Radioactive decay — alpha, beta, gamma — mechanisms: Exercises

This page is a self-testing ladder. Every problem builds on the parent mechanisms note. Try each one with the Solution callout collapsed, then open it. The rungs go from "just name it" (L1) up to "combine everything" (L5).

Before we start, here is the one picture that every problem leans on: what "conserved" means for a decay.

Figure — Radioactive decay — alpha, beta, gamma — mechanisms

Read it like a balance scale. The two numbers we watch are:

$Z$ = the number of protons (this is the charge, and it names the element).
$A$ = the number of nucleons = protons + neutrons (the mass number, the top-left number).

The rule from the parent note: on both sides of the arrow, the totals of $Z$ must match, and the totals of $A$ must match. That is all "conservation of charge and nucleon number" means. Keep this scale in your head for the whole page.

Level 1 — Recognition

Problem 1.1 (L1)

Name the decay type for each change:

(a) $A$ drops by 4, $Z$ drops by 2.
(b) $A$ stays the same, $Z$ rises by 1.
(c) $A$ and $Z$ both unchanged, a photon leaves.

Recall Solution 1.1

Use the mnemonic "Alpha 2-and-2, Beta plus-one to Z, Gamma changes nothing but the energy."

(a) $A - 4, Z - 2$ → an $_{2}^{4} He$ chunk left → alpha decay.
(b) $A$ fixed, $Z + 1$ → a neutron became a proton → beta-minus ( $β^{-}$ ).
(c) Nothing moves on the balance scale, only energy leaves as light → gamma decay.

Problem 1.2 (L1)

State the particle emitted, and its charge, for $α$ , $β^{-}$ , $β^{+}$ , and $γ$ .

Recall Solution 1.2

Decay	Emitted	Charge
$α$	$_{2}^{4} He$ nucleus	$+ 2 e$
$β^{-}$	electron $e^{-}$ + antineutrino $\overset{ν}{ˉ}_{e}$	$- e$ (neutrino: $0$ )
$β^{+}$	positron $e^{+}$ + neutrino $ν_{e}$	$+ e$ (neutrino: $0$ )
$γ$	photon	$0$

Level 2 — Application

Problem 2.1 (L2)

Complete the alpha decay and give the daughter's name-numbers: $_{88}^{226} Ra \to_{?}^{?} Y + α$

Recall Solution 2.1

WHAT: balance $A$ and $Z$ using the scale in the figure above. WHY: an alpha carries away $A = 4, Z = 2$ , so the daughter must hold the rest. $A_{Y} = 226 - 4 = 222, Z_{Y} = 88 - 2 = 86.$ $Z = 86$ is radon. So the daughter is $_{86}^{222} Rn$ . $_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} α .$

Problem 2.2 (L2)

$_{90}^{234} Th$ undergoes $β^{-}$ decay. Write the full equation, including the neutrino, and identify the daughter's $Z$ .

Recall Solution 2.2

WHAT: $β^{-}$ means $n \to p + e^{-} + \overset{ν}{ˉ}_{e}$ inside the nucleus. WHY: a neutron became a proton, so $Z$ rises by 1 while $A$ is untouched; the antineutrino balances lepton number. $_{90}^{234} Th \to_{91}^{234} Pa + e^{-} + \overset{ν}{ˉ}_{e} .$ $Z : 90 \to 91$ (protactinium), $A$ stays $234$ .

Problem 2.3 (L2)

For $_{92}^{238} U \to_{90}^{234} Th + α$ with $Q = 4.27 MeV$ , find the kinetic energy of the alpha.

Recall Solution 2.3

WHY this formula: the parent is at rest, so by momentum conservation the alpha and daughter fly apart with equal and opposite momenta. The lighter alpha therefore carries the larger share (see the split figure below). Use $K E_{α} = Q \cdot \frac{m _{Y}}{m _{Y} + m _{α}} \approx Q \cdot \frac{A - 4}{A} .$ $K E_{α} = 4.27 \times \frac{234}{238} = 4.20 MeV .$ The daughter keeps the tiny remainder, $Q - K E_{α} \approx 0.07 MeV$ .

Level 3 — Analysis

Problem 3.1 (L3)

$_{6}^{14} C$ decays by $β^{-}$ with $Q \approx 0.156 MeV$ . A student measures an emitted electron with kinetic energy $0.100 MeV$ . (a) Is this allowed? (b) How much energy did the antineutrino carry (ignore the tiny nitrogen recoil)? (c) Why can two different $^{14}$ C nuclei give electrons of different energies?

Recall Solution 3.1

(a) The endpoint (maximum electron KE) $\approx Q = 0.156 MeV$ . Since $0.100 < 0.156$ , it is allowed. (b) Energy is shared among three products; ignoring the negligible nitrogen recoil, $E_{\overset{ν}{ˉ}} = Q - K E_{e} = 0.156 - 0.100 = 0.056 MeV .$ (c) Beta decay is a three-body split (daughter + $e^{-}$ + $\overset{ν}{ˉ}_{e}$ ). With three ways to share a fixed $Q$ , the electron can get anything from $0$ up to $Q$ — a continuous spectrum, not a single line. That is exactly what the parent note's neutrino argument predicts.

Problem 3.2 (L3)

An excited daughter $_{28}^{60} Ni^{*}$ sits $E_{i} = 2.506 MeV$ above ground; it de-excites in two steps through a level at $1.332 MeV$ . Find the two gamma photon energies. (Ignore recoil.)

Recall Solution 3.2

WHY: a nuclear energy level gap is emitted as a photon of exactly $E_{γ} = E_{i} - E_{f}$ , the nuclear analog of an atom emitting light.

