E. Nuclear and quantum physics HL

Set up the evidence

In the Geiger–Marsden–Rutherford experiment, alpha particles were directed at a thin gold foil and detected around the foil. Most particles passed through without deflection, some were deflected, and a very small number scattered backwards.

Infer the nuclear model

The results show that an atom is mostly empty space. The rare large deflections require a small, dense, positively charged nucleus that contains most of the atom’s mass; the positive charge cannot be spread uniformly through the whole atom.

Keep the conclusion qualitative

For SL, focus on linking each observation to the model: many undeflected particles imply empty space, while rare back-scattering implies a concentrated repulsive centre. Do not treat the experiment as evidence that electrons occupy fixed-radius orbits.

Common trap

Do not say that all alpha particles are deflected. The dominant observation is that most pass through essentially undeflected; the large-angle events are rare but decisive.

Read the symbol

Nuclear notation is written as ${}^{A}_{Z}X$. The chemical symbol $X$ identifies the element, the proton number $Z$ is written below, and the nucleon number $A$ is written above.

Count the nucleus

The nucleus contains $Z$ protons and $N=A-Z$ neutrons. The number of electrons is not encoded by $A$ and $Z$; for a neutral atom it equals $Z$, while an ion has gained or lost electrons.

Compare nuclides

Atoms of the same element have the same $Z$. Isotopes have the same $Z$ but different $A$, so they contain different numbers of neutrons.

Common trap

Do not use the electron count as the proton number for an ion, and do not confuse $A$ with the number of neutrons. Subtract $Z$ from $A$ to find the neutron number.

Emission lines

An excited gas emits light at particular frequencies, producing bright spectral lines rather than a continuous spread of frequencies. Each line corresponds to a permitted energy difference between atomic states.

Absorption lines

When continuous light passes through a cooler gas, the atoms remove the same frequencies they can emit. The resulting dark lines therefore occur at specific, repeatable wavelengths.

Infer discrete levels

Because only particular photon energies are emitted or absorbed, the atom’s energy states are discrete rather than continuous. The spectrum is evidence for quantized atomic energy levels.

Common trap

Do not treat every visible line as a separate element without considering transitions. A spectrum is evidence of allowed energy differences; the pattern, not simply the number of lines, carries the information.

Emission

When an electron moves from a higher atomic energy level to a lower one, the atom emits one photon. The photon energy equals the level difference: $E_\gamma=\Delta E$.

Absorption

An atom can absorb a photon only when its energy matches an allowed upward transition. The electron then moves to the higher level, so the spectrum records the same allowed energy differences in reverse.

Read a transition diagram

For each downward arrow, calculate the energy gap between its initial and final levels. A larger gap produces a higher-frequency photon and a shorter wavelength; a smaller gap produces a lower-frequency photon and a longer wavelength.

Common trap

Do not use the absolute energy of one level as the photon energy. A photon is associated with the difference between two levels, and emission requires a downward transition.

Use the photon relation

Photon energy depends on frequency. For an atomic transition, first take the positive magnitude of the energy-level difference, then convert units consistently before finding frequency or wavelength.

Worked example — hydrogen transition

A transition with $E_\gamma=1.89\,\mathrm{eV}$ has energy $(1.89)(1.60\times10^{-19})=3.02\times10^{-19}\,\mathrm{J}$. Hence $f=E/h=(3.02\times10^{-19})/(6.63\times10^{-34})=4.56\times10^{14}\,\mathrm{Hz}$ and $\lambda=c/f=6.58\times10^{-7}\,\mathrm{m}$.

Connect energy to a level gap

For an atomic transition, use $E_\gamma=|E_i-E_f|$. Take the magnitude of the energy difference, then convert units consistently before finding frequency or wavelength.

Predict the wavelength

A larger energy gap gives a higher-frequency photon and a shorter wavelength. Therefore the longest wavelength comes from the smallest non-zero transition energy.

Common trap

Do not carry a negative sign from bound-state energies into photon energy. The photon energy is positive and equals the magnitude of the level difference.

Treat a spectrum as a fingerprint

Each element has a characteristic set of emission and absorption wavelengths because its allowed energy differences are unique. The pattern can therefore identify the chemical species producing or absorbing the light.

Use comparison evidence

Record the observed spectral lines and compare their wavelengths or frequencies with laboratory spectra of known elements. Matching several characteristic lines supports the identification.

Apply it to stars

Light from a star can contain absorption lines produced by cooler gases in its atmosphere. Comparing those lines with known spectra reveals which elements are present, even when the source cannot be sampled directly.

Common trap

Do not identify an element from one broad colour alone. The evidence is the set of matching spectral lines and their wavelengths.

Use the radius law

Nuclear radius $R$ grows as the cube root of nucleon number $A$. The constant $R_0=1.20\times10^{-15}\,\mathrm{m}$ represents the scale of a single-nucleon nucleus in this model.

Worked example — gold-197

For $A=197$, $R=(1.20\times10^{-15})(197)^{1/3}=6.98\times10^{-15}\,\mathrm{m}$. Since volume is proportional to $R^3\propto A$ while nuclear mass is also approximately proportional to $A$, the model predicts approximately constant nuclear density.

