4.4 Atomic structure — coverage pack

Answer

• Use atom radius divided by nucleus radius.
• $1.25\times10^4$ times

Method: Calculate $(1.0\times10^{-10})/(8.0\times10^{-15})=0.125\times10^5=1.25\times10^4$. The atom's radius is therefore $12\,500$ times the nucleus's radius.

Answer

• Protons are positive and are in the nucleus.
• Neutrons have no charge and are in the nucleus.
• Electrons are negative and occupy energy levels around the nucleus.
• Nearly all the mass is in the nucleus.

Method: Organise the description by particle. Place protons and neutrons together in the central nucleus, then place electrons at energy levels outside it. Finish by linking the proton and neutron masses to the concentration of atomic mass in the nucleus.

Answer

• On absorption, the electron can move to a higher energy level, further from the nucleus.
• On emission, the electron can move to a lower energy level, closer to the nucleus.

Method: Track the direction of energy transfer. Absorbing radiation raises the electron's energy, so it can move to a level further from the nucleus. Emitting radiation lowers its energy, so it can move to a level closer to the nucleus.

Answer

• Atomic number $=17$
• Mass number $=37$
• Number of electrons $=17$

Method: The atomic number equals the proton number, so it is $17$. Add protons and neutrons for the mass number: $17+20=37$. A neutral atom has equal proton and electron numbers, so it has $17$ electrons.

Answer

• $13$ protons
• $14$ neutrons
• $10$ electrons

Method: The lower number gives $13$ protons. The neutron number is $27-13=14$. A $3+$ ion has lost three electrons, so its electron number is $13-3=10$.

Answer

• Atomic number $=29$
• The mass-$63$ isotope has $34$ neutrons.
• The mass-$65$ isotope has $36$ neutrons.
• Both have $29$ protons, so they are the same element.
• Their different neutron numbers make them isotopes.

Method: A $2+$ ion has two fewer electrons than protons, so the proton number is $27+2=29$. Hence the atomic number is $29$. The neutron numbers are $63-29=34$ and $65-29=36$. Element identity depends on proton number, which is unchanged between the two atoms; only the neutron number differs.

Answer

• Plum pudding model, nuclear model, electrons at specific distances, evidence for the neutron

Method: The electron discovery produced the plum pudding model. Alpha scattering then produced the nuclear model. Bohr next placed electrons at specific distances, and Chadwick's neutron evidence came later.

Answer

• Most particles passing straight through shows that most of an atom is empty space.
• Large deflections of a small fraction show that the positive charge is concentrated.
• The deflections also show that most mass is concentrated in a tiny central region.
• This evidence supports a small charged nucleus rather than spread-out positive charge.

Method: Link each observation to one structural inference. Unchanged paths need very little matter in most of the atom. Rare, large changes of direction require a concentrated region capable of a strong interaction. Together these contradict diffuse positive charge and support the nuclear model.

Answer

• The nuclear model explained scattering by concentrating positive charge and mass at the centre.
• It did not yet specify electrons at fixed distances or energy levels.
• Bohr adapted it by placing electrons at specific distances, supported by agreement between calculations and observations.
• Later evidence showed nuclear positive charge is made from whole units called protons.
• Chadwick's work supplied evidence for neutral particles in the nucleus.
• Therefore the model remained provisional and changed when new evidence improved its explanations.

Method: Evaluate by separating what the first nuclear model explained from what later evidence added. It accounted for scattering, but Bohr's electron arrangement, the proton and the neutron were later refinements. The sequence shows that a scientific model can be useful without being complete and is revised when evidence supports a better description.

Answer

• (i) alpha
• (ii) gamma
• (iii) beta

Method: Match composition before using penetration: a helium nucleus is alpha, an electromagnetic wave from the nucleus is gamma, and the nuclear electron produced in a neutron-to-proton change is beta.

Answer

• $P$ is alpha.
• $Q$ is beta.
• $R$ is gamma.
• Alpha, $P$, has the greatest ionising power.

Method: Order the radiations by penetration. Card stopping the least penetrating identifies alpha; aluminium stopping the intermediate radiation identifies beta; substantial transmission until lead identifies gamma. Ionising power follows the reverse order, so alpha is greatest.

Answer

• Alpha would be absorbed over a very short distance and is strongly ionising.
• Beta has an intermediate range and may not penetrate to the target.
• Gamma is much more penetrating, so it can reach the target.
• Gamma is the least ionising of the three, reducing ionisation along the path for a given exposure.
• Gamma is the most suitable of the three, with dose and shielding still controlled.

Method: Test each option against both requirements. Alpha fails the penetration requirement and has high ionising power. Beta penetrates further but may still be absorbed before the target. Gamma best reaches a deep target and has lower ionising power, although exposure must still be managed because it passes through tissue.

Answer

• Mass number of the particle $=4$ and atomic number $=2$.
• ${}^{4}_{2}\alpha$

Method: Subtract daughter numbers from parent numbers: $218-214=4$ and $84-82=2$. The missing radiation is therefore an alpha particle, ${}^{4}_{2}\alpha$.

