \(m_{\text{nucleus}}\) is the nuclear mass.
• \(Z\) is the number of protons.
• \(m_e\) is the mass of one electron.

This relationship neglects the very small contribution from electronic binding energy.

The difference matters when working with nuclear masses.

For example, if you are calculating the mass defect of a nucleus, you must use the mass of the nucleus itself, not the mass of the entire neutral atom.

However, if you are given the mass of a pure sample and need to calculate the number of atoms, you normally use the isotope’s atomic molar mass.

A mass number of 210 also does not mean that the nucleus has an exact mass of \(210\ \text{u}\). The mass number is a count of nucleons, whereas atomic mass is a measured quantity.

Mass defect and binding energy

A bound nucleus has less mass than the total mass of its individual, separated protons and neutrons.

The difference is called the mass defect.

For a nucleus containing \(Z\) protons and \(A-Z\) neutrons:

\[ \Delta m= Zm_p+(A-Z)m_n-m_{\text{nucleus}} \]

where:

• \(m_p\) is the mass of a proton.
• \(m_n\) is the mass of a neutron.
• \(m_{\text{nucleus}}\) is the actual nuclear mass.
• \(A\) is the nucleon number.
• \(Z\) is the proton number.

The mass defect is related to nuclear binding energy through Einstein’s mass-energy relationship:

\[ \boxed{B=\Delta mc^2} \]

Here, \(B\) is the nuclear binding energy and \(c\) is the speed of light in a vacuum.

Binding energy is the energy required to separate a nucleus completely into its individual nucleons.

A larger total binding energy means that more energy is required to break the nucleus apart.

Released energy is not the same as binding energy

This is a common source of confusion.

Binding energy describes how tightly a particular nucleus is held together.

Released energy describes the difference in total rest-mass energy between the initial and final states of a particular nuclear reaction.

Using the convention that positive \(Q\) means energy is released:

\[ \boxed{Q=(m_{\text{initial}}-m_{\text{final}})c^2} \]

If the initial total rest mass is greater than the final total rest mass, the difference is transferred into other forms of energy.

For an alpha decay, the relationship between the total binding energies is:

\[ Q=B_{\text{daughter}}+B_{\alpha}-B_{\text{parent}} \]

The released energy depends on the change in total binding energy, not simply the binding energy of the original nucleus.

07Questions and solutions

Question 1

A radioactive sample initially contains \(8.00\times10^{12}\) unstable nuclei.

Its half-life is \(4.0\ \text{h}\).

Calculate the expected number of radioactive nuclei remaining after \(6.0\ \text{h}\).

Solution 1

Approximately \(2.83\times10^{12}\) radioactive nuclei remain.

The elapsed time is 1.5 half-lives, so the remaining fraction is approximately 0.3536.

Step 1: Calculate the number of half-lives.

\[ n=\frac{6.0}{4.0}=1.5 \]

Step 2: Apply the decay equation.

\[ N=N_0\left(\frac12\right)^n \]

\[ N=8.00\times10^{12}\times2^{-1.5} \]

Step 3: Calculate the remaining number.

\[ N=8.00\times10^{12}\times0.353553 \]

\[ \boxed{N=2.83\times10^{12}\text{ nuclei}} \]

Approximately 35.4% remains, meaning that 64.6% of the original parent nuclei have decayed.

The tempting mistake is to halve the sample twice and obtain 25%. That would require two complete half-lives, or 8 hours, rather than the 6 hours specified.

Question 2

A pure sample of a radioactive isotope has a mass of \(0.600\ \text{mg}\) and an isotopic molar mass of \(150\ \text{g mol}^{-1}\).

Its half-life is \(10.0\ \text{h}\).

After \(15.0\ \text{h}\), calculate:

(a) the expected number of radioactive nuclei remaining;

(b) the mass of the remaining radioactive parent isotope.

Use \(N_A=6.022\times10^{23}\ \text{mol}^{-1}\).

Solution 2

Approximately \(8.52\times10^{17}\) radioactive nuclei remain, with a remaining parent-isotope mass of \(0.212\ \text{mg}\).

The important extra step is converting the original sample mass into a number of atoms before applying the fractional half-life.

