Radioactive series — uranium, thorium, actinium: Exercises

Intuition How to use this page

These problems climb a ladder. L1 just checks you can read a series; L5 asks you to build one from scraps of information. Every problem has a complete solution hidden inside a collapsible callout — try first, then reveal. All the tools you need come from the parent note: the radioactive-series topic.

Two formulas do almost everything here. Let a chain run from a parent $(A_{1}, Z_{1})$ to a product $(A_{2}, Z_{2})$ , emitting $x$ alpha particles and $y$ beta $^{-}$ particles: $x = \frac{A _{1} - A _{2}}{4}, y = Z_{2} - Z_{1} + 2 x$ WHY these two: only $\alpha$ decay changes the mass number $A$ (by $- 4$ each), so the total mass lost, $A_{1} - A_{2}$ , must be exactly $4 x$ . And $\beta^-$ decay raises the atomic number $Z$ by $1$ without touching $A$ , while each $α$ lowers $Z$ by $2$ — so the net charge change $Z_{2} - Z_{1}$ equals $- 2 x + y$ , which rearranges to $y = Z_{2} - Z_{1} + 2 x$ .

Before you start, one picture of what a "series" actually looks like on the map of nuclei.

Figure — Radioactive series — uranium, thorium, actinium

Here the horizontal axis is $Z$ (number of protons) and the vertical axis is $A$ (total nucleons). An $α$ step is a long diagonal jump down-and-left ( $A$ drops 4, $Z$ drops 2); a $β^{-}$ step is a short move right ( $Z$ up 1, $A$ flat). Keep this picture in your head — every problem below is a walk on this grid.

Level 1 — Recognition

Recall Solution L1.1

WHAT we do: find $A mod 4$ — the remainder when $231$ is divided by $4$ . WHY: the parent note showed that $α$ and $β$ both leave $A mod 4$ unchanged, so this remainder $k$ is the permanent label of the series. $231 = 4 \times 57 + 3 \Rightarrow k = 3$ Answer: the actinium series, $4 n + 3$ . ✔

Recall Solution L1.2

From the parent-note table: parent $_{90}^{232} Th$ , end-product $_{82}^{208} Pb$ . Check the type: $232 = 4 \times 58$ , remainder $0$ , so it is the $4 n$ series. ✔

Level 2 — Application

Recall Solution L2.1

Alphas first: $x = \frac{A _{1} - A _{2}}{4} = \frac{238 - 206}{4} = \frac{32}{4} = 8$ . Why first: only $α$ moves $A$ , so the mass drop pins down $x$ with no ambiguity. Betas next: $y = Z_{2} - Z_{1} + 2 x = 82 - 92 + 2 (8) = - 10 + 16 = 6$ . Why the $+ 2 x$ : the 8 alphas alone would drag $Z$ down to $92 - 16 = 76$ ; but the real end is $Z = 82$ , so we must have climbed back up by $6$ — that climb is 6 betas. Answer: $8 α$ , $6 β^{-}$ . ✔

Recall Solution L2.2

$x = \frac{235 - 207}{4} = \frac{28}{4} = 7$ . $y = 82 - 92 + 2 (7) = - 10 + 14 = 4$ . Answer: $7 α$ , $4 β^{-}$ . ✔

Level 3 — Analysis

Recall Solution L3.1

WHAT we do: apply the two changes forward instead of solving backward.

Mass number: only alphas move it. $A = 232 - 4 (3) = 232 - 12 = 220$ .
Atomic number: alphas drop it by $2$ each, betas raise it by $1$ each. $Z = 90 - 2 (3) + 1 (2) = 90 - 6 + 2 = 86$ Answer: $_{86}^{220} Rn$ ( $Z = 86$ is radon). ✔ Picture: on the band-of-stability grid this is 3 long down-left jumps and 2 short rightward steps from thorium.

Recall Solution L3.2

Membership check: $214 = 4 \times 53 + 2$ , so $k = 2$ — yes, the uranium series. ✔ Remaining alphas: $x = \frac{214 - 206}{4} = \frac{8}{4} = 2$ . Remaining betas: $y = Z_{2} - Z_{1} + 2 x = 82 - 82 + 2 (2) = 0 + 4 = 4$ . Why $y$ can exceed the net $Z$ change of zero: both start and end sit at $Z = 82$ , but the 2 alphas dipped $Z$ down to $78$ , so 4 betas were needed to return to 82. Answer: $2 α$ and $4 β^{-}$ remain. ✔

Level 4 — Synthesis

Recall Solution L4.1

WHAT we do: run the balance equations backward to find the parent.