Answer

• $A=131$
• $Z=54$

Method: Mass number balances as $131=A+0$, so $A=131$. Atomic number balances as $53=Z+(-1)$, so $Z=54$. Beta-minus emission therefore leaves mass number unchanged and raises the daughter's atomic number by one.

Answer

• After alpha emission: mass number $236$ and atomic number $94$.
• Each beta-minus emission leaves mass number unchanged.
• Two beta-minus emissions raise atomic number from $94$ to $96$.
• Final nucleus: ${}^{236}_{96}N$ for any suitable daughter symbol $N$.

Method: Alpha emission changes $(A,Z)$ from $(240,96)$ to $(236,94)$. Each beta-minus emission changes $Z$ by $+1$ without changing $A$. Two beta emissions therefore give $(236,96)$. The final element name is not needed; balanced numbers are sufficient.

Answer

• $640\rightarrow320\rightarrow160$ is two half-lives.
• Half-life $=5$ hours

Method: The activity halves twice in the $10$-hour interval. Divide the total time by two: $10/2=5$ hours.

Answer

• Initial net count rate $=400$ counts per minute.
• Final net count rate $=50$ counts per minute.
• $400\rightarrow200\rightarrow100\rightarrow50$ is three half-lives.
• Half-life $=6$ minutes

Method: Remove background from both readings: $420-20=400$ and $70-20=50$. The net count rate halves three times over $18$ minutes. Therefore the half-life is $18/3=6$ minutes.

Answer

• Initial net count rate $=800$ counts per minute.
• Final net count rate $=100$ counts per minute.
• $800\rightarrow400\rightarrow200\rightarrow100$ is three half-lives.
• Half-life $=12/3=4$ minutes.
• The decay of each individual nucleus is random.
• The number of decays recorded in equal short intervals therefore fluctuates around an expected value.

Method: Subtract background to obtain $830-30=800$ and $130-30=100$ counts per minute. This is three halvings in $12$ minutes, so the half-life is $4$ minutes. Half-life describes the statistical behaviour of many nuclei; random individual decays make short repeat counts vary.

Answer

• The instrument is irradiated.
• It does not become radioactive.

Method: The source only exposes the instrument to radiation; no radioactive atoms are transferred. This is irradiation, and the irradiated instrument does not itself become a radioactive source.

Answer

• Outside the body, alpha has a very short range and is stopped by skin.
• If inhaled, the source remains close to living cells.
• Alpha's high ionising power can then cause substantial cell damage.
• Prevent inhalation or spread, for example by using sealed containment and protective handling.

Method: Keep the radiation properties fixed and change only source position. Skin shields an external alpha source, but inhalation removes that shielding and creates continuing internal exposure. A precaution should prevent transfer of the radioactive material, such as sealed containment.

Answer

• Other scientists can check how exposure and illness were measured.
• They can identify uncontrolled variables or alternative causes.
• They can check the analysis and whether the sample is sufficient.
• The study can be repeated or compared with independent evidence.
• A reproducible association supports a conclusion more strongly than one unreviewed result and still does not by itself prove causation.

Method: Treat the reported association as evidence to test, not an automatic cause. Peer reviewers inspect the procedure, controls, sample and analysis. Publication also permits repetition. Agreement across well-controlled studies makes the conclusion more reliable; weaknesses or failure to reproduce it reduce confidence.

Answer

• Cosmic rays: natural
• Radioactive rock: natural
• Weapons-test fallout: man-made

Method: Cosmic radiation arrives from space and radioactivity occurs naturally in rock. Fallout is produced by human nuclear weapons testing, so it is man-made.

Answer

• Greater altitude can increase exposure to cosmic rays.
• Granite can contain more radioactive material than the other ground.
• Both locations receive real natural background radiation.
• Therefore non-zero readings without a test source are expected, not automatically detector error.

Method: Use one location factor from above the site and one from below it: cosmic-ray exposure can rise with altitude, and local geology can change rock radiation. These sources exist without any test source, so a non-zero count is physically expected.

Answer

• Cave-guide dose $=0.288\,\mathrm{mSv}$
• Airline-worker dose $=0.432\,\mathrm{mSv}$
• The airline worker's dose is $0.144\,\mathrm{mSv}$ greater.
• $0.432\,\mathrm{mSv}=4.32\times10^{-4}\,\mathrm{Sv}$

Method: Multiply weekly dose by $48$: $0.006\times48=0.288\,\mathrm{mSv}$ and $0.009\times48=0.432\,\mathrm{mSv}$. The difference is $0.432-0.288=0.144\,\mathrm{mSv}$. Divide millisieverts by $1000$ to obtain $0.000432\,\mathrm{Sv}=4.32\times10^{-4}\,\mathrm{Sv}$.

Answer

• The isotope with the $24$-year half-life
• Its activity falls much more slowly, so radioactive material persists for longer.

Method: A longer half-life means fewer successive halvings occur in a fixed time. The $24$-year isotope therefore remains radioactive over a much longer storage period.

Answer

• Their initial hazards from activity and radiation type are similar under the stated conditions.
• $A$ loses activity much faster because its half-life is short.
• $A$ therefore becomes a much smaller hazard relatively soon.
• $B$ retains significant activity for many years and needs longer secure storage.

