Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: nuclear physics and radioactivity (203-SN3-RE)

This sheet does not redo the nuclear physics course, it lists what loses marks on test day. The chapter has one peculiarity that explains almost all its errors: every interesting quantity is a DIFFERENCE of the order of a thousandth of a unit, computed between two six-decimal numbers that look very much alike. Premature rounding destroys the result without leaving a trace.

It is written for CEGEP students in the Natural Science program, and for anyone approaching radioactivity with a table of atomic masses in front of them.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

In nuclear physics you never weigh a nucleus, you weigh a DIFFERENCE: keep every decimal, and always ask whether the table gives atomic masses or nuclear masses.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

The mass defect and the curve that decides everything

• Binding energy with ATOMIC masses: $B=[Z\,m(^{1}\mathrm{H})+N\,m_{n}-m(^{A}_{Z}\mathrm{X})]c^{2}$ . Write $m(^{1}\mathrm{H})$ and not $m_{p}$ , because the $Z$ electrons then cancel exactly on both sides.
• The stability criterion is $\dfrac{B}{A}$ , never $B$ . The curve peaks at $8.79$ MeV per nucleon around iron-56: everything to the left gains by FUSING, everything to the right gains by undergoing FISSION.
• Conversions: $1$ u $=931.494$ MeV/c $^{2}=1.66054\times 10^{-27}$ kg. Table masses: hydrogen-1 $1.007825$ u, neutron $1.008665$ u, electron $0.000549$ u.
• Reaction energy $Q=[m_{\text{initial}}-m_{\text{final}}]c^{2}$ in atomic masses, EXCEPT for $\beta^{+}$ , where you must subtract $2m_{e}c^{2}=1.022$ MeV.

It is

B/A

that decides:

8.79

MeV per nucleon for iron at the peak,

7.57

for uranium. To the left of the peak fusion releases energy, to the right fission does.

The valley of stability: position decides the mode

• Stable nuclei follow $N\approx Z$ up to calcium, then become enriched in neutrons: $126$ neutrons for $82$ protons in lead-208. Without this surplus, the repulsion between protons would win.
• Above the valley, too many neutrons: $\beta^{-}$ , which changes a neutron into a proton. Below, too many protons: $\beta^{+}$ or electron capture.
• Beyond lead, $Z>82$ , no nucleus is stable: $\alpha$ emission carries away four nucleons at once.

Position relative to the valley decides the mode: above it

\beta^{-}

, below it

\beta^{+}

, beyond lead

\alpha

. Uranium-238 is already outside the stable zone.

Decay: three quantities not to confuse

• $N(t)=N_{0}e^{-\lambda t}$ with $\lambda=\dfrac{\ln 2}{T_{1/2}}$ . The constant $\lambda$ must be in s $^{-1}$ for the activity to come out in becquerels: converting the half-life to seconds is the very first step.
• Activity $A=\lambda N$ , in becquerels, one decay per second. $1$ Ci $=3.7\times 10^{10}$ Bq.
• The time to fall to a fraction $f$ is $T_{1/2}\dfrac{\ln(1/f)}{\ln 2}$ : $6.64$ half-lives to reach 1%, $9.97$ to reach one thousandth. These two numbers come up constantly.
• Secular equilibrium, when $T_{1}\gg T_{2}$ : the ACTIVITIES become equal and the numbers of atoms are in the ratio $\dfrac{N_{2}}{N_{1}}=\dfrac{T_{2}}{T_{1}}$ .

On a logarithmic scale the decay is a straight line: 1% after

6.64

half-lives, one thousandth after

9.97

, whatever the half-life.

