Intuition The big pictureHeavy nuclei like 238 U ^{238}\text{U} 238 U one jump. So they decay step-by-step through a chain of unstable daughters , each emitting an α \alpha α β \beta β stable isotope of lead . This whole staircase of decays is called a radioactive (disintegration) series .
WHY a series and not one decay? An α \alpha α A A A β \beta β A = 238 A=238 A = 238 A ≈ 206 A\approx206 A ≈ 206 α \alpha α
Definition Why exactly four seriesAn α \alpha α A A A β − \beta^- β − A A A every member of one chain has a mass number that differs from the parent by a multiple of 4 . Hence A m o d 4 A \bmod 4 A mod 4 conserved along a chain. Since A m o d 4 ∈ { 0 , 1 , 2 , 3 } A \bmod 4 \in \{0,1,2,3\} A mod 4 ∈ { 0 , 1 , 2 , 3 } four possible families.
Series
Type
Parent
Half-life of parent
Final stable nuclide
Thorium 4 n 4n 4 n 90 232 Th ^{232}_{90}\text{Th} 90 232 Th 1.4 × 10 10 1.4\times10^{10} 1.4 × 1 0 10 82 208 Pb ^{208}_{82}\text{Pb} 82 208 Pb
Neptunium 4 n + 1 4n+1 4 n + 1 93 237 Np ^{237}_{93}\text{Np} 93 237 Np 2.1 × 10 6 2.1\times10^{6} 2.1 × 1 0 6 83 209 Bi ^{209}_{83}\text{Bi} 83 209 Bi
Uranium (radium) 4 n + 2 4n+2 4 n + 2 92 238 U ^{238}_{92}\text{U} 92 238 U 4.5 × 10 9 4.5\times10^{9} 4.5 × 1 0 9 82 206 Pb ^{206}_{82}\text{Pb} 82 206 Pb
Actinium 4 n + 3 4n+3 4 n + 3 92 235 U ^{235}_{92}\text{U} 92 235 U 7.0 × 10 8 7.0\times10^{8} 7.0 × 1 0 8 82 207 Pb ^{207}_{82}\text{Pb} 82 207 Pb
Intuition Why is the Neptunium (4 n + 1 4n+1 4 n + 1 237 Np ^{237}\text{Np} 237 Np ∼ 2.1 × 10 6 \sim2.1\times10^6 ∼ 2.1 × 1 0 6 tiny compared to the age of the Earth (4.5 × 10 9 4.5\times10^9 4.5 × 1 0 9 survive and are still found today.
Derivation How to get number ofα \alpha α β \beta β ( A 1 , Z 1 ) (A_1, Z_1) ( A 1 , Z 1 ) ( A 2 , Z 2 ) (A_2, Z_2) ( A 2 , Z 2 ) x x x y y y − ^- − Mass-number balance (only α \alpha α A A A − 4 -4 − 4 A 1 = A 2 + 4 x ⇒ x = A 1 − A 2 4 A_1 = A_2 + 4x \;\Rightarrow\; \boxed{x = \dfrac{A_1 - A_2}{4}} A 1 = A 2 + 4 x ⇒ x = 4 A 1 − A 2 Atomic-number balance (α \alpha α − 2 -2 − 2 β − \beta^- β − + 1 +1 + 1 Z 1 = Z 2 + 2 x − y ⇒ y = Z 2 − Z 1 + 2 x Z_1 = Z_2 + 2x - y \;\Rightarrow\; \boxed{y = Z_2 - Z_1 + 2x} Z 1 = Z 2 + 2 x − y ⇒ y = Z 2 − Z 1 + 2 x WHY the + 2 x − y +2x-y + 2 x − y removing x x x Z Z Z 2 x 2x 2 x y y y Z Z Z y y y Z 2 − Z 1 Z_2-Z_1 Z 2 − Z 1
Worked example Example 1 — Uranium series:92 238 U → 82 206 Pb ^{238}_{92}\text{U} \to {}^{206}_{82}\text{Pb} 92 238 U → 82 206 Pb Find x x x x = 238 − 206 4 = 32 4 = 8 x = \dfrac{238-206}{4} = \dfrac{32}{4} = 8 x = 4 238 − 206 = 4 32 = 8 Why this step? Only α \alpha α Find y y y y = Z 2 − Z 1 + 2 x = 82 − 92 + 16 = 6 y = Z_2 - Z_1 + 2x = 82 - 92 + 16 = 6 y = Z 2 − Z 1 + 2 x = 82 − 92 + 16 = 6 Why this step? The 8 alphas alone would lower Z Z Z 92 − 16 = 76 92-16=76 92 − 16 = 76 raise it back to 82, which takes 82 − 76 = 6 82-76 = 6 82 − 76 = 6 Answer: 8 α 8\alpha 8 α 6 β − 6\beta^- 6 β − 238 = 4 n + 2 238 = 4n+2 238 = 4 n + 2
Worked example Example 2 — Thorium series:90 232 Th → 82 208 Pb ^{232}_{90}\text{Th} \to {}^{208}_{82}\text{Pb} 90 232 Th → 82 208 Pb x = 232 − 208 4 = 6 x = \dfrac{232-208}{4} = 6 x = 4 232 − 208 = 6 Why? Mass drops by 24, that's 24 / 4 = 6 24/4=6 24/4 = 6 y = 82 − 90 + 2 ( 6 ) = 82 − 90 + 12 = 4 y = 82 - 90 + 2(6) = 82 - 90 + 12 = 4 y = 82 − 90 + 2 ( 6 ) = 82 − 90 + 12 = 4 Why? 6 alphas push Z Z Z 90 − 12 = 78 90-12=78 90 − 12 = 78 4 4 4 Answer: 6 α 6\alpha 6 α 4 β − 4\beta^- 4 β − 232 = 4 ( 58 ) ⇒ 4 n 232 = 4(58) \Rightarrow 4n 232 = 4 ( 58 ) ⇒ 4 n
Worked example Example 3 — Actinium series:92 235 U → 82 207 Pb ^{235}_{92}\text{U} \to {}^{207}_{82}\text{Pb} 92 235 U → 82 207 Pb x = 235 − 207 4 = 28 4 = 7 x = \dfrac{235-207}{4} = \dfrac{28}{4} = 7 x = 4 235 − 207 = 4 28 = 7 y = 82 − 92 + 14 = 4 y = 82 - 92 + 14 = 4 y = 82 − 92 + 14 = 4 Answer: 7 α 7\alpha 7 α 4 β − 4\beta^- 4 β − 235 = 4 ( 58 ) + 3 ⇒ 4 n + 3 235 = 4(58)+3 \Rightarrow 4n+3 235 = 4 ( 58 ) + 3 ⇒ 4 n + 3
Worked example Example 4 — Forecast-then-Verify (Neptunium series)Forecast: 93 237 Np → 83 209 Bi ^{237}_{93}\text{Np} \to {}^{209}_{83}\text{Bi} 93 237 Np → 83 209 Bi α , β \alpha,\beta α , β before computing.