Infer the density

Nuclear volume scales as $R^3$, so $V\propto A$. Since nuclear mass is approximately proportional to $A$, the mass per unit volume is approximately constant across nuclei.

Scale carefully

If $A$ changes by a factor of $k$, radius changes by $k^{1/3}$, not by $k$. The density remains approximately unchanged in this model.

Common trap

Do not assume a nucleus with eight times the nucleon number has eight times the radius. It has twice the radius and approximately the same density.

Start with Rutherford scattering

At moderate energies, alpha-particle scattering can be modelled as electrostatic repulsion from a concentrated positive nucleus. The predicted deflections follow the Rutherford picture.

Read the high-energy deviation

At sufficiently high alpha-particle energies, the particles can approach more closely and the observed scattering departs from the electrostatic prediction. This provides evidence that the nucleus has a finite size and that a short-range strong interaction becomes relevant.

State what the evidence supports

The deviation is evidence about the nuclear scale and the interaction at very small separation. It is not evidence about the size of the alpha particle or the weak force.

Common trap

Do not continue applying pure Coulomb scattering after the experiment has entered the regime where the alpha particle probes the nuclear force.

Use energy conservation

For a head-on alpha particle, the initial kinetic energy is converted into electric potential energy as the particle approaches the positive nucleus. At the turning point, the radial kinetic energy is zero.

Set the energies equal

At the turning point the radial kinetic energy is zero, so the initial kinetic energy equals the electric potential energy of the repulsive alpha-particle–nucleus system. Both positive charges must be included.

Worked example — alpha particle toward gold

For $E_k=5.0\,\mathrm{MeV}=8.0\times10^{-13}\,\mathrm{J}$, $q_\alpha=2e$ and $q_N=79e$, $r_{\min}=k(2e)(79e)/E_k=4.5\times10^{-14}\,\mathrm{m}$. This is a turning-point distance, not automatically the nuclear radius.

Check the turning point

At closest approach the alpha particle has momentarily stopped moving toward the nucleus, then reverses. A larger initial kinetic energy gives a smaller closest-approach distance.

Common trap

Do not use the charge of gold alone: the interaction contains both $q_\alpha$ and $q_{nucleus}$. Also do not leave energy in MeV while using $k$ in SI units.

Use the hydrogen levels

In the Bohr model for hydrogen, $n=1,2,3,\ldots$ is the principal quantum number. Bound-state energies are negative and approach zero as $n$ increases; the equation is not the general spectrum formula for multi-electron atoms.

Worked example — fifth level

For $n=5$, $E_5=-13.6/5^2=-0.544\,\mathrm{eV}$. In joules this is $(-0.544)(1.60\times10^{-19})=-8.70\times10^{-20}\,\mathrm{J}$. Keep the negative sign for the bound level; use a positive energy difference for a photon.

Find a transition energy

For a transition between levels, calculate $\Delta E=|E_i-E_f|$. Emission occurs for a downward transition and absorption for an upward transition; the photon then obeys $E_\gamma=hf=hc/\lambda$.

Compare levels

The gaps are not equally spaced. A transition involving low $n$ can have a larger energy difference than one involving high $n$, so compare the actual level values rather than relying on the visual spacing of an unscaled sketch.

Common trap

Do not omit the negative sign while identifying the level, but do use the positive magnitude of the difference when calculating photon energy.

Apply the angular-momentum condition

The Bohr model permits only integer values of $n=1,2,3,\ldots$. The electron's orbital angular momentum is therefore quantized rather than continuously variable.

Worked example — $n=4$

For $n=4$, $L=4h/(2\pi)=4(6.63\times10^{-34})/(2\pi)=4.22\times10^{-34}\,\mathrm{kg\,m^2\,s^{-1}}$. An intermediate value is not an allowed Bohr-orbit angular momentum.

Connect quantization to energy

Only selected radii, speeds and total energies are allowed. The electron cannot occupy an intermediate orbit energy in this model, which explains discrete atomic levels and line spectra.

Use ratios efficiently

For hydrogen, combining the quantization condition with the electrostatic circular-orbit model gives $r_n\propto n^2$ and $v_n\propto 1/n$. If $r_2/r_1=4$, then $v_2/v_1=1/2$.

Common trap

Do not say that quantization fixes only the radius. The condition restricts angular momentum and leads to discrete allowed energies; do not treat $mvr$ as an arbitrary continuous value.

Read the observations

When monochromatic light illuminates a metal, electrons may be emitted. Increasing intensity increases the emission rate, but for fixed frequency it does not increase the maximum kinetic energy of the emitted electrons.

Use the photon model

Light transfers energy in individual photons. One photon interacts with one electron, so photon frequency sets the energy available per interaction, while intensity changes the number of photons arriving per second.

Identify the evidence

The intensity–energy distinction and the existence of a threshold frequency cannot be explained by a simple continuous wave-energy model. They support the particle nature of light.

Common trap

Do not say that brighter light makes each photoelectron more energetic. At fixed frequency, it produces more emitted electrons, not a larger maximum kinetic energy.

