GENERAL CHEMISTRY · NUCLEAR REACTIONS

Close both ledgers.
Then choose the model.

Learn all five official topics through nuclide, A/Z, mass–energy, decay, radiation, and terminology ledgers, then retrieve with twenty original five-choice questions.

Start the lessons ↓Open all problems →

5guided lessons

20practice questions

15study objectives

5choices per item

A repeatable nuclear-chemistry routine

Inventory. Conserve. Model. Bound.

01Inventory the system
Name every nuclide, particle, mass convention, time basis, unit, process clue, and source location supplied.
02Conserve both ledgers
Track mass number A and atomic number Z independently; keep ionic charge in the separate electron inventory.
03Select the model
Use the stated equation direction, compatible mass–energy relation, single-radionuclide decay model, radiation profile, or process definition.
04Bound the conclusion
Check total versus per-particle basis, time and energy units, activity versus dose, internal versus external source, and whether the evidence is unique.

Five prerequisite-aware lessons

Let the nuclear inventory choose the claim.

Each lesson pairs two semantic ledgers with a fully worked example and closed-note retrieval before multiple-choice practice.

LESSON 1 · 23 MIN

Study + retrieve

Balance the two-number ledger

Read nuclide and particle notation, inventory protons, neutrons, and electrons, and balance decay or bombardment equations by conserving mass number and atomic number independently.

ESSENTIAL QUESTIONWhat belongs in the nuclear ledger, which side contains each participant, and do both A and Z close?

Nuclide and equation ledgers

Nuclide and common-particle inventory
Entry	Mass ledger A	Atomic ledger Z	Boundary
ᴬZX	p + n	p	Ion charge changes electrons only
α / ⁴₂He	4	2	Two protons + two neutrons
β⁻ / β⁺	0	−1 / +1	Not a stored nuclear electron
γ / n / p	0 / 1 / 1	0 / 0 / 1	Use the stated participant

Two-number equation audit
Check	Reactants	Products
Mass number	Σ coefficient × A	Must equal reactant total
Atomic number	Σ coefficient × Z	Must equal reactant total
Direction	Captured or bombarding particle	Emitted particle
Identity	Use the balanced Z to name the element

NUCLEAR CHEMISTRY LEDGERS · CAPTIONS INCLUDED

Inventory before reacting

For ᴬZX, A counts protons plus neutrons and Z counts protons; neutrons equal A − Z. Ionic charge changes the separate electron inventory, never the nuclear A/Z ledger.

A = protons + neutrons
Ion charge changes electrons, not Z

Conserve A and Z independently

Add upper-left mass numbers and lower-left atomic numbers on each side, including coefficients. The missing pair identifies a particle or a daughter element even when the element symbols change.

ΣAreactants = ΣAproducts
ΣZreactants = ΣZproducts

Preserve process direction

An emitted particle belongs among products; a captured or bombarding particle belongs among reactants. Close the ledger without inventing neutrinos, intermediates, or mechanisms absent from the stated convention.

Emission: product side
Capture: reactant side

Worked example

Complete ¹⁴₇N + ⁴₂He → ¹⁷₈O + ?.

1
Reactants total A = 18 and Z = 9.
2
The known oxygen product contributes A = 17 and Z = 8, leaving A = 1 and Z = 1.
3
The missing ¹₁p is a proton; including it makes both sides total A = 18 and Z = 9.

ConclusionThe two-number difference—not a familiar chemical formula—identifies the missing proton and verifies the bombardment equation.

Close the notes first

Retrieve the ledger.

01What changes when a neutral atom becomes a 1− ion?

Its electron count increases by one; A, Z, protons, and neutrons do not change.

Ionic charge is an extranuclear electron ledger.

02What pair represents an alpha particle in a nuclear equation?

A = 4 and Z = 2, written ⁴₂He or α.

An alpha particle contains two protons and two neutrons.

03Where is a captured neutron written?

On the reactant side as ¹₀n.

Capture adds the particle to the starting nuclear inventory.

LESSON 2 · 25 MIN

Study + retrieve

Keep mass and energy on one basis

Calculate mass defect and binding energy from a consistent supplied mass convention, convert compatible units, and distinguish total binding energy from binding energy per nucleon.