Decay chains: the daughter catches up with the parent's activity

• A daughter of half-life $T_{2}$ born from a parent of half-life $T_{1}$ : $\dfrac{dN_{2}}{dt}=\lambda_{1}N_{1}-\lambda_{2}N_{2}$ , the production by the parent minus its own disappearance.
• If $T_{1}\gg T_{2}$ and the daughter has just been removed, its activity rises according to $A_{2}=A_{1}\left(1-e^{-\lambda_{2}t}\right)$ : it is the half-life of the DAUGHTER that sets the rise, not that of the parent.
• It takes $6.64\,T_{2}$ to reach 99% of equilibrium: $25.4$ days for radon-222, $T_{2}=3.82$ days, born from freshly purified radium.

The activity of the daughter (orange) catches up with that of the parent (blue): half of it in one daughter half-life, 99% in

6.64

. After that the two stay equal: this is secular equilibrium.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. An iodine-131 source, with a half-life of $8.02$ days, is left to sit for $16$ days.

2 marks, and in a laboratory a regulatory violation

What not to write

Twice eight is sixteen, so the source is completely extinguished and can be thrown out with ordinary waste.

What to write

After two half-lives, $\left(\dfrac{1}{2}\right)^{2}=25$ % of the initial activity remains. It takes $6.64$ half-lives, that is $53$ days, to get down to 1%.

After two half-lives

25

% still remains (green); the curve never reaches the axis and passes 1% after

53

days. The red line, extinguished at

16

days, is the wrong answer.

Why: Decay is MULTIPLICATIVE, not additive: each half-life divides what remains by two, so the activity never reaches zero in a finite time. The control reflex is to check that every duration that is a multiple of the half-life gives a power of two: a quarter, an eighth, a sixteenth.

2. Calculating the binding energy of iron-56 from a table of masses.

1 to 3 marks, and an error that survives rereading

What not to write

$B=[26\,m_{p}+30\,m_{n}-m(^{56}\mathrm{Fe})]c^{2}$ , with $m_{p}=1.007276$ u, which gives $479.0$ MeV.

What to write

$B=[26\,m(^{1}\mathrm{H})+30\,m_{n}-m(^{56}\mathrm{Fe})]c^{2}$ , with $m(^{1}\mathrm{H})=1.007825$ u, which gives $492.3$ MeV.

Why: Tables publish ATOMIC masses, electrons included, because that is what a spectrometer measures. The subtracted term contains the $26$ electrons of iron: they must therefore be counted on the left too, which is what the hydrogen atom does. With bare protons, you miss $26\times 0.000549$ u, that is $13.30$ MeV and $2.7$ %: plausible, hence invisible.

3. Calculating the reaction energy of the $\beta^{+}$ decay of sodium-22 to neon-22.

2 marks, systematically

What not to write

$Q=[m(^{22}\mathrm{Na})-m(^{22}\mathrm{Ne})]c^{2}=2.843$ MeV.

What to write

$Q=[m(^{22}\mathrm{Na})-m(^{22}\mathrm{Ne})-2m_{e}]c^{2}=2.843-1.022=1.821$ MeV.

The difference of atomic masses,

2.843

MeV (red), forgets the extra electron and the positron created: we pay

2m_{e}c^{2}=1.022

MeV, leaving

Q=1.821

MeV (green).

Why: The electron count does not balance by itself here: on the left the sodium atom carries $11$ , on the right neon carries only $10$ , and a positron must ALSO be created. Hence a deficit of two electron masses, that is $1.022$ MeV, which appears neither for $\alpha$ nor for $\beta^{-}$ . Physical consequence: a nucleus whose energy excess is below $1.022$ MeV cannot emit a positron, and only electron capture is left to it.

4. Comparing the stability of uranium-238 and iron-56.

The whole reasoning question, 3 marks

What not to write

Uranium has a binding energy of $1802$ MeV against $492$ MeV for iron: it is therefore the more tightly bound.

What to write

I compare the binding energies PER NUCLEON: $\dfrac{1802}{238}=7.57$ MeV for uranium against $\dfrac{492}{56}=8.79$ MeV for iron. Iron is the more tightly bound.