Define the threshold

The threshold frequency $f_0$ is the minimum photon frequency that can eject an electron from a particular metal. At threshold, the photon has just enough energy to equal that metal's work function $\Phi$, leaving zero maximum kinetic energy.

Worked example — threshold frequency

For $\Phi=2.70\,\mathrm{eV}=(2.70)(1.60\times10^{-19})=4.32\times10^{-19}\,\mathrm{J}$, $f_0=\Phi/h=(4.32\times10^{-19})/(6.63\times10^{-34})=6.52\times10^{14}\,\mathrm{Hz}$. Brighter light below this frequency still ejects no electrons.

Explain the intensity result

Below $f_0$, each photon has too little energy to overcome the work function. Increasing intensity supplies more low-energy photons, but it does not make any one photon energetic enough, so no electrons are emitted.

Keep the metal fixed

Threshold frequency depends on the metal’s work function. Two metals illuminated by the same radiation can behave differently because their electron-binding energies differ.

Common trap

Do not explain the threshold in terms of total light energy accumulated over time. The interaction is photon-by-photon.

Track the energy budget

One photon transfers energy $hf$ to one electron. The work function is spent releasing the electron; any remainder is its maximum kinetic energy. Use joules or electronvolts consistently.

Worked example — 420 nm light

For $\lambda=420\,\mathrm{nm}$, photon energy is $hc/\lambda=2.96\,\mathrm{eV}$. With $\Phi=2.0\,\mathrm{eV}$, $E_{k,\max}=2.96-2.0=0.96\,\mathrm{eV}$. The result is positive, so emission occurs; a negative calculated remainder would mean no emission.

Use wavelength when given

Because $f=c/\lambda$, write $E_{k,max}=\frac{hc}{\lambda}-\Phi$. Keep $hc/\lambda$ and $\Phi$ in the same energy unit before subtracting.

Find maximum speed

Once $E_{k,max}$ is known, use $E_{k,max}=\frac12m_ev_{max}^2$, so $v_{max}=\sqrt{2E_{k,max}/m_e}$.

Common trap

Do not add the work function to the kinetic energy. The work function is the energy already spent escaping the surface.

Read the diffraction pattern

A beam of particles can produce diffraction or interference patterns after passing through a suitable crystal or narrow structure. The pattern is evidence that the particles have wave-like behaviour.

Use the experiment as evidence

Electron-diffraction experiments demonstrate wave properties of electrons. This complements the photon evidence from the photoelectric effect: matter and radiation can each show both particle-like and wave-like behaviour.

Connect to wavelength

The wave description is quantified by the de Broglie wavelength $\lambda=h/p$. A shorter wavelength generally requires a larger momentum.

Common trap

Rutherford alpha scattering is evidence for the nuclear structure of the atom, not the clearest evidence for matter waves. Use particle diffraction or interference when the question asks for wave properties of electrons.

Hold both descriptions

Quantum objects can show particle-like and wave-like behaviour. The observed property depends on the experiment: photoelectric emission and Compton scattering reveal particle-like transfers, while diffraction reveals wave-like behaviour.

Do not combine classical pictures blindly

Wave-particle duality is not a claim that an object is simultaneously a classical wave and a classical particle. It is a quantum description in which different measurements reveal complementary aspects.

Use scale carefully

For macroscopic objects the de Broglie wavelength is extremely small because momentum is large, so wave effects are not normally detectable. This is a practical limit, not a loss of the relation $\lambda=h/p$.

Common trap

Do not use “wave-particle duality” as an explanation without naming the observation it explains. Match the experiment to the property it demonstrates.

Use the de Broglie relation

Every moving particle has a wavelength inversely proportional to its momentum $p$. For non-relativistic motion, momentum may be written $mv$; choose the momentum relationship supported by the given data.

Worked example — moving electron

For $m_e=9.11\times10^{-31}\,\mathrm{kg}$ and $v=5.0\times10^6\,\mathrm{m\,s^{-1}}$, $p=mv=4.56\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}$. Hence $\lambda=h/p=(6.63\times10^{-34})/(4.56\times10^{-24})=1.5\times10^{-10}\,\mathrm{m}$, comparable with atomic spacing.

Choose the momentum form

For non-relativistic motion, use $p=mv$, so $\lambda=h/(mv)$. If kinetic energy is given, use $E_k=p^2/(2m)$ and $p=\sqrt{2mE_k}$.

Scale with accelerating voltage

For an electron accelerated from rest through potential $V$, $eV=E_k$, so $p\propto\sqrt V$ and $\lambda\propto1/\sqrt V$. Quadrupling $V$ halves the wavelength.

Common trap

Do not use $h/E_k$ as the wavelength. The denominator is momentum, not kinetic energy.

Model the collision

In Compton scattering, a photon transfers energy and momentum to an electron. The scattered photon has a changed direction and wavelength, while the electron recoils.

Use the evidence

The measured wavelength shift is evidence that photons carry momentum as well as energy. A wave-only model does not account for the collision-like transfer in the same way.

Track conservation laws

Analyse the photon–electron event using conservation of energy and momentum. The photon’s lost energy becomes kinetic energy of the recoiling electron, with the remaining photon energy determining its new wavelength.