ESSENTIAL QUESTIONDo the free and bound masses describe the same particle inventory, and what energy basis does the requested unit represent?

Mass-defect and energy ledgers

Consistent free-versus-bound mass conventions
Convention	Free inventory	Bound inventory
Nuclear masses	Zmₚ + (A − Z)mₙ	m(nucleus)
Neutral atomic masses	Zm(¹H) + (A − Z)mₙ	m(neutral atom)
Mass defect	Δm = free mass − bound mass
Do not mix	Bare proton	Neutral atom without correction

Binding-energy unit and basis audit
Input	Relationship	Output or basis
Δm in kg	E = Δmc²	J for stated system
Δm in u	Multiply by supplied MeV/u	MeV for stated system
Total energy	Divide by A	Energy per nucleon
Comparison	Use normalized evidence	Not an exact half-life

NUCLEAR CHEMISTRY LEDGERS · CAPTIONS INCLUDED

Choose one mass convention

With nuclear masses use free protons plus neutrons minus nucleus mass. With neutral atomic masses, the common introductory shortcut uses hydrogen-atom masses plus neutrons minus neutral-atom mass so electron inventories cancel consistently.

Free inventory − bound inventory
Do not mix proton and neutral-atom masses

Match mass to energy units

A kilogram mass defect with c in meters per second gives joules through E = Δmc². A mass defect in u can use a compatible supplied equivalence such as 931.5 MeV per u.

kg + m/s → J
u × MeV/u → MeV

Normalize only when asked

Total binding energy describes the entire nucleus; dividing by A gives energy per nucleon. The normalized value supports tightness comparisons but does not reveal an exact half-life or unique decay path.

Per nucleon = total ÷ A
Tight binding is not a decay clock

Worked example

For deuterium, use m(¹H) = 1.007825 u, mₙ = 1.008665 u, and m(²H) = 2.014102 u. Find Δm and binding energy with 931.5 MeV/u.

1
Use the neutral-atomic convention on both sides: one hydrogen atom plus one neutron minus one neutral deuterium atom.
2
Δm = 1.007825 + 1.008665 − 2.014102 = 0.002388 u.
3
E = (0.002388 u)(931.5 MeV/u) = 2.224 MeV for the deuterium nucleus.

ConclusionThe positive defect is free-system mass minus bound-system mass; the compatible u-to-MeV factor closes the unit ledger.

Close the notes first

Retrieve the ledger.

01Why should a bare-proton mass not be mixed with a neutral-atom mass?

The two sides then contain different electron inventories.

A mass defect comparison must represent the same particles before and after binding.

02What energy unit follows directly from kilograms and c in meters per second?

Joules.

kg·m²/s² is the joule.

03How is binding energy per nucleon calculated?

Divide total binding energy by mass number A.

A counts the nucleus’s nucleons.

LESSON 3 · 24 MIN

Study + retrieve

Model the parent and the clock

Predict parent–daughter changes, connect half-life with decay constant, and calculate remaining fraction, elapsed time, or activity within a stated single-radionuclide model.

ESSENTIAL QUESTIONWhich A/Z change identifies the daughter, and are the time, amount, and activity quantities compatible with one decay model?

Daughter-change and decay-clock ledgers

Parent-to-daughter A/Z changes
Mode	ΔA	ΔZ	Evidence boundary
Alpha	−4	−2	Emits ⁴₂He
Beta-minus	0	+1	Emits ⁰₋₁e
Positron / EC	0	−1	A/Z alone cannot distinguish
Gamma	0	0	Identity unchanged

Exponential amount and activity model
Quantity	Relationship	Unit or condition
Decay constant	λ = ln 2/t₁⁄₂	Inverse time
Remaining fraction	N/N₀ = e⁻ˡᵃᵐᵇᵈᵃᵗ	Compatible time units
Repeated halves	N/N₀ = (1/2)ᵗ⁄ᵗ¹⁄₂	Closed single-nuclide model
Activity	A = λN	Bq = transformations/s

NUCLEAR CHEMISTRY LEDGERS · CAPTIONS INCLUDED

Apply the stated daughter change

Alpha changes A by −4 and Z by −2; beta-minus changes Z by +1; positron emission or electron capture changes Z by −1; gamma changes neither ledger.