Method: The given equal activity and radiation type make the initial comparison fair. Then apply half-life: many $3$-hour halvings occur in a short storage time, but very little of a $40$-year half-life passes. Thus $A$ declines rapidly while $B$ remains a persistent hazard.

Answer

• $C$ has the shorter half-life and so decays more rapidly initially for equal numbers of nuclei.
• Its activity and associated hazard also fall rapidly.
• $D$ decays more slowly initially.
• $D$ persists as a radioactive contaminant for much longer.
• Total risk also depends on radiation type, amount, exposure route and whether the material is removed or contained.

Method: Equal numbers allow a qualitative decay-rate comparison: the population with the shorter half-life loses a larger fraction per second, so $C$ is initially more active but fades quickly. $D$ has the longer-lived waste hazard. Do not declare one universally more dangerous because ionising power, penetration, contamination route and control measures are not specified.

Answer

• Gamma can penetrate out of the body to an external detector.
• Gamma is less ionising than alpha or beta, so it causes less cell damage for a comparable exposure.

Method: A tracer must be detected without surgery, which requires radiation able to leave the body. Gamma is penetrating and relatively weakly ionising, giving the two linked advantages.

Answer

• Ionisation can damage or kill tumour cells, controlling or destroying the unwanted tissue.
• Radiation reaching healthy tissue can damage healthy cells and may increase later health risks.
• A focused beam and controlled exposure can increase tumour dose while limiting healthy-tissue dose.
• Treatment is justified if the benefit of controlling the likely tumour growth outweighs the controlled risk to healthy tissue.

Method: Link the intended effect and side effect to ionisation. Radiation can kill unwanted tumour cells but can also harm nearby healthy cells. Then use the supplied consequence of no treatment and a control measure, such as careful aiming and limited exposure, to reach a balanced judgement.

Answer

• $K$ is the best candidate.
• $K$ emits penetrating gamma radiation that can reach an external detector.
• $K$ remains active for the four-hour investigation.
• Its $6$-hour half-life limits how long the patient remains a source after the test.
• $J$'s alpha radiation is poorly penetrating and strongly ionising, and its long half-life prolongs contamination risk.
• $L$ is detectable but its $30$-year half-life exposes the patient and creates a disposal hazard for unnecessarily long.

Method: Apply three criteria: detectability, sufficient duration and rapid removal of hazard. Alpha from $J$ is unsuitable for external detection and is strongly ionising internally. Both $K$ and $L$ emit detectable gamma, but $K$ lasts long enough for four hours and then decays relatively quickly. $L$ persists for decades without adding diagnostic benefit.

Answer

• A large unstable nucleus absorbs a neutron.
• Two or three neutrons are emitted.
• Gamma rays are emitted; energy or kinetic energy is also an acceptable second product.

Method: Begin with neutron absorption by the large unstable nucleus. After splitting, list products beyond the two daughter nuclei: emitted neutrons, gamma radiation and released kinetic energy.

Answer

• Generation 2: $3$ fissions
• Generation 3: $9$ fissions
• Generation 4: $27$ fissions

Method: Multiply by three in each generation because every fission supplies three successful neutrons: $1\times3=3$, then $3\times3=9$, then $9\times3=27$.

Answer

• The diagram shows an incoming neutron absorbed by a large unstable nucleus.
• It shows that nucleus splitting into two smaller nuclei and emitting two neutrons.
• Each emitted neutron is shown travelling to a different unstable nucleus and causing another fission.
• A controlled reactor prevents excess neutrons from causing further fissions, for example by absorbing them.
• Keeping about one continuing neutron per fission prevents the reaction rate and energy release from multiplying.

Method: Draw the process as a branching sequence: one neutron enters a large nucleus, the nucleus splits, and two outgoing neutron arrows lead to two further nuclei that split. Label the smaller nuclei and released neutrons. For control, explain that absorbing enough neutrons leaves about one successful neutron per fission, so successive generations do not grow.

Answer

• light
• heavier

Method: Fusion begins with two light nuclei and combines them into a nucleus heavier than either starting nucleus.

Answer

• Fusion joins two light nuclei to form a heavier nucleus.
• Fission begins with one large unstable nucleus.
• Fission splits it into two smaller nuclei of roughly equal size.
• Both processes release energy.

Method: Describe each process by its input and output pattern. Fusion changes two light nuclei into a heavier one; fission changes one large nucleus into two smaller ones. Finish with the shared outcome that energy is released.

Answer

• Two light nuclei are joining, which is the defining pattern of fusion.
• The joined product is a heavier nucleus than either reactant.
• The heavier nucleus having less mass shows that not all starting mass remains as nuclear mass.
• Some of the mass has been converted into energy.
• The emitted radiation carries energy away from the reaction.

Method: Use the change in number and size of nuclei to identify fusion: two light reactants become one heavier nucleus. Then apply the specified mass-energy principle. The small reduction in total mass corresponds to energy released from the process, and the observed radiation is one way that energy leaves.