Common trap

Do not describe Compton scattering as simple reflection. The photon transfers energy and momentum, so its wavelength generally changes.

Follow the energy transfer

After Compton scattering, the photon has transferred energy to the electron. Its final energy is lower than its initial energy.

Convert energy to wavelength

Since $E=hc/\lambda$, lower photon energy means larger wavelength. Therefore the scattered photon has a longer wavelength than the incident photon.

Keep the direction of change

The wavelength shift is zero only for no energy transfer. A stronger transfer to the electron produces a larger positive $\Delta\lambda$, subject to the scattering geometry.

Common trap

Do not infer a shorter wavelength from a lower photon energy. Energy and wavelength are inversely related.

Use the Compton equation

The wavelength shift is the scattered wavelength minus the incident wavelength. The electron is treated as initially at rest, and $\theta$ is the photon's scattering angle.

Worked example — $30^\circ$ scattering

Using $h/(m_ec)=2.426\times10^{-12}\,\mathrm{m}$, $\Delta\lambda=(2.426\times10^{-12})(1-\cos30^\circ)=3.25\times10^{-13}\,\mathrm{m}$. The shift is positive and depends on angle, not on the incident wavelength.

Check the limits

For $\theta=0^\circ$, $\Delta\lambda=0$. For back-scattering $\theta=180^\circ$, the shift is maximal at $2h/(m_ec)$. The shift is always non-negative for the usual scattering geometry.

Solve for angle

If $\Delta\lambda$ is given, rearrange to $\cos\theta=1-\frac{m_ec\Delta\lambda}{h}$, then take the inverse cosine and check that the result is physically allowed.

Common trap

Do not use the scattered wavelength itself as $\Delta\lambda$. The equation requires the difference $\lambda_f-\lambda_i$ and the scattering angle.

Define an isotope

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They share chemical identity but can have different physical properties.

Read the numbers

The proton number $Z$ stays fixed within an element. Different isotopes have different nucleon numbers $A$, so their neutron numbers $N=A-Z$ differ.

Common trap

Do not define isotopes only as atoms with different A and Z. The same proton number is essential; otherwise the atoms are different elements.

Define mass defect

A bound nucleus has less mass than the separated protons and neutrons that form it. The missing mass is the mass defect $\Delta m$, associated with the energy released when the nucleus forms.

Convert mass to binding energy

Use $E_b=\Delta mc^2$. If $\Delta m$ is in unified atomic mass units, the convenient conversion is approximately $931.5\,\mathrm{MeV}/c^2$ per u, giving energy directly in MeV.

Worked example — mass defect

If separated nucleons have total mass $4.0320\,\mathrm{u}$ and the nucleus has mass $4.0015\,\mathrm{u}$, then $\Delta m=0.0305\,\mathrm{u}$. Hence $E_b=(0.0305)(931.5)=28.4\,\mathrm{MeV}$. The positive result is the energy needed to separate the nucleus, and the same energy magnitude was released when it formed.

Interpret the sign

Binding energy is the energy required to separate the nucleons completely, and the same amount is released when the bound nucleus forms. It is positive as a required or released energy magnitude.

Common trap

Do not multiply a mass difference in u by c² again after using 931.5 MeV per u; that conversion already includes the mass–energy relation.

Read the curve

Binding energy per nucleon rises for light nuclei, reaches a broad maximum for medium-mass nuclei, then decreases gradually for very heavy nuclei. The curve compares average nuclear stability per nucleon, not total binding energy.

Predict energy release

Fusion of light nuclei can move products upward toward the maximum. Fission of very heavy nuclei can also move products upward. In either case, the increase in binding energy per nucleon corresponds to released energy.

Sketch the trend

Show a rise from the light-nucleus region, a maximum between roughly A=50 and A=100, and a slow decline for larger A. Exact numerical values are not required for the qualitative graph.

Common trap

Do not claim that the heaviest nucleus is most stable simply because it has the largest total binding energy. Use binding energy per nucleon to compare stability.

Use E=mc²

A change in rest mass corresponds to energy through $E=mc^2$. In a nuclear reaction, compare the total mass before and after to find the mass converted into released or absorbed energy.

Worked example — energy from a mass decrease

For $\Delta m=2.0\times10^{-12}\,\mathrm{kg}$, $\Delta E=(2.0\times10^{-12})(3.00\times10^8)^2=1.8\times10^5\,\mathrm{J}$. A smaller total rest mass of the products means this energy is released.

Compare energy yields

Energy released per reaction is proportional to mass converted. Energy released per unit mass also depends on the converted fraction: divide the energy from one reaction by the mass of fuel involved.

Track the system

Mass–energy equivalence applies to the mass difference of the defined reaction system. Do not compare only the total mass of the reactants without accounting for products.

Common trap

Do not confuse a large energy per reaction with a large energy per unit mass. The question’s denominator determines the comparison.

Describe the force

The strong nuclear force is attractive between nucleons at nuclear separations and has a very short range. It can bind protons and neutrons despite the electrostatic repulsion between protons.

Explain stability

At short distances the strong force can dominate, while the electromagnetic force is repulsive and long range. A stable nucleus requires the attractive nuclear interaction to overcome proton repulsion within the nucleus.