Alpha: −4, −2
β⁻: 0, +1 · β⁺/EC: 0, −1

Connect λ and half-life

For one radionuclide, λ = ln 2/t₁⁄₂ and N/N₀ = e⁻ˡᵃᵐᵇᵈᵃᵗ. Preserve one time unit between elapsed time, half-life, and the reciprocal unit of λ.

λ has inverse-time units
One half-life leaves one-half

Separate amount, activity, and dose

Activity A = λN counts decays per time, so activity and parent amount fall by the same fraction when λ is constant. Activity alone does not determine absorbed or biologically weighted dose.

1 Bq = 1 decay/s
Activity is not Gy or Sv

Worked example

A radionuclide has an 8.0-day half-life. What fraction remains after 20 days?

1
The elapsed time is 20/8.0 = 2.5 half-lives.
2
Use the continuous repeated-half model: N/N₀ = (1/2)²⋅⁵.
3
The remaining fraction is 0.1768, or 17.7%.

ConclusionRadioactive decay removes the same fraction per half-life, not the same absolute number of nuclei.

Close the notes first

Retrieve the ledger.

01How does alpha decay change A and Z?

A decreases by 4 and Z decreases by 2.

The emitted alpha particle carries A = 4 and Z = 2.

02What is λ when t₁⁄₂ is known?

λ = ln 2/t₁⁄₂.

The exponential reaches one-half when λt = ln 2.

03Can activity in Bq be converted directly to absorbed dose in Gy?

No, not without additional energy, geometry, absorption, and sample information.

Activity counts transformations; dose measures deposited energy per mass.

LESSON 4 · 21 MIN

Study + retrieve

Identify the particle and its reach

Identify common nuclear particles from notation, compare qualitative ionization and penetration, and infer only those decay pathways determined by the supplied A/Z change.

ESSENTIAL QUESTIONWhat A, Z, and charge does the particle contribute, and what can its radiation profile actually establish?

Particle identity and radiation ledgers

Particle symbols, nuclear contribution, and charge
Particle	Symbol	A / Z	Electric charge
Alpha	⁴₂He	4 / 2	+2
Beta-minus / positron	⁰₋₁e / ⁰₊₁e	0 / −1 or +1	−1 / +1
Gamma	⁰₀γ	0 / 0	0
Neutron / proton	¹₀n / ¹₁p	1 / 0 or 1	0 / +1

Introductory ionization, penetration, and shielding
Radiation	Ionization	Penetration	Initial shielding cue
Alpha	High	Low	Paper or skin; internal source matters
Beta	Moderate	Moderate	Plastic or thin metal
Gamma	Lower per track	High	Dense, thick material
Exact protection	Requires energy, material, thickness, geometry, and exposure data

NUCLEAR CHEMISTRY LEDGERS · CAPTIONS INCLUDED

Read identity from the ledger

Alpha, beta-minus, positron, gamma, neutron, and proton have characteristic A/Z pairs. Electric charge and nuclear atomic-number contribution are related in the common symbols but remain conceptually distinct ledgers.

β⁻ = ⁰₋₁e · β⁺ = ⁰₊₁e
n = ¹₀n · p = ¹₁p

Compare interaction profiles

Alpha is strongly ionizing and weakly penetrating, beta is intermediate, and gamma is highly penetrating. Introductory shielding choices follow type, but exact protection also depends on energy, thickness, and geometry.

Alpha: high ionization, low penetration
Gamma: dense, thick shielding

Admit underdetermination

Parent and daughter A/Z changes can identify alpha or beta-minus patterns, but ΔA = 0 and ΔZ = −1 fits both positron emission and electron capture. Particle placement or added evidence is required to choose uniquely.

Infer from ΔA and ΔZ
β⁺ versus EC needs more evidence

Worked example

Compare an external gamma source with an inhaled alpha emitter using only introductory qualitative evidence.

1
Gamma radiation is more penetrating, so an external source can reach tissue through the body surface and generally calls for dense, thick shielding.
2
Alpha radiation has low external penetration but high ionization over a short path.
3
Once an alpha emitter is inhaled, skin no longer provides the same separation, so low penetration does not imply low internal hazard.

