Atomic identity and nuclear change: count what is conserved

MCAT content library / Chemistry & physics / Atomic identity and nuclear change: count what is conserved

Proton number identifies an element, electron count sets its charge, and nuclear energy adds a separate mass-accounting question.

About 20 minutes, including the worked examples and checks.

Useful foundations

• Work, energy, and power: keep the system account consistent

On this page

• Separate identity, nucleon count, and electrical charge
• Nuclear binding involves a different interaction
• Track what each radiation type carries away
• Worked application: complete a specified decay chain
• Use mass defect to calculate binding energy
• Check your reasoning
• Continue learning
• Sources

Separate identity, nucleon count, and electrical charge

The atomic number Z counts protons and identifies an element. The mass number A counts protons plus neutrons, so the neutron count is N = A − Z. Neither number tells you the electron count unless the atom’s charge is also specified. These distinctions connect the nucleus, isotopes, and nuclear transformations in AAMC category 4E.¹

Isotopes share Z but differ in N and therefore A. Ions differ in electron count from the corresponding neutral atom. Removing two electrons makes a 2+ ion without removing two protons or two nucleons. In charge units of the positive elementary charge, q = Z − number of electrons. Thus an atom with 12 protons has 12 electrons when neutral and 10 electrons as a 2+ ion.²

Card	Z	A	Charge	Neutrons	Electrons
P	12	25	0	13	12
Q	12	26	0	14	12
R	12	25	2+	13	10

P and Q are different isotopes of the same element. P and R contain the same isotope in different ionization states. Q and R differ in both isotope and charge. An observation of a changed mass-to-charge ratio would therefore need additional information before it could be assigned to isotope substitution alone.

A is an integer count, not a measured mass in kilograms or atomic mass units. The listed atomic weight of a natural element can be noninteger because it averages isotope masses using their abundances. Even the mass of one pure isotope is not generally exactly A atomic mass units. Binding energy contributes to that difference.

Nuclear binding involves a different interaction

Positively charged protons repel electrically. The short-range attractive nuclear interaction can bind protons and neutrons together at nuclear distances. Neutrons participate in nuclear binding without adding proton–proton electrical repulsion. This helps explain why heavier stable nuclei generally require more neutrons than protons, although a neutron-to-proton ratio alone is not a complete stability test.³

Ordinary chemical bonds involve electronic arrangements between atoms. Forming a molecule does not usually convert one element into another. A nuclear transformation can change proton number, neutron number, or nuclear energy. Nuclear and chemical binding should therefore not be treated as two names for the same bond.

Track what each radiation type carries away

In the nuclear notation ^A_ZX, the upper value is nucleon count and the lower value is proton number. An alpha particle is a helium-4 nucleus containing two protons and two neutrons. Its emission lowers the daughter nucleus’s A by four and Z by two. A beta-minus particle is an electron, but its emission is a nuclear conversion rather than the removal of a pre-existing orbital electron.⁴

Event	Emitted radiation	Daughter A	Daughter Z
Alpha emission	Helium-4 nucleus	A − 4	Z − 2
Beta-minus emission	Electron and antineutrino	A	Z + 1
Beta-plus emission	Positron and neutrino	A	Z − 1
Gamma emission	Photon	A	Z

For beta-minus decay, a neutron converts into a proton while an electron and an antineutrino are emitted. The daughter gains one unit of nuclear charge; the electron carries one negative unit away, so total electric charge is conserved. In beta-plus decay, a proton converts into a neutron with positron and neutrino emission. These weak-interaction processes retain the total nucleon count. A beta particle’s mass number is written as zero because it is not a nucleon, not because an electron or positron has zero mass.⁵

A gamma photon carries energy and momentum from an excited nucleus. The daughter has the same proton and neutron counts but lower nuclear energy. Its exact rest mass consequently changes slightly even though the integer mass number does not. Gamma emission must not be mistaken for the loss of a proton or for ordinary ion formation.

Worked application: complete a specified decay chain

The following generic chain is an original bookkeeping exercise; it does not claim that a named real isotope follows this decay route. Start with a nucleus having A = 150 and Z = 60. Suppose it emits an alpha particle, then undergoes beta-minus decay, then emits a gamma photon.

1. After alpha emission, A = 146 and Z = 58. The neutron count is 146 − 58 = 88.
2. After beta-minus decay, A remains 146 while Z rises to 59. One neutron has converted to a proton, so N falls to 87.
3. After gamma emission, A = 146, Z = 59, and N = 87 remain unchanged. Nuclear energy has decreased.

The full chain emitted one alpha particle and one electron, plus an antineutrino and a photon. The final nuclear charge is 59 positive units. Adding the alpha particle’s two positive units and the electron’s one negative unit gives 59 + 2 − 1 = 60, matching the starting charge. The final nucleus’s 146 nucleons plus the alpha particle’s four give the starting 150. These checks catch an incorrect beta sign immediately.

Use mass defect to calculate binding energy

The rest mass of a bound nucleus is lower than the sum of the rest masses of its separated, stationary constituent nucleons. Define the positive mass defect as Δm = Zm_p + Nm_n − m_nucleus. Its binding energy is E_b = Δmc²: the energy required to separate that nucleus into those free nucleons. Conversely, forming it can release that energy.³

For an original illustrative 20-nucleon example, suppose the sum of free nucleon masses is 20.160 u and the bound nuclear mass is 20.000 u. The mass defect is 0.160 u. Using the supplied conversion 1 u·c² = 931.5 MeV gives E_b = 149.04 MeV per nucleus, or 149.04/20 ≈ 7.45 MeV per nucleon. The input masses are simplified instructional values, not a measured isotope record.

Keep the mass convention consistent. If a problem supplies a neutral atomic mass rather than a nuclear mass, its electrons must be accounted for consistently; subtracting it from bare proton and neutron masses mixes different systems. Likewise, total binding energy and binding energy per nucleon answer different comparisons. An illustrative 40-nucleon nucleus with Δm = 0.300 u has 279.45 MeV total but only about 6.99 MeV per nucleon.

Neither the larger total binding energy nor this average alone specifies a radioactive half-life or proves that every possible decay channel is forbidden. A proposed transformation must also satisfy its complete energy and conservation conditions. Use A and Z to identify the products, and use consistent masses to evaluate an energy change.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Continue learning

Related lessons

• Work, energy, and power: keep the system account consistent
• Electromagnetic light: wavelength, photons, and polarization

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Sources

1. AAMC: C/P content category 4E

2. OpenStax University Physics 3: Properties of Nuclei

3. OpenStax Chemistry 2e: Nuclear Structure and Stability

4. OpenStax Chemistry 2e: Nuclear Equations

5. OpenStax University Physics 3: Nuclear Reactions