Keep the range distinction

The strong force does not act as a long-range force between separated nuclei. Its short range is why increasing nuclear size makes stability more difficult.

Common trap

Do not call the strong force repulsive between nucleons in the binding explanation, and do not confuse it with the weak nuclear interaction.

Treat each nucleus independently

Radioactive decay is spontaneous and random: the exact nucleus and instant of decay cannot be predicted. For a large sample, however, the fraction decaying per unit time follows a stable statistical law.

Separate random from law-like

Random decay does not mean the activity is random noise. The expected number of decays is predictable from the number of undecayed nuclei and the decay constant.

Apply the statistical model

You cannot identify which nucleus will decay next. But if two large samples contain the same nuclide and the second has twice as many undecayed nuclei, its expected activity is twice as large. Individual unpredictability and ensemble predictability coexist.

Common trap

Do not claim that randomness prevents prediction of half-life or activity. It prevents prediction of an individual decay, not the ensemble behaviour.

Alpha decay

Alpha decay emits a ${}^{4}_{2}\mathrm{He}$ nucleus. The parent’s nucleon number decreases by 4 and proton number decreases by 2.

Beta decay

In beta-minus decay, a neutron becomes a proton and an electron is emitted, so $A$ is unchanged and $Z$ increases by 1. In beta-plus decay, a proton becomes a neutron and a positron is emitted, so $A$ is unchanged and $Z$ decreases by 1.

Gamma decay

Gamma emission changes the nucleus from an excited state to a lower energy state. Neither $A$ nor $Z$ changes.

Common trap

Do not change A during beta decay, and do not treat gamma emission as a change of element.

Balance alpha decay

Write ${}^{A}_{Z}X\rightarrow{}^{A-4}_{Z-2}Y+{}^{4}_{2}\mathrm{He}$. Check both A and Z on the two sides.

Balance beta decay

For beta-minus use $^{A}_{Z}X\rightarrow{}^{A}_{Z+1}Y+{}^{0}_{-1}e+\bar{\nu}_e$. For beta-plus use $^{A}_{Z}X\rightarrow{}^{A}_{Z-1}Y+{}^{0}_{+1}e+\nu_e$. Gamma emission adds $^{0}_{0}\gamma$ after an excited daughter.

Balance a reaction

Conserve total nucleon number and charge. For uranium-235 absorbing a neutron and producing xenon-140 and strontium-94, the remaining nucleon number identifies the emitted neutrons.

Common trap

Do not omit the neutrino or antineutrino when the syllabus asks for a complete beta-decay equation, and do not balance A while leaving charge unbalanced.

Identify the neutral leptons

A neutrino $\nu_e$ and an antineutrino $\bar{\nu}_e$ are neutral, extremely low-mass leptons. They interact very weakly with matter, so they are difficult to detect directly.

Choose the correct particle

Beta-minus decay emits an electron and an electron antineutrino: $n\rightarrow p+e^-+\bar{\nu}_e$. Beta-plus decay emits a positron and an electron neutrino: $p\rightarrow n+e^++\nu_e$ (inside a nucleus).

Check lepton number

An electron has lepton number $+1$, so the accompanying antineutrino has $-1$. A positron has $-1$, so the accompanying neutrino has $+1$. Each beta reaction therefore keeps the initial total lepton number at zero.

Common trap

Do not swap the beta partners: $\beta^-$ pairs with $\bar{\nu}_e$, while $\beta^+$ pairs with $\nu_e$. The continuous beta spectrum is treated separately in the HL objective E.3.18.

Alpha radiation

Alpha particles are heavy and doubly charged. They interact strongly with matter, so they are highly ionizing but have low penetration and a short range in air.

Beta radiation

Beta particles are much lighter and singly charged. They are moderately ionizing and more penetrating than alpha particles, but can be deflected by electric and magnetic fields.

Gamma radiation

Gamma photons are neutral and travel at the speed of light in vacuum. They are weakly ionizing compared with alpha and beta, but have the greatest penetration.

Common trap

Do not rank penetration and ionization in the same order. The usual qualitative order is alpha > beta > gamma for ionization and gamma > beta > alpha for penetration.

Define activity

Activity is the number of nuclear decays per unit time, measured in becquerels: one Bq is one decay per second. As the number of undecayed nuclei falls, activity falls.

Use half-life steps

After each half-life, half of the remaining nuclei survive: $N=N_0(1/2)^n$, where $n=t/T_{1/2}$ is the number of half-lives elapsed.

Track count rate

If detector efficiency and background are unchanged, count rate is proportional to activity. Apply the same half-life scaling to the net count rate.

Common trap

Do not halve the original amount repeatedly without using the remaining amount, and do not confuse count rate with the number of nuclei when background is present.

Use integer half-lives

If activity changes from $A_0$ to $A$, use $A/A_0=(1/2)^n$ to find the number of half-lives $n$. For example, a fall to one-eighth means three half-lives.

Worked example — integer half-lives

A net count rate falls from $640\,\mathrm{s^{-1}}$ to $80\,\mathrm{s^{-1}}$ in 18 h. Since $80/640=1/8=(1/2)^3$, three half-lives elapsed. Therefore $T_{1/2}=18/3=6.0\,\mathrm{h}$.