Ranked by total

B

(red), uranium looks

3.7

times more tightly bound than iron; ranked by

B/A

(green), iron wins,

8.79

against

7.57

MeV per nucleon.

Why: The total binding energy grows trivially with the size of the nucleus, there are simply more bonds to count. It is the shape of the $B/A$ curve, with its peak at iron, that makes both fusion and fission exothermic depending on the side you are on, and it alone.

5. Ranking the three radiations by danger.

2 marks on a course question, and a complete misreading in radiation protection

What not to write

$\alpha$ is the least dangerous since a sheet of paper stops it, whereas $\gamma$ goes through lead.

What to write

For EXTERNAL exposure, yes, the epidermis is enough to stop $\alpha$ . For INTERNAL contamination, $\alpha$ is by far the most dangerous, with a quality factor of $20$ against $1$ for $\beta$ and $\gamma$ .

Outside, the dead layer of the skin stops the

\alpha

; inhaled, it deposits all its energy within a few tens of micrometres of living tissue, hence its quality factor of

20

Why: Low penetrating power is exactly what makes $\alpha$ formidable once inhaled: it deposits all its energy within a few tens of micrometres of living tissue. A danger is never judged without its exposure scenario, and inhaled radon is the typical example.

6. Comparing the activity of two grams of technetium-99m and two grams of uranium-238.

2 marks, and an inability to handle any source problem

What not to write

Same mass, so same activity: the activity depends only on the quantity of radioactive material.

What to write

$A=\lambda N$ , and $\lambda=\dfrac{\ln 2}{T_{1/2}}$ : for equal masses, the activity is inversely proportional to the half-life. Technetium, at $6.0$ hours against $4.5$ billion years, is billions of times more active.

For equal masses,

2

g of technetium-99m emit

3.9\times 10^{17}

Bq,

2

g of uranium-238

2.5\times 10^{4}

Bq (logarithmic scale): the half-life decides, not the mass.

Why: What distinguishes two isotopes is not their mass but their $\lambda$ , and the half-lives encountered span thirty-six orders of magnitude, from the nanosecond to $10^{19}$ years. No other quantity in the program has such a span, and this is why one microgram of iodine-131 is already worth $4.60$ GBq.

Which method to choose

Which decay mode, and which calculation

First identify what the problem gives: a nucleus to balance, a table of masses, a half-life, or the mass of a sample.

The three displacements in the neutron versus proton plane:

\alpha

steps back diagonally by two squares,

\beta^{-}

goes down to the right,

\beta^{+}

goes up to the left.

If The nucleus is HEAVY, beyond lead or so → $\alpha$ decay: $A$ loses 4, $Z$ loses 2. Reaction energy directly from the difference of atomic masses.

Example: $^{238}_{92}\mathrm{U}\to\,^{234}_{90}\mathrm{Th}+^{4}_{2}\mathrm{He}$ , $Q=4.271$ MeV.
If The nucleus has TOO MANY NEUTRONS, it is above the valley of stability → $\beta^{-}$ decay: a neutron becomes a proton, $Z$ increases by 1, $A$ unchanged, plus an ANTIneutrino. $Q$ by direct difference of atomic masses.

Example: $^{14}_{6}\mathrm{C}\to\,^{14}_{7}\mathrm{N}+e^{-}+\bar{\nu}_{e}$ , $Q=0.156$ MeV.
If The nucleus has TOO MANY PROTONS and an energy excess above $1.022$ MeV → $\beta^{+}$ decay: $Z$ decreases by 1, plus a neutrino. $Q$ by difference of masses MINUS $2m_{e}c^{2}$ .

Example: $^{22}_{11}\mathrm{Na}\to\,^{22}_{10}\mathrm{Ne}+e^{+}+\nu_{e}$ , $Q=1.821$ MeV.
If The nucleus has too many protons but an energy excess BELOW $1.022$ MeV → Electron capture, the only route open: the nucleus swallows an electron from its $K$ shell. Same daughter nucleus as $\beta^{+}$ , zero threshold, and a characteristic X-ray signals it.