Step 1: Convert the initial mass into grams.

\[ m=0.600\times10^{-3}\ \text{g} \]

\[ m=6.00\times10^{-4}\ \text{g} \]

Step 2: Calculate the initial number of moles.

\[ n_{\text{mol}}=\frac{m}{M} \]

\[ n_{\text{mol}}= \frac{6.00\times10^{-4}}{150} \]

\[ n_{\text{mol}}=4.00\times10^{-6}\ \text{mol} \]

Step 3: Calculate the initial number of radioactive nuclei.

\[ N_0=n_{\text{mol}}N_A \]

\[ N_0=(4.00\times10^{-6})(6.022\times10^{23}) \]

\[ N_0=2.409\times10^{18}\text{ nuclei} \]

Step 4: Calculate the number of half-lives elapsed.

\[ n=\frac{15.0}{10.0}=1.5 \]

Step 5: Calculate the remaining nuclei.

\[ N=2.409\times10^{18}\times2^{-1.5} \]

\[ \boxed{N=8.52\times10^{17}\text{ nuclei}} \]

Step 6: Calculate the remaining mass of the parent isotope.

The parent isotope’s mass follows the same remaining fraction as its number of nuclei.

\[ m_{\text{remaining}}=0.600\times0.353553 \]

\[ \boxed{m_{\text{remaining}}=0.212\ \text{mg}} \]

The trap here is confusing the remaining parent-isotope mass with the total mass of the sample. The daughter products still exist, even though the original radioactive isotope has decreased.

Question 3

A small radioactive sample initially contains 40 unstable nuclei.

After half a half-life, a student uses the exponential decay equation and obtains:

\[ N=40\times2^{-0.5}=28.3 \]

The student argues that this result must be incorrect because a sample cannot contain 28.3 nuclei.

Another student argues that the answer must be 28 because the number of nuclei must be rounded down.

Who is correct?

Explain the physical meaning of the calculated result. Then calculate the expected total number of surviving nuclei if 1000 identical, independent samples are observed after the same elapsed time.

Solution 3

Neither student is completely correct. The value 28.3 is a valid statistical prediction, but it is not an exact count for an individual sample.

Step 1: Interpret the calculated result.

After half a half-life:

\[ N_{\text{expected}}=40\times2^{-0.5} \]

\[ N_{\text{expected}}=28.28 \]

This is the expected number of surviving nuclei.

An individual sample must contain a whole number of nuclei. Its actual result will depend on which nuclei happen to decay.

There is no physical rule requiring the result to be rounded down to 28. It could be 27, 28, 29, or another integer.

Step 2: Calculate the expected number across 1000 samples.

Each sample has an expected 28.284 surviving nuclei.

Therefore:

\[ N_{\text{total, expected}}=1000\times28.284 \]

\[ \boxed{N_{\text{total, expected}}\approx28\,284} \]

The actual total will fluctuate, but it is expected to be close to this value.

Step 3: Explain why the distinction matters.

The decay equation describes the statistical behaviour of a population.

For a small sample, random fluctuations can be noticeable. For a very large sample, these fluctuations become relatively small compared with the total number of nuclei.

The tempting approach is to interpret the equation as a deterministic prediction for every individual sample. Radioactive decay does not work that way.

Question 4

A hypothetical radioactive isotope undergoes beta-minus decay.

A sample initially contains \(5.00\times10^{12}\) parent nuclei and has a half-life of \(3.0\ \text{h}\).

After \(7.5\ \text{h}\):

(a) calculate the expected number of parent nuclei remaining;

(b) calculate the expected number of parent nuclei that have decayed;

(c) assuming the daughter isotope is stable and was initially absent, calculate the number of daughter nuclei produced;

(d) write a general nuclear equation for beta-minus decay and demonstrate conservation of nucleon number and nuclear charge.

Solution 4

Approximately \(8.84\times10^{11}\) parent nuclei remain, and \(4.12\times10^{12}\) parent nuclei have decayed. The same number of stable daughter nuclei has been produced.

The calculation uses the half-life relationship, while the nuclear equation explains what happens during each individual decay.

Step 1: Calculate the number of half-lives elapsed.

\[ n=\frac{7.5}{3.0}=2.5 \]

Step 2: Calculate the remaining parent nuclei.

\[ N=N_0\left(\frac12\right)^{2.5} \]

\[ N=5.00\times10^{12}\times0.176777 \]

\[ \boxed{N=8.84\times10^{11}\text{ nuclei}} \]

Step 3: Calculate the number of parent nuclei that have decayed.