Find the half-life

Once $n$ is known, divide the elapsed time by $n$: $T_{1/2}=t/n$. This is often quicker and clearer than starting with the exponential form.

Check the direction

A decay interval must reduce activity or count rate. If the calculated half-life or number of half-lives implies growth, revisit the ratio.

Common trap

Do not call a drop to one-eighth “one half-life”; half-life is the time for one factor of one-half.

Separate sample and background

A detector count rate can include decays from the sample plus background radiation. The measured rate is $R_{measured}=R_{sample}+R_{background}$.

Subtract before analysing

Estimate the background count rate with the source absent or from the long-time plateau, then calculate $R_{net}=R_{measured}-R_{background}$. Use the net rate for half-life comparisons.

Worked example — subtract, decay, restore

A detector reads $260\,\mathrm{Bq}$ with a $20\,\mathrm{Bq}$ background. The initial net rate is $240\,\mathrm{Bq}$. After four half-lives it is $240/16=15\,\mathrm{Bq}$, so the detector reads $15+20=35\,\mathrm{Bq}$.

Interpret a non-zero limit

If the measured rate approaches a non-zero constant, the remaining signal may be background radiation or a systematic detector contribution. The sample activity itself may have continued toward zero.

Common trap

Do not fit a half-life directly to a count rate that still contains background; the offset distorts the decay curve.

Use nuclear stability as evidence

Protons repel electrically, yet stable nuclei exist. This requires an additional attractive interaction between nucleons that is strong enough at nuclear distances.

Use scattering evidence

At high energies, deviations from Rutherford scattering show that the electrostatic model is incomplete at close range. The change is evidence for the strong interaction becoming relevant.

State the evidence precisely

Evidence supports a short-range strong force; it does not by itself provide a complete potential-energy curve or a long-range attraction between nuclei.

Common trap

Do not use “the nucleus is stable” as a complete explanation. State which observed fact requires an attractive force and how its range differs from electromagnetic repulsion.

Light stable nuclei

For small proton numbers, stable nuclei tend to have similar numbers of neutrons and protons, so $N\approx Z$.

Heavy stable nuclei

As $Z$ increases, proton–proton electromagnetic repulsion grows. Stable heavy nuclei therefore need extra neutrons to add strong-force binding without adding proton repulsion, so $N>Z$.

Read the stability band

The line of stable nuclides bends above $N=Z$ at larger $Z$. Nuclei on either side can decay toward the band, often through beta decay.

Common trap

Do not say every stable nucleus has more neutrons than protons. The approximation $N\approx Z$ is useful for light nuclei.

Read the heavy-nucleus trend

Above approximately $A\approx60$, binding energy per nucleon is broadly similar but slowly decreases as nucleon number increases. The increasing proton repulsion makes very heavy nuclei less tightly bound per nucleon.

Use the approximation carefully

“Approximately constant” does not mean identical for every nuclide. Use the trend to compare regions and to explain why fission of very heavy nuclei can release energy.

Common trap

Do not turn the broad plateau into a new maximum at large A. The main maximum is in the medium-mass region, followed by a gradual decline.

Use nuclear spectra

Alpha and gamma radiation can contain discrete energies. Since $E=hf$, fixed photon frequencies correspond to fixed energy differences between nuclear states.

Infer nuclear quantization

A line spectrum means the nucleus changes between allowed, discrete energy levels rather than a continuous range. Different transitions produce different alpha or gamma energies.

Use multiple routes

If two decay routes lead to the same final state, their energy relationships can reveal shared intermediate nuclear levels. Treat the routes as evidence about the level structure.

Common trap

Do not infer continuous nuclear energies from a continuous beta spectrum; beta continuity has a different explanation involving the neutrino.

Read the beta spectrum

Beta particles from one radioactive transition are emitted with a continuous range of kinetic energies, from nearly zero up to a maximum.

Use energy sharing

The beta particle and neutrino share the decay energy in variable proportions. The neutrino therefore explains why the beta particle does not always receive one fixed energy.

Common trap

Do not attribute the continuous spectrum to a continuous set of nuclear levels. Alpha and gamma line spectra show the contrasting discrete-level behaviour.

Use the exponential law

The number of undecayed nuclei after time $t$ is $N=N_0e^{-\lambda t}$. The same factor applies to the remaining mass when each daughter product is stable and the sample starts pure.

Worked example — arbitrary time

For $N_0=1.0\times10^{10}$, $\lambda=0.0126\,\mathrm{s^{-1}}$ and $t=60\,\mathrm{s}$, $N=(1.0\times10^{10})e^{-(0.0126)(60)}=4.70\times10^9$ nuclei. About $5.30\times10^9$ parent nuclei have decayed.

Find daughter amount

If every parent decay produces one daughter nucleus, the number formed is $N_{daughter}=N_0-N$. Define whether the question asks for remaining parent or accumulated daughter before substituting.

Control units

Use seconds when $\lambda$ is in $\mathrm{s^{-1}}$. Convert minutes, days or years before evaluating the exponential.

