Question bank — Radioactive decay — alpha, beta, gamma — mechanisms

True — the alpha carries off 2 protons + 2 neutrons, so $A\to A-4$ and $Z\to Z-2$ by nucleon and charge conservation, no exceptions.

True — a neutron becomes a proton inside the same nucleus, so the total nucleon count is unchanged; only $Z$ shifts by $+1$.

False — $\gamma$ emission only sheds energy; $A$ and $Z$ are both unchanged, so it is the same isotope of the same element.

False — $Q<0$ means the products are heavier than the parent, so the decay costs energy and is forbidden spontaneously; time cannot supply that energy.

False — the electron is created at the instant $n\to p+e^{-}+{\bar{\nu }}_e$; no electron sits in the nucleus beforehand.

Mostly true but subtle — kinematics fixes a single central $K{E}_{\alpha }$ (a discrete line, see figure s03), but real lines have tiny finite width from recoil broadening and thermal Doppler motion of the emitting atoms.

False — the daughter must recoil to conserve momentum, so $K{E}_{\alpha }=Q{m}_Y/(m_Y+m_{\alpha })$, slightly less than $Q$.

False — higher alpha energy means a thinner/lower Coulomb barrier to tunnel through, giving a much shorter half-life (Geiger–Nuttall).

False — $\gamma$ is a two-body de-excitation between fixed nuclear levels, so $E_{\gamma }=E_i-E_f$ gives discrete lines.

False — in ${\beta }^{+}$ a proton becomes a neutron, so $Z\to Z-1$; it is ${\beta }^{-}$ that raises $Z$.

False — electron capture (K-capture) does the same $Z\to Z-1$ by swallowing an inner orbital electron, $p+e^{-}\to n+{\nu }_e$, and is the only option when $Q<2m_e{c}^2$ (too little energy to create a positron).

Wrong particle — ${\beta }^{-}$ emits an antineutrino ${\bar{\nu }}_e$; the ${\nu }_e$ (matter neutrino) accompanies ${\beta }^{+}$, as required by lepton-number bookkeeping.

Error — $Z$ drops by 2, so Thorium has $Z=90$; the alpha took away 2 units of charge.

Error of motive — it was forced by the observed continuous beta spectrum, which two-body kinematics cannot produce; it also restores energy, momentum and angular-momentum conservation.

Reversed — $\gamma$ is emitted when an excited daughter $Y^{\ast }$ loses energy dropping to a lower level, $E_{\gamma }=E_i-E_f>0$.

Error — the ${}_2^4$He clump has an unusually high binding energy (~7.07 MeV/nucleon), which is exactly what makes $Q$ positive; ejecting a lone proton usually gives $Q<0$.

Error — the created proton adds $+e$ to the nucleus, exactly cancelling the $-e$ of the electron; net charge is conserved.

Reversed — being heavy and doubly charged, alphas ionise strongly and stop almost immediately (paper stops them); $\gamma$ penetrates deepest.

Error — it can instead hand the energy directly to an inner orbital electron, ejecting it (internal conversion); the energy leaves as a monoenergetic conversion electron, not a photon.

Beta is a three-body decay (daughter + electron + ${\bar{\nu }}_e$), so the released energy is shared in any proportion; alpha is two-body, so kinematics pins one fixed energy (figure s02 vs s03).

The number of ways to share energy (phase space) grows then falls across the range, and this statistical weighting — multiplied by the Fermi function, a Coulomb correction that pulls emitted electrons back / pushes positrons away — sculpts the observed curve.

Classically the alpha lacks the energy to climb the Coulomb barrier (figure s01); quantum mechanics gives it a small probability to pass through the wall, which is the whole mechanism.

With the parent at rest, momentum conservation gives $0=p_{\alpha }-p_Y$, so $p_{\alpha }=p_Y\equiv p$; since $KE=p^2/2m$ and both share the same $p$, kinetic energy scales as $1/m$, so the heavy daughter ($m_Y\gg m_{\alpha }$) gets the small slice while $K{E}_{\alpha }=Q{m}_Y/(m_Y+m_{\alpha })$.

Beta decay converts a quark flavour (neutron↔proton) and creates leptons — processes only the weak force mediates; alpha is governed by nuclear + electromagnetic forces on an intact clump.

The daughter frequently lands in an excited nuclear level, not the ground state, and then de-excites by shedding the leftover energy as a photon.

$Q=(m_{\text{parent}}-m_{\text{products}})c^2$ with $c$ the speed of light; only if the parent is heavier can freed rest-mass-energy appear as kinetic energy — otherwise nature would have to supply energy.

${}^{14}\text{C}$ is neutron-rich, so converting a neutron to a proton (${\beta }^{-}$, $Z:6\to 7$) moves it toward the line of stability.

No — with zero freed energy there is no kinetic energy for the products, so the transition is on the threshold and not spontaneous.

There is no daughter — such a "decay" would leave nothing behind, so it is not an alpha decay; alpha decay needs a residual ${}_{Z-2}^{A-4}Y$ nucleus.

The endpoint $\approx Q$, reached in the limiting case where the antineutrino carries away essentially zero energy.

Only electron capture, not ${\beta }^{+}$: making a positron costs at least $2m_e{c}^2$ of rest energy (electron created in daughter + positron emitted), so below that threshold the nucleus must instead absorb a $K$-shell electron, $p+e^{-}\to n+{\nu }_e$.

An inner-shell vacancy is left, so outer electrons cascade down emitting characteristic X-rays (or Auger electrons) — a fingerprint distinguishing EC from ${\beta }^{+}$.

Yes — momentum conservation forces a tiny recoil, so $E_{\gamma }=(E_i-E_f)$ minus a small recoil term; the recoil is usually negligible but never exactly zero.

If enough energy is available it can emit a nucleon or even a small cluster, but for ordinary excited daughters the competition is between $\gamma$-emission and internal conversion (both leave $A,Z$ unchanged).

It would break — the electron has lepton number $+1$ with no partner, so the ${\bar{\nu }}_e$ (lepton number $-1$) is required to keep the total at zero.

No — a free neutron ${\beta }^{-}$-decays ($n\to p+e^{-}+{\bar{\nu }}_e$) with $Q>0$ and a ~15-minute mean life, exactly the same mechanism as bound-neutron beta decay.