The number decayed is the initial number minus the number remaining.

\[ N_{\text{decayed}}=N_0-N \]

\[ N_{\text{decayed}} =5.00\times10^{12}-8.84\times10^{11} \]

\[ \boxed{N_{\text{decayed}}=4.12\times10^{12}} \]

Step 4: Determine the number of daughter nuclei produced.

Each parent nucleus produces one daughter nucleus in this beta-minus decay.

Since the daughter is stable and initially absent:

\[ \boxed{N_{\text{daughter}}=4.12\times10^{12}} \]

Step 5: Write the beta-minus nuclear equation.

The general equation is:

\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \]

Check nucleon number:

\[ A=A+0+0 \]

Check nuclear charge:

\[ Z=(Z+1)-1+0 \]

Both conservation laws are satisfied.

The common misconception is that the atomic number must remain unchanged because the nucleon number remains unchanged. These are different quantities. In beta-minus decay, a neutron becomes a proton, so the atomic number increases by one while the nucleon number stays constant.

Question 5

An initially pure sample contains \(4.00\times10^{15}\) neutral polonium-210 atoms.

Assume that:

• Polonium-210 undergoes alpha decay directly into stable lead-206.
• Its half-life is \(138\ \text{days}\).
• There are initially no lead-206 or helium nuclei in the sample.
• The alpha particles and daughter nuclei remain inside a rigid, perfectly insulated container.
• All energy released by the decay is retained inside the container and eventually becomes internal energy.
• No mass-energy crosses the container boundary.

After \(207\ \text{days}\), determine:

(a) the expected number of polonium-210 nuclei remaining and the number that have decayed;

(b) the expected numbers of lead-206 and helium nuclei produced, and the total number of nuclei inside the container;

(c) the total nucleon number and nuclear charge before and after decay;

(d) the energy released per decay and the total energy released;

(e) whether the total mass-energy inside the container decreases in proportion to the number of remaining polonium nuclei.

For part (d), use these illustrative, rounded atomic masses:

Use \(1\ \text{u}\,c^2=931.5\ \text{MeV}\) and \(1\ \text{MeV}=1.602\times10^{-13}\ \text{J}\).

Solution 5

After 207 days, approximately \(1.41\times10^{15}\) polonium-210 nuclei remain, while \(2.59\times10^{15}\) have decayed.

The container contains approximately \(6.59\times10^{15}\) nuclei in total, and its total mass-energy remains constant because all the released energy is retained inside.

This problem combines radioactive decay, nuclear conservation laws, mass defect, and energy conservation. The important trap is assuming that a decrease in the number of parent nuclei means the same proportion of the sample’s total mass-energy has disappeared.

Step 1: Calculate the number of half-lives elapsed.

\[ n=\frac{207}{138}=1.50 \]

The remaining fraction is:

\[ f=2^{-1.50} \]

\[ f=0.353553 \]

Step 2: Calculate the number of polonium-210 nuclei remaining.

\[ N_{\mathrm{Po}}=N_0f \]

\[ N_{\mathrm{Po}} =4.00\times10^{15}\times0.353553 \]

\[ \boxed{N_{\mathrm{Po}}=1.41\times10^{15}} \]

Step 3: Calculate the number of parent nuclei that have decayed.

\[ N_{\text{decayed}}=N_0-N_{\mathrm{Po}} \]

\[ N_{\text{decayed}} =4.00\times10^{15}-1.41\times10^{15} \]

\[ \boxed{N_{\text{decayed}}=2.59\times10^{15}} \]

Step 4: Determine the number of daughter and helium nuclei.

The nuclear equation is:

\[ {}^{210}_{84}\mathrm{Po} \rightarrow {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\mathrm{He} \]

Every polonium-210 decay produces one lead-206 nucleus and one helium-4 nucleus.

Therefore:

\[ N_{\mathrm{Pb}}=2.59\times10^{15} \]

\[ N_{\mathrm{He}}=2.59\times10^{15} \]

Step 5: Calculate the total number of nuclei inside the container.

The total includes the remaining parent nuclei, daughter nuclei, and helium nuclei.

\[ N_{\text{total}} =N_{\mathrm{Po}}+N_{\mathrm{Pb}}+N_{\mathrm{He}} \]

\[ N_{\text{total}} =1.41\times10^{15} +2.59\times10^{15} +2.59\times10^{15} \]

\[ \boxed{N_{\text{total}}=6.59\times10^{15}} \]

This is a surprising result if you think radioactive decay simply removes nuclei.

