A particular radioactive nucleus has a half-life of 10 years. A sample nucleus has existed for 30 years without decaying. Compared to a freshly created nucleus of the same isotope, what is the probability that this 'old' nucleus decays in the next second?
AHigher — the old nucleus has built up energy and is now overdue for decay
BLower — the fact that it survived 30 years suggests it is an unusually stable specimen
CIdentical — each nucleus has the same constant decay probability per unit time regardless of age
DIt depends on the decay mode — alpha decay is age-independent, but beta decay is not
Radioactive decay is a quantum process: each nucleus has a fixed, age-independent probability λ of decaying per unit time. A nucleus that has 'survived' 30 years is not 'overdue' — there is no internal clock counting down, no built-up pressure, and no memory of past survival. This is the quantum analogue of a fair coin: heads on the last flip tells you nothing about the next flip. The exponential decay law follows directly from this constant-probability assumption. Options A and B both commit the 'gambler's fallacy' at the nuclear level — a common misconception that the probabilistic nature of decay eliminates.
Question 2 Multiple Choice
Why is alpha decay the dominant spontaneous decay mode for the heaviest nuclei (mass number A > 200), rather than single-proton or single-neutron emission?
AAlpha particles have the highest charge and therefore the greatest ability to tunnel through the Coulomb barrier
BAlpha particles are the largest particles that can escape the nucleus; heavier fragments are always trapped
CThe alpha particle (⁴He) is doubly magic and extraordinarily tightly bound, so emitting it releases more energy (Q > 0) than emitting individual nucleons for these heavy nuclei
DHeavy nuclei are neutron-rich, and alpha emission is the only way to simultaneously remove both protons and neutrons
The key is the Q-value. The alpha particle is one of the most tightly bound nuclei per nucleon (doubly magic: Z=2, N=2). For very heavy nuclei on the right side of the binding energy curve, total binding energy increases when the heavy nucleus loses four nucleons as an alpha particle — the daughter is more tightly bound per nucleon than the parent, so Q > 0. Emitting a single proton or neutron yields a much smaller energy release because the individual nucleon is not specially stabilized. Option D has some truth (heavy nuclei are often above the stability line), but misidentifies the mechanism — alpha decay is favored by energetics, not by the need to remove equal numbers of each nucleon type.
Question 3 True / False
Two radioactive samples of different isotopes both contain exactly 10²⁴ atoms. They will necessarily have the same activity (decays per second).
TTrue
FFalse
Answer: False
Activity A = λN, where λ is the decay constant and N is the number of atoms. Two samples with the same N but different decay constants λ will have dramatically different activities. For example, ¹⁴C (t₁/₂ ≈ 5,730 years, so λ is very small) and ²¹⁰Po (t₁/₂ = 138 days, λ is much larger) would show orders-of-magnitude different activity from the same number of atoms. A nuclide with a shorter half-life has a larger λ and therefore higher activity per atom. Equal atom counts tell you nothing about equal decay rates.
Question 4 True / False
A nuclear decay is spontaneous if and only if the Q-value is positive — meaning the total mass-energy of the products is less than that of the parent nucleus.
TTrue
FFalse
Answer: True
The Q-value is defined as Q = [M(parent) − Σ M(products)]c², representing the energy released. If Q > 0, energy is released and the decay is energetically allowed to proceed spontaneously. If Q < 0, the products would have more mass-energy than the parent, which violates conservation of energy — the decay cannot proceed without an external energy input. Q > 0 is a necessary condition for spontaneous decay (though a positive Q doesn't guarantee a fast decay — some decays are energetically allowed but kinetically inhibited by a large Coulomb barrier, as in alpha decay from heavy nuclei).
Question 5 Short Answer
Explain why the exponential decay law N(t) = N₀ exp(−λt) follows from the assumption that each nucleus has a constant, age-independent probability of decaying per unit time.
Think about your answer, then reveal below.
Model answer: If each nucleus has a constant probability λ of decaying per unit time (independent of age), then in a small interval dt, the number of decays from a population of N nuclei is dN = −λN dt. Rearranging: dN/N = −λ dt. Integrating both sides from 0 to t gives ln(N/N₀) = −λt, which exponentiates to N(t) = N₀ exp(−λt). The exponential form is a direct mathematical consequence of the constant-probability-per-unit-time assumption — nothing more is required. The half-life t₁/₂ = ln(2)/λ is the time at which N = N₀/2, found by setting exp(−λt) = 1/2.
This derivation reveals that the exponential law is not a complicated empirical fact but follows from a single simple assumption: memorylessness (each nucleus decays independently with fixed probability per unit time). This is the continuous-time analogue of a geometric distribution. Understanding this connection helps explain why radioactive decay is used in dating — the constant λ means the exponential clock runs at a fixed rate regardless of temperature, pressure, or chemical environment.