Mass: $A_{1} = A_{2} + 4 x = 209 + 4 (7) = 209 + 28 = 237$ .
Charge: from $Z_{1} = Z_{2} + 2 x - y$ (rearranged from $y = Z_{2} - Z_{1} + 2 x$ ): $Z_{1} = 83 + 2 (7) - 4 = 83 + 14 - 4 = 93$ Answer: the parent is $_{93}^{237} Np$ — the neptunium series. Check: $237 = 4 \times 59 + 1$ , so $4 n + 1$ . ✔ Why this series is absent in nature: its longest-lived member $^{237} Np$ has half-life $\sim 2.1 \times 1 0^{6}$ yr, tiny beside Earth's $4.5 \times 1 0^{9}$ yr — it decayed away long ago.

Recall Solution L4.2

WHAT we do: count total emissions ( $2 α$ , $2 β^{-}$ ) and reverse them; order does not matter for the net $A, Z$ .

Mass: $A_{X} = 224 + 4 (2) = 224 + 8 = 232$ .
Charge: $Z_{X} = Z_{end} + 2 x - y = 88 + 2 (2) - 2 = 88 + 4 - 2 = 90$ . Answer: $X =_{90}^{232} Th$ (thorium). ✔ Does order matter? For the final $A$ and $Z$ , no — addition is commutative, so $2 α + 2 β$ in any order gives the same net shift. Order only matters for which intermediate nuclides you pass through, not the endpoints. This is the Soddy–Fajans bookkeeping applied stepwise.

Level 5 — Mastery

Recall Solution L5.1

Step 1 — pin the endpoint from the $4 n + k$ label. The three lead endpoints are $^{206},^{207},^{208} Pb$ (all $Z = 82$ ). We don't yet know which — we will let the numbers decide. Step 2 — use the alpha count to link masses. $x = 6 \Rightarrow A_{P} = A_{Pb} + 24$ . Step 3 — use the beta count to link charges. $y = 4 = 82 - Z_{P} + 2 (6) \Rightarrow Z_{P} = 82 + 12 - 4 = 90$ . So $Z_{P} = 90$ (thorium), and thorium's only long-lived natural parent is $A = 232$ . Step 4 — get the endpoint. $A_{Pb} = A_{P} - 24 = 232 - 24 = 208$ . Answer: $P =_{90}^{232} Th$ , endpoint $_{82}^{208} Pb$ , thorium series ( $232 = 4 n$ ). ✔ Consistency check on $k$ : parent $232 mod 4 = 0$ and $208 mod 4 = 0$ — same label, as required.

Recall Solution L5.2

Both parents have $Z = 92$ (uranium). Endpoints are lead, $Z_{Pb} = 82$ . Series A — solve for $A$ : the beta equation $6 = 82 - 92 + 2 x$ gives $2 x = 16$ , $x = 8$ (matches). Mass: $A_{A} = A_{Pb,A} + 4 (8)$ . The charge balance already forces uranium, and the standard uranium endpoint has $A_{Pb,A} = 206$ , so $A_{A} = 206 + 32 = 238$ . → $_{92}^{238} U \to_{82}^{206} Pb$ . Series B: $4 = 82 - 92 + 2 x \Rightarrow x = 7$ (matches). $A_{B} = A_{Pb,B} + 28$ ; with endpoint $^{207} Pb$ , $A_{B} = 207 + 28 = 235$ . → $_{92}^{235} U \to_{82}^{207} Pb$ . Different series? $238 mod 4 = 2$ (uranium/ $4 n + 2$ ) vs $235 mod 4 = 3$ (actinium/ $4 n + 3$ ). Different $k$ ⇒ genuinely different families, even though both parents are uranium isotopes. ✔ This is exactly how U–Pb dating uses two independent clocks ( $^{238} U \to^{206} Pb$ and $^{235} U \to^{207} Pb$ ) that must agree.

Answer key (quick check)

#	Answer
L1.1	$4 n + 3$ (actinium)
L1.2	$^{232} Th \to^{208} Pb$ , $4 n$
L2.1	$8 α, 6 β^{-}$
L2.2	$7 α, 4 β^{-}$
L3.1	$_{86}^{220} Rn$
L3.2	$2 α, 4 β^{-}$
L4.1	$_{93}^{237} Np$ , $4 n + 1$
L4.2	$_{90}^{232} Th$
L5.1	$^{232} Th \to^{208} Pb$
L5.2	$^{238} U \to^{206} Pb$ and $^{235} U \to^{207} Pb$

Connections

Parent topic — the theory behind every problem here.
Alpha decay — the only step in the $x$ -formula.
Beta decay — the source of the $+ 2 x$ correction in $y$ .
Group displacement law (Soddy–Fajans) — stepwise $Z$ -bookkeeping used in L3/L4.
Half-life and decay constant — why Np (L4.1) is gone from nature.
Radioactive dating — the twin U–Pb clocks of L5.2.
Nuclear stability and band of stability — the grid the decays walk on.