Each alpha decay transforms one nucleus into two nuclei. The total number of nuclei therefore increases, even though the number of radioactive parent nuclei decreases.

Step 6: Check conservation of nucleon number.

Initially, each polonium-210 nucleus contains 210 nucleons.

The initial total nucleon number is:

\[ A_{\text{initial}}=210N_0 \]

\[ A_{\text{initial}}=210(4.00\times10^{15}) \]

\[ A_{\text{initial}}=8.40\times10^{17} \]

After decay:

\[ A_{\text{final}} =210N_{\mathrm{Po}}+206N_{\mathrm{Pb}}+4N_{\mathrm{He}} \]

Since the numbers of daughter and helium nuclei are equal:

\[ A_{\text{final}} =210N_{\mathrm{Po}}+210N_{\text{decayed}} \]

\[ A_{\text{final}}=210N_0 \]

\[ \boxed{A_{\text{final}}=8.40\times10^{17}} \]

The total nucleon number is conserved.

Step 7: Check conservation of nuclear charge.

Initially:

\[ Z_{\text{initial}}=84N_0 \]

\[ Z_{\text{initial}}=84(4.00\times10^{15}) \]

\[ Z_{\text{initial}}=3.36\times10^{17} \]

After decay:

\[ Z_{\text{final}} =84N_{\mathrm{Po}}+82N_{\mathrm{Pb}}+2N_{\mathrm{He}} \]

Since each decay produces one lead nucleus and one helium nucleus:

\[ Z_{\text{final}} =84N_{\mathrm{Po}}+84N_{\text{decayed}} \]

\[ Z_{\text{final}}=84N_0 \]

\[ \boxed{Z_{\text{final}}=3.36\times10^{17}} \]

The total nuclear charge is also conserved.

Step 8: Calculate the energy released per alpha decay.

First, calculate the difference between the initial and final atomic masses.

\[ \Delta m=209.9829-205.9745-4.0026 \]

\[ \Delta m=0.0058\ \text{u} \]

Using mass-energy equivalence:

\[ Q=\Delta mc^2 \]

\[ Q=0.0058\times931.5 \]

\[ \boxed{Q=5.40\ \text{MeV per decay}} \]

Convert this into joules:

\[ Q=5.40\times1.602\times10^{-13} \]

\[ Q=8.65\times10^{-13}\ \text{J per decay} \]

The positive value means that energy is released during each alpha decay.

Step 9: Calculate the total energy released.

There are approximately \(2.59\times10^{15}\) decays.

Therefore:

\[ E_{\text{total}}=N_{\text{decayed}}Q \]

\[ E_{\text{total}} =(2.59\times10^{15})(8.65\times10^{-13}) \]

\[ \boxed{E_{\text{total}}\approx2.24\times10^3\ \text{J}} \]

Approximately 2.24 kJ of energy is released.

Step 10: Explain what happens to the total mass-energy.

This is the conceptual hinge of the question.

A tempting prediction is that because only 35.4% of the original polonium nuclei remain, only 35.4% of the original sample’s mass-energy remains.

That is incorrect.

The parent nuclei have transformed into lead nuclei and helium nuclei. The material has not simply disappeared.

The sum of the nuclear rest masses decreases because energy has been released. However, the problem states that all this energy remains inside the insulated container.

It is eventually transferred into internal energy, such as thermal energy.

Therefore, the total mass-energy of the closed system remains constant.

If the energy instead escaped from the container, the system would lose mass-energy equivalent to:

\[ \Delta m=\frac{E}{c^2} \]

\[ \Delta m= \frac{2.24\times10^3}{(3.00\times10^8)^2} \]

\[ \boxed{\Delta m=2.49\times10^{-14}\ \text{kg}} \]

This is the mass-equivalent of the energy that escaped. It is not the mass of all the parent nuclei that decayed.

The crucial distinction is that radioactive decay changes the identity and rest mass of nuclei, while mass-energy is conserved when the complete system is considered.

The next step after these calculations is understanding radioactive decay chains, where an unstable daughter nucleus can undergo further decay. That requires tracking the parent and daughter populations separately rather than assuming every nucleus produced remains in the sample indefinitely.

For broader practice with repeated halving, activity, and detector readings, see Radioactive Half-Life: HSC Physics Calculations Explained.