SCIENCE · INVESTIGATE AND EXPLAIN

Conserve nucleon number and charge in nuclear change

Complete nuclear reaction equations while distinguishing necessary conservation from energetic possibility.

Learning outcome: Complete nuclear reaction equations while distinguishing necessary conservation from energetic possibility.

Starting recap

A is nucleon number and Z is charge number of a nucleus; an electron has A 0 and charge number−1.

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Try

In 238/92U→234/90Th+?, infer the omitted particle from two separate ledgers.

Open the experiment

Predict

Can a candidate conserve charge while violating nucleon count?

Your prediction and reason

Experiment

Check each offered candidate in alpha, beta− and gamma cases. Retain one failed and one passed pair of ledgers.

Balance nuclear mass number and charge

Worked teaching model · synthetic values · Conservation of nucleon number and electric charge is necessary, not sufficient for a possible reaction. Beta− includes an electron and antineutrino; no reaction rates, branching probabilities, reaction construction or radiation handling.

Reaction: 0=alpha,1=beta−,2=gamma,3=fusion,4=fission: 0 case Candidate omitted product: 0 choice

Reset keeps your prediction, captures and practice history.

Quantity	Calculated value	Unit
Initial mass number	238.0000	integer accounting
Final mass number	238.0000	integer accounting
Initial charge number	92.0000	integer accounting
Final charge number	92.0000	integer accounting

Both accounting checks pass.

238/92 U → 234/90 Th + ? Candidate: alpha (A4,Z2). Initial-minus-final residuals: A=0, Z=0. Charge and nucleon counts alone do not prove energy feasibility or branching probability.

The candidate selector uses these four entries for the selected reaction:

alpha (A4,Z2)
neutron (A1,Z0)
proton (A1,Z1)
gamma (A0,Z0)

Model boundary: Conservation of nucleon number and electric charge is necessary, not sufficient for a possible reaction. Beta− includes an electron and antineutrino; no reaction rates, branching probabilities, reaction construction or radiation handling.

All investigations are virtual. Do not obtain radioactive sources, expose eyes to bright beams, or build high-voltage apparatus. Notebook captures preserve selected recipes, not physical measurements.

Both accounting checks pass.

238/92 U → 234/90 Th + ? Candidate: alpha (A4,Z2). Initial-minus-final residuals: A=0, Z=0. Charge and nucleon counts alone do not prove energy feasibility or branching probability.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Two ledgers, not one guess

Nuclear notation uses mass number A to count nucleons and atomic number Z to count protons. For the offered reactions, compare total A and total electric charge before and after. In 238/92U→234/90Th+?, the missing values are A=4 and Z=2, consistent with an alpha particle. A neutron preserves charge here but fails the mass-number ledger.

Beta− decay requires more care. In 14/6C→14/7N+electron+antineutrino, A stays 14 and nuclear Z increases by one. The electron has charge number−1, so final total charge is 7−1=6. The antineutrino has no electric charge or nucleon number but is essential to the full physical account, including energy, momentum and lepton number. A beta electron is created in the weak transformation; it is not an atomic electron that was stored in the nucleus.

Excitation energy is not a new element

The gamma case changes an excited nuclear state to a lower state without changing A or Z. A photon carries energy and momentum but zero nucleon number and zero electric charge. Identical nuclide labels therefore do not imply no physical change: the internal energy has changed.

The supplied gamma example is a simplified selected transition. A real nuclide may have competing branches, internal conversion and multiple states. Passing the two ledgers does not establish a particular branch or probability.

Diagnose an incorrect proposal

Suppose a learner proposes a proton as the missing alpha-decay product. The final mass number is 234+1=235 rather than 238, and charge is 90+1=91 rather than 92. State both failures rather than saying only that the answer looks wrong. In the beta case, proposing a positron would give 7+1=8, also violating charge conservation for this specified direction.

Transfer: a new alpha equation

For 210/84Po→daughter+4/2He, the daughter has A=206 and Z=82. The arithmetic determines its nuclide label once an element table is consulted. It does not establish the energy released or show that this decay is the only possible mode.

Necessary is not sufficient

A balanced reaction can still need energy input, be forbidden by other constraints or be extremely unlikely. Mass-energy and momentum must also balance. This exercise asks for symbolic evidence only. Never obtain, handle or investigate radioactive sources as a home activity; real radiological work requires qualified facilities and supervision.

Source and authorship

Reference for the physical relationship. Source checked 30 October 2026. The explanations, numerical cases, questions and diagrams here are original; reference material is not copied. Exact h,c,e follow NIST SI definitions.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why beta− increases the daughter nuclear charge and why conservation alone cannot determine a decay rate.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Compare matter wavelength with an observational scale Next: Predict an ensemble without assigning a lifetime to one nucleus

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