Common trap

Do not use $N_0e^{-\lambda t}$ for daughter amount directly; it gives the parent nuclei remaining.

Define lambda

The decay constant $\lambda$ is the probability per unit time that an individual undecayed nucleus will decay, in the small-time interval sense. Its unit is inverse time.

Use the approximation

When $\lambda\Delta t$ is very small, $\lambda\Delta t$ approximates the probability that a particular nucleus decays during $\Delta t$. The exact exponential law applies over longer intervals.

Separate lambda from activity

$\lambda$ describes a property of the nuclide. Activity $A$ describes the whole sample and depends on how many nuclei remain: $A=\lambda N$.

Common trap

Do not call lambda the number of decays per second of the whole sample; that is activity.

Use the activity relation

Activity is the decay rate: $A=\lambda N$. Combining this with the decay law gives $A=\lambda N_0e^{-\lambda t}$.

Worked example — number of nuclei

If $A=2.5\times10^5\,\mathrm{Bq}$ and $\lambda=1.8\times10^{-6}\,\mathrm{s^{-1}}$, then $N=A/\lambda=(2.5\times10^5)/(1.8\times10^{-6})=1.4\times10^{11}$ undecayed nuclei.

Find N first

For a sample mass $m$, find the number of nuclei using $N=(m/M)N_A$ before multiplying by $\lambda$. Use the isotopic molar mass and consistent units.

Track time dependence

Activity falls with the same exponential factor as the number of undecayed nuclei. If $t=0$, use $A_0=\lambda N_0$.

Common trap

Do not multiply lambda by sample mass directly. Convert mass to a number of nuclei first.

Use the half-life relation

Half-life and decay constant are related by $T_{1/2}=\frac{\ln2}{\lambda}$. A larger decay constant means a shorter half-life.

Worked example — convert time first

For $T_{1/2}=6.0\,\mathrm{h}=2.16\times10^4\,\mathrm{s}$, $\lambda=0.693/(2.16\times10^4)=3.21\times10^{-5}\,\mathrm{s^{-1}}$. The inverse-second unit matches a probability rate.

Convert units first

If half-life is given in days, hours or years but lambda is required in $\mathrm{s^{-1}}$, convert the time to seconds before dividing $\ln2$ by it.

Check the scale

The product $\lambda T_{1/2}$ should equal approximately 0.693. Use this as a quick unit and order-of-magnitude check.

Common trap

Do not use $1/\lambda$ as the half-life; it is the characteristic time and differs by the factor $\ln2$.

Model fission

A heavy nucleus can split into two lighter nuclei after absorbing a neutron, or spontaneously in an unstable state. The products have a greater binding energy per nucleon than the original heavy nucleus.

Track the release

The increase in total binding energy appears as kinetic energy of the fission products, neutron energy and radiation. The mass of the products is slightly smaller, with the mass difference converted to energy.

Worked example — use binding energy per nucleon

For $^{235}\mathrm{U}$ splitting into $^{89}\mathrm{Kr}$ and $^{144}\mathrm{Ba}$, use $B=A(B/A)$. With values $7.59$, $8.72$ and $8.27\,\mathrm{MeV}$ per nucleon, $E=[89(8.72)+144(8.27)]-235(7.59)=1.83\times10^2\,\mathrm{MeV}$. The products are more tightly bound, so this positive difference is released.

Understand fissile material

Enrichment increases the fraction of uranium-235 relative to uranium-238, making a sustained fission process more feasible.

Common trap

Do not say energy is created from nothing. It comes from the mass defect and the change in nuclear binding energy.

Start the chain

A fission event can emit neutrons. If one of them causes another fission, the process becomes a chain reaction.

Control the multiplication

A self-sustaining reactor requires, on average, one effective neutron from each fission to cause the next fission. Neutrons can instead escape, be absorbed by control rods, or be absorbed without causing fission.

Explain moderation

Fast neutrons are slowed by collisions with a moderator because low-energy neutrons have a higher probability of causing the relevant fission in this reactor model.

Common trap

Do not say every emitted neutron continues the chain. Losses and absorption determine whether the reaction dies out, stays critical or grows.

Follow neutrons, energy and radiation

Each reactor component controls a different part of the process. The moderator changes neutron energy; control rods change how many neutrons remain available; the heat exchanger moves thermal energy; shielding reduces radiation reaching people.

Track the energy path

Nuclear energy becomes kinetic energy of fission products, then internal energy of coolant, kinetic energy of steam and turbine, and finally electrical energy from the generator. The heat exchanger transfers energy; it does not create or regulate the fission reaction.

Common trap

Both moderator and control rods interact with neutrons, but their jobs differ: the moderator slows them, whereas control rods remove some by absorption.

Identify the products and the hazard

Fission produces two medium-mass fragments, free neutrons, radiation and energy. Many fragments are neutron-rich and radioactive; their decay produces ionizing radiation and continues to release thermal energy after the chain reaction stops.

Judge the management problem

A long half-life does not automatically mean a greater activity: for the same number of nuclei, a longer half-life means a smaller decay constant. Waste decisions must consider amount, radiation type, activity, heat, containment and timescale together.