ConclusionRadiation hazard depends on both particle behavior and source location; penetration rank alone is not a complete risk rank.

Close the notes first

Retrieve the ledger.

01What particle is written ⁰₊₁e?

A positron, or beta-plus particle.

It has zero nucleon count and a +1 charge/atomic-number contribution.

02Which of alpha, beta, and gamma is generally most penetrating?

Gamma.

Gamma photons interact less strongly per path than charged alpha or beta particles in the introductory comparison.

03Does ΔA = 0 and ΔZ = −1 identify one unique decay mode?

No; positron emission and electron capture share that daughter change.

Their particle placement differs even though the parent–daughter A/Z difference matches.

LESSON 5 · 22 MIN

Study + retrieve

Name the process and the measured quantity

Use nuclear vocabulary precisely, distinguish fission, fusion, transmutation, and chain reactions, and keep activity, dose, and radioisotope-use claims within supplied evidence.

ESSENTIAL QUESTIONIs the claim about nuclide identity, nuclear process, transformation rate, deposited energy, biological weighting, or a practical isotope use?

Process and radiation-quantity ledgers

Nuclide and nuclear-process vocabulary
Term	Required evidence	Do not substitute
Isotopes	Same Z, different A	Same A alone
Fission	Heavy nucleus splits	Fusion
Fusion	Light nuclei combine	Any energy release
Chain reaction	Products induce more events	Several unrelated events

Activity, dose, and use boundaries
Unit or use	Quantity	Boundary
Bq / Ci	Activity	1 Ci = 3.7 × 10¹⁰ Bq
Gy	Absorbed energy per mass	1 Gy = 1 J/kg
Sv	Biologically weighted dose	Needs weighting evidence
Tracer	Detectable labeled pathway	Activity alone does not give dose

NUCLEAR CHEMISTRY LEDGERS · CAPTIONS INCLUDED

Name the nuclear actors

A nuclide is one proton–neutron inventory; isotopes share Z but differ in A. A radionuclide is unstable in the stated context, the parent precedes a decay step, and the daughter follows it.

Same Z, different A → isotopes
Parent → decay → daughter

Classify the event evidence

Fission splits a heavy nucleus, fusion combines light nuclei, and transmutation changes one element into another. A chain reaction additionally requires released particles capable of inducing further events.

Split heavy → fission
Combine light → fusion

Keep radiation quantities distinct

Bq and Ci measure activity, Gy measures absorbed energy per mass, and Sv represents biologically weighted dose. A tracer use follows detectability and chemistry; it does not make activity a dose unit.

1 Ci = 3.7 × 10¹⁰ Bq
Bq/Ci ≠ Gy/Sv

Worked example

Classify a heavy-nucleus split that releases neutrons able to trigger more splits, then name what 3.7 × 10¹⁰ events per second represents.

1
A heavy nucleus splitting is fission.
2
Released neutrons that can induce additional fissions provide chain-reaction evidence.
3
A transformation rate of 3.7 × 10¹⁰ s⁻¹ is activity: 3.7 × 10¹⁰ Bq, equal to 1 Ci.

ConclusionProcess, propagation, and measured quantity are separate claims; each needs its own evidence.

Close the notes first

Retrieve the ledger.

01How are isotopes related?

They have the same atomic number Z but different mass numbers A.

They are nuclides of the same element with different neutron counts.

02What extra evidence turns repeated fissions into a chain reaction?

Released particles, commonly neutrons, can induce additional fissions.

Several unrelated fissions do not by themselves establish propagation.

03What do Gy and Sv represent?

Gy is absorbed energy per mass; Sv is biologically weighted dose.

Neither is a unit of decay rate.

All twenty Nuclear Reactions problems

Conserve the evidence before selecting it.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

20 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions are draft and have not been calibrated to the official score scale. Use each explanation to repair the exact A/Z, direction, mass convention, energy basis, decay clock, activity, radiation, or terminology decision.

DAT TRAIN does not claim topic quotas because the official manual does not publish them. Chemical Kinetics owns empirical chemical rate-law selection; this domain applies first-order mathematics only to radioactive decay. Activity does not become absorbed or weighted dose without additional evidence.