Example: Potassium-40, which decays in both directions, toward argon-40 and toward calcium-40.
If The problem gives a half-life and a sample mass → Calculate $N$ from the mole, then $\lambda=\dfrac{\ln 2}{T_{1/2}}$ IN SECONDS, then $A=\lambda N$ . For a duration, prefer $2^{-t/T_{1/2}}$ to the exponential, it is safer to do in your head.

Example: $1.00$ μg of iodine-131: $N=4.60\times 10^{15}$ , $\lambda=1.00\times 10^{-6}$ s $^{-1}$ , $A=4.60$ GBq.

The daughter nucleus is often produced in an EXCITED state: it then de-excites by $\gamma$ emission, which changes neither $A$ nor $Z$ and is therefore not a transmutation.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Calculating a binding energy or a reaction energy

When to use it: As soon as a table of masses is supplied and an energy in MeV is asked for.

1 Write the complete nuclear equation and balance it in $A$ and $Z$ , neutrino included.
2 Check that the masses supplied are ATOMIC, and say so explicitly in one sentence.
3 Count the electrons on both sides, and derive the correction term: zero for $\alpha$ and $\beta^{-}$ , equal to $2m_{e}$ for $\beta^{+}$ .
4 Calculate the mass difference keeping ALL the decimals supplied, without rounding along the way.
5 Convert using $931.494$ MeV/c $^{2}$ per atomic mass unit, and give the result with three or four significant figures.

Concluding sentence

“Since the masses supplied are atomic, the $Z$ electrons appear on both sides and cancel: $Q=[m_{X}-m_{Y}-m_{\alpha}]c^{2}$ , with no correction term.”

The trap: Rounding the masses to four decimals before subtracting: the difference sought is often $0.0046$ u, and rounding destroys it entirely.

Marking: 1 mark for the balanced equation, 1 mark for justifying the electron count, 2 marks for the calculation, 1 mark for the conversion and the significant figures.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The numbers $A$ and $Z$ balance

Add the mass numbers on the left and on the right, then the atomic numbers. The two sums must be equal, without exception.

$238=234+4$ and $92=90+2$ : the uranium-238 equation is correct.
The reaction energy is POSITIVE

A spontaneous decay releases energy: a negative $Q$ signals a sign error, a parent-daughter inversion, or a forgotten $\beta^{+}$ correction term.

$Q=-1.0$ MeV for a $\beta^{+}$ almost always means that $1.022$ MeV was added instead of subtracted.
$B/A$ lies between $7$ and $9$ MeV

Except for the very light nuclei, the binding energy per nucleon of any nucleus falls within this narrow range. A result outside it is wrong.

$B/A=479/56=8.55$ MeV looks acceptable and is nevertheless wrong: it is the bare-proton calculation. The right value is $8.790$ .
The number of half-lives elapsed is consistent

Divide the duration by the half-life BEFORE any calculation. Each whole unit divides the activity by two, which can be checked in your head.

$24.1$ days for a half-life of $8.02$ days make $3.01$ half-lives, hence a division by eight: $4.60$ GBq becomes $0.57$ GBq.

The typical problem, taken apart

The alpha decay of uranium-238, all the way to the recoil of the thorium

Atomic masses in u: $^{238}\mathrm{U}$ $238.050788$ ; $^{234}\mathrm{Th}$ $234.043601$ ; $^{4}\mathrm{He}$ $4.002602$ . Write the equation, calculate the reaction energy, then share this energy between the $\alpha$ particle and the recoiling nucleus.

The situation: uranium-238 at rest splits into a thorium-234 nucleus and an

\alpha

particle that fly off in opposite directions; the sharing of

Q

between them remains to be found.