Common trap

Do not assume shutting down the chain reaction makes spent fuel immediately safe. Unstable fission products continue to decay after neutron-induced fission has stopped.

Balance inward and outward effects

A stable main-sequence star is in hydrostatic equilibrium: inward gravitational force is balanced by outward thermal or radiation pressure. The star’s radius remains approximately stable while the balance holds.

Link the balance to fusion

Fusion in the core releases energy. The energy transported outward produces thermal and radiation pressure that resists gravitational collapse.

Predict imbalance

If the outward pressure falls, gravity compresses the star and raises core temperature; if pressure grows, the star expands until a new balance is reached.

Common trap

Do not write only “gravity balances pressure” without directions. State that gravity acts inward and thermal/radiation pressure acts outward, with fusion supplying the energy.

Use fusion as a stellar source

In stellar fusion, light nuclei combine to form more tightly bound nuclei. The mass difference is released as energy, which powers the star and supports its pressure balance.

Worked example — deuterium–tritium fusion

For $^2\mathrm{H}+{}^3\mathrm{H}\rightarrow{}^4\mathrm{He}+n$, the binding energies are $2(1.11)=2.22\,\mathrm{MeV}$, $3(2.83)=8.49\,\mathrm{MeV}$ and $4(7.07)=28.28\,\mathrm{MeV}$. Therefore $E=28.28-2.22-8.49=17.57\,\mathrm{MeV}$. The product is more tightly bound, so energy is released.

Follow nucleosynthesis

Fusion in stars can build elements up to iron through successive reactions. Elements heavier than iron are mainly formed in explosive environments and neutron-capture processes rather than ordinary core fusion.

Compare fusion and fission

Fusion can offer high energy per mass and potentially fewer long-lived waste products, but it requires extreme temperature and confinement conditions.

Common trap

Do not say all heavy elements are made by fusion in ordinary stars. The pathway changes around iron.

Use high temperature

Nuclei are positively charged and repel electrically. High core temperature gives them enough kinetic energy and speed to approach despite this repulsion.

Use high density

High density puts more nuclei into a given volume, increasing the collision frequency and the probability of close encounters.

Reach the strong-force range

Fusion requires nuclei to approach closely enough for the attractive strong interaction to act and for a bound product to form.

Common trap

Do not use surface temperature or star size as the direct fusion condition. The relevant evidence is high core temperature and density.

Start when core hydrogen is depleted

Reduced core fusion lowers outward pressure, so gravity contracts and heats the core. Hydrogen shell fusion then expands the outer layers: lower-mass stars become red giants, while high-mass stars become red supergiants.

Connect mass to lifetime

A more massive main-sequence star has more fuel, but its fusion rate and luminosity rise much more strongly. It therefore uses core hydrogen faster and has a shorter main-sequence lifetime.

Keep the path conditional

Mass controls the pathway; not every star becomes a supernova, and not every supernova leaves a black hole. A planetary nebula is expelled gas from a red giant, not a planet-forming stage.

Read the axes

An HR diagram compares luminosity with surface temperature. Temperature usually decreases from left to right, so the hottest stars are on the left.

Use constant-radius lines

From $L=4\pi R^2\sigma T^4$, a constant-radius line links luminosity and temperature. At the same temperature, greater luminosity means greater radius; at the same luminosity, the hotter star has the smaller radius.

Common trap

Do not read the temperature axis as increasing to the right, and do not identify a white dwarf from temperature alone; its low luminosity is also essential.

Use the Earth’s orbit

Observe a nearby star from opposite sides of Earth’s orbit at different times of year. Its apparent position shifts against distant background stars; half of the total angular shift is the parallax angle $p$.

Calculate distance

When $p$ is measured in arcseconds, distance in parsecs is $d=1/p$. For $p=0.25$ arcsec, $d=4$ pc, which can then be converted to light-years.

Know the range

Parallax is a geometric distance method and is most useful for relatively nearby stars. It does not use a star’s spectrum or brightness directly.

Common trap

Do not use the full annual position shift as p if the diagram shows the total displacement from one side of Earth’s orbit to the other.

Model a star as a spherical black body

Its luminosity $L$ is the total power radiated from surface area $4\pi R^2$ at absolute surface temperature $T$. This approximation connects observable luminosity and spectrum-derived temperature to radius.

Form the ratio

$\frac{R_1}{R_2}=\sqrt{\frac{L_1}{L_2}\left(\frac{T_2}{T_1}\right)^4}$. A hotter star can have a smaller radius at the same luminosity, while a very luminous cool star must be large.

Worked example — Canopus

For $L=10700L_\odot=4.12\times10^{30}\,\mathrm{W}$, $T=7400\,\mathrm{K}$ and $\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$, $R=\sqrt{L/(4\pi\sigma T^4)}=4.4\times10^{10}\,\mathrm{m}$. The large radius explains high luminosity despite a moderate surface temperature.

Check powers

Temperature enters to the fourth power and radius enters squared. Keep the temperature ratio in the inverse order shown before taking the square root.

Common trap

Do not use $R\propto L/T$; the correct scaling is $R\propto\sqrt L/T^2$.