Step 1: $2.506 \to 1.332$ : $E_{γ 1} = 2.506 - 1.332 = 1.174 MeV$ .
Step 2: $1.332 \to 0$ : $E_{γ 2} = 1.332 - 0 = 1.332 MeV$ . Two discrete lines — this is why gamma spectra map the level diagram.

Level 4 — Synthesis

Problem 4.1 (L4)

Given atomic masses $m (^{226} Ra) = 226.025410 u$ , $m (^{222} Rn) = 222.017578 u$ , $m (^{4} He) = 4.002603 u$ , and $1 u = 931.494 MeV / c^{2}$ : find $Q$ for the alpha decay, then $K E_{α}$ .

Recall Solution 4.1

Step 1 — mass defect. The parent must be heavier than the products for decay to run (mass defect released as energy): $Δ m = 226.025410 - (222.017578 + 4.002603) = 0.005229 u .$ Step 2 — convert to energy ( $Q = Δ m c^{2}$ ): $Q = 0.005229 \times 931.494 = 4.871 MeV .$ Step 3 — alpha's share (mass-ratio split): $K E_{α} = Q \cdot \frac{222}{226} = 4.871 \times 0.98230 = 4.785 MeV (\approx 4.79 MeV) .$ Positive $Q$ ⇒ the decay is spontaneous, and the alpha carries almost all of it.

Problem 4.2 (L4)

The parent note says more energetic alphas have much shorter half-lives (Geiger–Nuttall). Explain, using tunnelling through the Coulomb barrier, why a slightly higher $K E_{α}$ produces a dramatically shorter half-life. (Qualitative.)

Recall Solution 4.2

WHAT: the alpha is trapped behind a repulsive Coulomb wall; classically it can't escape, but quantum mechanics gives it a small tunnelling probability $P$ . WHY sensitivity: a higher-energy alpha starts higher up the barrier, so the slice of barrier it must cross is thinner and lower. The tunnelling probability depends exponentially on that barrier width/height. A small drop in barrier thickness ⇒ a large jump in $P$ ⇒ many more escapes per second. WHAT it looks like: in the figure below, raising the alpha's energy line shortens the shaded barrier it must pierce. Because $P$ sits inside an exponent, half-life $\sim 1/ P$ can fall by orders of magnitude for a modest energy rise — exactly the steep Geiger–Nuttall trend.

Level 5 — Mastery

Problem 5.1 (L5)

$_{83}^{212} Bi$ can decay by two branches: (i) $α$ to $_{81}^{208} Tl$ , and (ii) $β^{-}$ to $_{84}^{212} Po$ . Both daughters then continue to stable $_{82}^{208} Pb$ . (a) For each branch, write the next decay ( $^{208} Tl \to Pb$ and $^{212} Po \to Pb$ ) as $α$ or $β^{-}$ , checking $A, Z$ . (b) Confirm both branches reach the same final nucleus.

Recall Solution 5.1

Branch (i): $_{83}^{212} Bi α_{81}^{208} Tl$ . Now $_{81}^{208} Tl \to_{82}^{208} Pb$ : $A$ fixed, $Z : 81 \to 82$ ⇒ $β^{-}$ . ✓ Branch (ii): $_{83}^{212} Bi β^{-}_{84}^{212} Po$ . Now $_{84}^{212} Po \to_{82}^{208} Pb$ : $A : 212 \to 208$ ( $- 4$ ), $Z : 84 \to 82$ ( $- 2$ ) ⇒ $α$ . ✓ (b) Both routes end at $_{82}^{208} Pb$ (stable lead). Net change from Bi either way: $Δ A = - 4, Δ Z = - 1$ (one $α$ + one $β^{-}$ in either order). The order of $α$ and $β^{-}$ differs, the destination does not — a classic branching in a decay chain.

Problem 5.2 (L5)

A $^{60}$ Co source undergoes $β^{-}$ then two gammas (from Problem 3.2). Using the decay law $N = N_{0} e^{- λ t}$ with half-life $T_{1/2} = 5.27 yr$ , what fraction of the source remains after $10.54 yr$ ? Which emission (the $β$ or the $γ$ s) sets this clock?

Recall Solution 5.2

WHY the exponential: each nucleus decays with a fixed probability per second $λ$ , independent of age, so the population falls as $e^{- λ t}$ . Step 1 — count half-lives: $t = 10.54 yr = 2 \times 5.27 yr$ , so exactly 2 half-lives. Step 2 — fraction left: $\frac{N}{N _{0}} = (\frac{1}{2})^{2} = \frac{1}{4} = 0.25.$ Check via $λ = ln 2/ T_{1/2}$ : $e^{- λ t} = e^{- (l n 2/5.27) (10.54)} = e^{- 2 l n 2} = 0.25$ . ✓ Which clock: the slow, rate-limiting step is the $β^{-}$ decay (half-life years). The gammas follow essentially instantly (sub-nanosecond de-excitation) once the beta has happened — they don't set the timescale.

Recall One-line self-check (open only after all problems)

Alpha split ::: $K E_{α} = Q m_{Y} / (m_{Y} + m_{α})$ , daughter recoils with the rest. Beta spectrum ::: continuous 0→ $Q$ because three bodies share the energy. Gamma energy ::: $E_{γ} = E_{i} - E_{f}$ , discrete lines. Q from masses ::: $Q = (m_{parent} - m_{products}) c^{2}$ , use atomic masses. Chain destination ::: fixed by total $Δ A, Δ Z$ , not by the order of decays.