Step 1

$^{238}_{92}\mathrm{U}\to\,^{234}_{90}\mathrm{Th}+^{4}_{2}\mathrm{He}$ , with $238=234+4$ and $92=90+2$

Why

We balance BEFORE touching the masses. The two conservation laws are enough to identify the daughter nucleus, and an unbalanced equation loses all the following marks even if the calculation is right.

Step 2

Electron count: $92$ on the left, $90+2=92$ on the right. No correction term.

Why

This is the sentence the marker expects, and it takes one line. It justifies the direct use of atomic masses, and it proves you did not apply the rule out of habit.

Step 3

$\Delta m=238.050788-234.043601-4.002602=0.004585$ u, so $Q=0.004585\times 931.494=4.271$ MeV

Why

A difference of $0.0046$ between three six-decimal numbers: this is where the whole result is decided. Rounding to $238.05$ , $234.04$ and $4.00$ would give $0.01$ u, that is $9.3$ MeV, more than double.

Step 4

Conservation of momentum, initial nucleus at rest: $p_{\alpha}=p_{\mathrm{Th}}$ , so $K=\dfrac{p^{2}}{2m}$ is shared in INVERSE proportion to the masses

Why

The energy is not shared equally nor in proportion to the masses: it is the equality of the momenta that fixes the sharing, and it favours the lighter one. The reasoning is the same as for the recoil of a rifle.

Step 5

$K_{\alpha}=4.271\times\dfrac{234}{238}=4.199$ MeV and $K_{\mathrm{Th}}=4.271\times\dfrac{4}{238}=0.0718$ MeV

Why

The $\alpha$ particle carries away $98.3$ % of the energy. This is what allows the energy of $\alpha$ particles to be tabulated as if it were $Q$ , but a problem that explicitly asks for the recoil expects both values.

Step 6

Relativistic check: $\dfrac{K_{\alpha}}{m_{\alpha}c^{2}}=\dfrac{4.199}{3727}=1.1\times 10^{-3}$

Why

The ratio is of the order of a thousandth, so the nonrelativistic treatment is justified to better than $0.1$ %. This line is often worth half a mark and shows that you know why you may write $K=\dfrac{p^{2}}{2m}$ .

The conclusion, written out

$Q=4.271$ MeV, of which $4.199$ MeV for the $\alpha$ particle and $0.0718$ MeV for the recoiling thorium. The sum gives $Q$ back, which is the final check.

The classic mistake on this problem: Giving all the energy to the $\alpha$ particle when the problem explicitly asks for the recoil, or worse, sharing $Q$ equally between the two fragments: conservation of momentum fixes the sharing, not equality.

Learn by heart

• $1$ u $=931.494$ MeV/c $^{2}$ . Hydrogen-1: $1.007825$ u. Neutron: $1.008665$ u. Electron: $0.000549$ u, that is $0.511$ MeV.
• $B=[Z\,m(^{1}\mathrm{H})+N\,m_{n}-m_{\text{atom}}]c^{2}$ with ATOMIC masses. The stability criterion is $\dfrac{B}{A}$ , not $B$ .
• $\alpha$ : $A-4$ , $Z-2$ . $\beta^{-}$ : $Z+1$ , antineutrino. $\beta^{+}$ : $Z-1$ , neutrino, and $Q$ reduced by $1.022$ MeV. $\gamma$ : nothing changes.
• $\lambda=\dfrac{\ln 2}{T_{1/2}}$ , $A=\lambda N$ , $N(t)=N_{0}2^{-t/T_{1/2}}$ . It takes $6.64$ half-lives to fall to 1%.
• Secular equilibrium: the activities become equal and $\dfrac{N_{2}}{N_{1}}=\dfrac{T_{2}}{T_{1}}$ . Hence $0.34$ g of radium per tonne of uranium.
• Fission: $200$ MeV per nucleus, that is about $3.2$ kg of uranium-235 per day for a plant of $1000$ MW electrical.
• Quality factors: $1$ for $\beta$ and $\gamma$ , $20$ for $\alpha$ . Average natural exposure, $2$ to $3$ mSv per year.

