Atomic nucleus and radioactivity

Add the masses of free protons and neutrons and compare the sum with the mass of a nucleus made from the same particles. The nucleus is lighter, although the number of nucleons is unchanged. We will connect this result with energy and distinguish a predictable composition from an unpredictable decay time.

An atom's positive charge is concentrated in a small nucleus surrounded by electrons. Protons are positively charged and neutrons are electrically neutral; both are called nucleons. Ordinary nuclei with several nucleons contain both kinds, but hydrogen-1 provides an important limiting case: one proton and no neutrons. Determine the composition of the particular nucleus; do not assume that every atom must contain a neutron by definition.

To count the particles, use Z for the number of protons, N for the number of neutrons and the mass number A=Z+N. Thus N=A-Z. These dimensionless integers are not measured masses. Z identifies the element; nuclei with the same Z but different N are isotopes. Adding a neutron increases A by 1 without changing the element. Adding or removing an electron changes the atom's charge, but not these nuclear numbers.

Read ^{4}_{2}\mathrm{He} using the same counting rule: four nucleons, including two protons, so the other two are neutrons. This is an alpha particle, a helium nucleus. In alpha decay it carries away A=4 and Z=2, reducing the corresponding numbers of the residual nucleus. Checking the totals before and after demonstrates conservation of nucleon number and electric charge. The particle written here has no electron shell.

The notation ^{0}_{-1}\mathrm{e} represents the electron emitted in β⁻ decay: zero nucleons and a negative charge measured in units of |e|. A neutron in the nucleus becomes a proton, emitting an electron and an antineutrino. A is therefore unchanged, while Z increases by 1. The electron was not hidden inside as another nucleon: it is produced in the transformation. Omitting the antineutrino would leave both the products and the energy distribution incompletely described.

An excited nucleus can release energy without changing its numbers of protons and neutrons. The emitted photon is written ^{0}_{0}\gamma: it carries neither nucleons nor electric charge. A and Z stay the same, while the nucleus moves to a lower energy state. Comparison with alpha and beta decay shows why saying only that a nucleus emitted radiation is insufficient to predict a change of element: the type of transformation matters.

Measured masses are conveniently expressed in the atomic mass unit u, also written а.е.м. It equals one twelfth of the mass of a free, neutral carbon-12 atom at rest in its ground state; 1 u≈1,66·10⁻²⁷ кг. This is a mass unit, whereas A is a particle count. For some nuclei the numbers may be close, but the mass number cannot replace the precise mass: the small mass difference is exactly what is needed to find binding energy.

Use the tabulated masses m_p\approx1{,}00728\,u, m_n\approx1{,}00866\,u, m_e\approx0{,}00055\,u. The subscripts p, n and e identify the proton, neutron and electron. The proton and neutron have similar, but unequal, masses. Rounding both to 1 u too early may distort or erase the small difference being sought. In mass-defect calculations, keep enough digits before subtracting, then round the result to match the precision of the data.

Tables often give the mass m_ат of a neutral atom, while the binding-energy formula needs the nuclear mass. The atom contains Z electrons: neglecting their binding energy gives m_{\text{яд}}\approx m_{\text{ат}}-Zm_e. Use the same unit for every mass. This is an approximation because the electron shell is also bound and contributes to the energy. An ion has a different number of electrons; subtracting Z electron masses without checking neutrality would be wrong.

Forming a bound nucleus from free nucleons releases energy, leaving the resulting system with less rest energy. Its mass is therefore smaller than the sum of the free-particle masses; the difference is called the mass defect Δm. Separating the same particles completely requires the binding energy. No nucleons have vanished: their state of interaction has changed. Taking apart joined pieces is a useful analogy for the energy cost, but nuclear binding has its own physical origin.

First find the mass of the free particles, then subtract the nuclear mass: \Delta m=Zm_p+Nm_n-m_{\text{яд}}. Here N still means the neutron count in one nucleus. For a state bound against complete separation into nucleons, Δm>0. At Δm=0 there is no corresponding reduction in rest energy. All terms must use the same units and refer to the nucleus, rather than accidentally including the electrons of a neutral atom.

The energy required to separate a nucleus completely is E_{\text{св}}=\Delta m c^2. With Δm in kilograms and c≈3·10⁸ м/с, the result is in joules. Doubling the mass defect doubles the binding energy; the enormous factor c² gives even a small mass a substantial energy equivalent. This does not say that matter disappears: it compares the total energies of two states of the same particles. Binding energy per nucleon is useful for comparison, but alone does not determine whether a particular decay is possible.

If Δm is expressed in atomic mass units, use E_{\text{св}}=\Delta m\cdot931\,\frac{\text{МэВ}}{\text{а.е.м.}} directly. MeV means a million electronvolts: 1 МэВ≈1,60·10⁻¹³ Дж. The mass units cancel, leaving energy; this is the same Δmc² relation. The factor 931 is rounded, so the answer is approximate. Do not label the result of this multiplication as joules: each choice of units must remain consistent.

An unstable nucleus can spontaneously change into another nucleus or state, emitting particles or radiation. This is radioactive decay. The exact decay time of a particular nucleus cannot be predicted in advance; identical preparation does not give every nucleus the same schedule. For a large collection of nuclei of one kind, however, the average fraction remaining can be predicted. A statistical law therefore describes the population reliably without predicting each individual event exactly.

The half-life T is the time after which, on average, half the original undecayed nuclei of a given kind remain. After the next T, half of that remaining group survives; another half of the original number is not subtracted. A nucleus that survived one half-life is therefore not obliged to decay immediately afterwards. T describes population statistics; individual nuclei do not contain identical timers set to T.

For an initial number N₀, the law is N=N_0\,2^{-t/T}. N now counts the original nuclei still remaining in the sample, not the neutrons in one nucleus. At t=0 the result is N₀; at t=T it is half, and at t=2T one quarter. Express t and T in the same units so that the exponent is dimensionless. Doubling the initial number at the same elapsed time doubles the expected remainder, without changing the surviving fraction.

The model assumes identical nuclei of one kind, with no new parent nuclei being added. It counts only that kind of nucleus: decay products may stay in the sample, so the formula does not give its total particle count or total mass. Random departures from the mean are especially noticeable for small populations. At large t, the expected N can be fractional: this is an average over many repetitions, not a physical fraction of a nucleus in one container.

A sample initially contains 1000 identical unstable nuclei. The half-life is 5 h. How many original nuclei are expected to remain after 10 h?

The time t = 10 ч and half-life T = 5 ч use the same unit. Their ratio t/T = 10/5 = 2 shows that two half-lives have elapsed.

After the first half-life, we expect 1000/2 = 500 original undecayed nuclei.

After the second, take half the remainder: 500/2 = 250. The formula gives the same result: N = 1000·2⁻² = 250.

This is the mean expected number. The exact remainder in one sample can differ randomly; decayed nuclei become products rather than disappearing on a shared timetable.

On average, 250 original undecayed nuclei remain, one quarter of 1000. This is a statistical expectation, not an exact count guaranteed in every experiment.

Nucleon count, atomic mass and nuclear mass are different quantities; distinguishing them is essential when calculating mass defect.

Binding energy corresponds to the lower mass of the bound state and must be calculated with consistent units.

The half-life law predicts the fraction of original nuclei remaining in a population, not the lifetime of one nucleus or the total mass of the products.

You have completed the course: connect observations to a model, check its conditions, and only then calculate.