Frequently asked questions

Should I use the mass of the proton or that of the hydrogen atom?

That of the hydrogen atom, because tables publish atomic masses, electrons included. The subtracted term already contains the Z electrons of the nucleus under study: counting them on the left too, through the hydrogen atom, makes them cancel exactly. Using the bare proton underestimates the binding energy by about three percent.

Why must 1.022 MeV be subtracted in a beta-plus decay?

Because the electron count does not balance by itself. The parent atom carries one more electron than the daughter atom, and a positron must also be created: that makes two electron masses missing, that is 1.022 megaelectronvolts. This correction term appears neither for alpha nor for beta-minus decay, where the electrons cancel.

How long does it take for a radioactive source to be extinguished?

Never completely, since the decay is exponential. Each half-life divides the activity by two: a quarter remains after two half-lives and an eighth after three. In practice a source is considered negligible after about seven half-lives, which corresponds to one percent of the initial activity.

Why is uranium less stable than iron even though its binding energy is larger?

Because the right criterion is the binding energy per nucleon, not the total energy. A large nucleus simply has more bonds to count. Iron-56 is at the peak of the curve with 8.79 megaelectronvolts per nucleon, against 7.57 for uranium-238: each nucleon of uranium is held less firmly, hence the possibility of fission.

Which is the most dangerous radiation, alpha, beta or gamma?

It depends entirely on the exposure scenario. Outside the body, alpha is the least dangerous because the skin stops it, and gamma the most. In internal contamination, inhaled or ingested, alpha becomes by far the most dangerous: it deposits all its energy within a few tens of micrometres of tissue, hence its quality factor of twenty.

How do I calculate the activity of a sample in becquerels?

First calculate the number of atoms from the mass and the molar mass, using Avogadro's number. Then calculate the decay constant as the natural logarithm of two divided by the half-life expressed IN SECONDS. The activity is the product of the two. Forgetting to convert the half-life to seconds is the most common mistake.

Practise it

Corrected exercises: Nuclear physics and radioactivity, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: nuclear physics and radioactivity (203-SN3-RE)

The essentials

The mass defect and the curve that decides everything

The valley of stability: position decides the mode

Decay: three quantities not to confuse

Decay chains: the daughter catches up with the parent's activity

The mistakes that cost marks

1. An iodine-131 source, with a half-life of 8.028.028.02 days, is left to sit for 161616 days.

2. Calculating the binding energy of iron-56 from a table of masses.

3. Calculating the reaction energy of the β+\beta^{+}β+ decay of sodium-22 to neon-22.

4. Comparing the stability of uranium-238 and iron-56.

5. Ranking the three radiations by danger.

6. Comparing the activity of two grams of technetium-99m and two grams of uranium-238.

Which method to choose

Which decay mode, and which calculation

How the answer is expected to be written

Calculating a binding energy or a reaction energy

Check before you hand in

The typical problem, taken apart

The alpha decay of uranium-238, all the way to the recoil of the thorium

Learn by heart

Frequently asked questions

Should I use the mass of the proton or that of the hydrogen atom?

Why must 1.022 MeV be subtracted in a beta-plus decay?

How long does it take for a radioactive source to be extinguished?

Why is uranium less stable than iron even though its binding energy is larger?

Which is the most dangerous radiation, alpha, beta or gamma?

How do I calculate the activity of a sample in becquerels?

Corrected exercises: Nuclear physics and radioactivity, 203-SN3-RE

See also

Looking for a 203-SN3-RE tutor in Montreal?

1. An iodine-131 source, with a half-life of $8.02$ days, is left to sit for $16$ days.

3. Calculating the reaction energy of the $\beta^{+}$ decay of sodium-22 to neon-22.