Mass drop = 237 − 209 = 28 ⇒ = 237-209 = 28 \Rightarrow = 237 − 209 = 28 ⇒ 7 α 7\alpha 7 α Z Z Z 7 α 7\alpha 7 α 93 − 14 = 79 93-14=79 93 − 14 = 79 83 83 83 4 β 4\beta 4 β Verify: x = 28 / 4 = 7 x=28/4=7 x = 28/4 = 7 y = 83 − 93 + 14 = 4 y = 83-93+14 = 4 y = 83 − 93 + 14 = 4 7 α 7\alpha 7 α 4 β − 4\beta^- 4 β − 4 n + 1 4n+1 4 n + 1
Common mistake "I'll just subtract atomic numbers to get beta count."Why it feels right: β \beta β Z Z Z y = Z 2 − Z 1 y = Z_2 - Z_1 y = Z 2 − Z 1 Why it's wrong: the α \alpha α also lower Z Z Z 2 2 2 Z Z Z Fix: always compute x x x y = Z 2 − Z 1 + 2 x y = Z_2 - Z_1 + 2x y = Z 2 − Z 1 + 2 x + 2 x +2x + 2 x
Common mistake "The end product is always206 Pb ^{206}\text{Pb} 206 Pb Why it feels right: lead is the famous stable endpoint and U-238 is the famous series. Why it's wrong: each series ends on a different stable nuclide — 206 Pb ^{206}\text{Pb} 206 Pb 207 Pb ^{207}\text{Pb} 207 Pb 208 Pb ^{208}\text{Pb} 208 Pb 209 Bi ^{209}\text{Bi} 209 Bi Fix: remember k k k 4 n + k 4n+k 4 n + k A m o d 4 A \bmod 4 A mod 4
β \beta β Why it feels right: an electron is leaving the nucleus, surely something about mass changes. Why it's wrong: the electron comes from a neutron→proton conversion; total nucleon count is unchanged, so A A A Fix: only α \alpha α A A A A m o d 4 A \bmod 4 A mod 4
Recall Try these before reading answers
Why are there exactly four radioactive series?
Which natural quantity is conserved along a chain?
Why is the neptunium series absent in nature?
Give the parent and stable end-product of each natural series.
Recall Feynman: explain to a 12-year-oldImagine a giant, wobbly Jenga tower (a heavy atom). It's too tall to stand, so it can't just become a small steady tower in one move. Instead it drops blocks one at a time : sometimes a big chunky block of 4 (that's an alpha), sometimes a tiny invisible flick that just changes the color of a block but not the height (that's a beta). It keeps dropping until it becomes a short, rock-solid tower — that's lead , which never falls again. Because the big blocks are always exactly 4, the height of every tower in one family always differs by multiples of 4 — that's why there are only 4 different families of towers!
Mnemonic Remember the series labels & endpoints"Uncle Tom Acts Naturally" → U ranium (4 n + 2 4n+2 4 n + 2 206 ^{206} 206 T horium (4 n 4n 4 n 208 ^{208} 208 A ctinium (4 n + 3 4n+3 4 n + 3 207 ^{207} 207 N eptunium (4 n + 1 4n+1 4 n + 1 209 ^{209} 209 2-7-8 as you read U-Ac-Th — and the odd one out (Np) ends on Bismuth .
Alpha decay — the only step that changes A A A − 4 -4 − 4 4 n + k 4n+k 4 n + k Beta decay — raises Z Z Z A A A + 2 x +2x + 2 x Half-life and decay constant — why 237 ^{237} 237 238 ^{238} 238 Group displacement law (Soddy–Fajans) — the per-step bookkeeping behind these chains.Radioactive dating — U–Pb and Th–Pb clocks use these series' stable endpoints.Nuclear stability and band of stability — why heavy nuclei must shed mass at all.
Why are there exactly four radioactive series? Because
α \alpha α changes
A A A by 4 and
β \beta β by 0, so
A m o d 4 A \bmod 4 A mod 4 is conserved; it can be 0,1,2,3 → four families.
What quantity is conserved along a decay chain? A m o d 4 A \bmod 4 A mod 4 (the value of
k k k in
A = 4 n + k A = 4n+k A = 4 n + k ).
Parent and end-product of the uranium series? 92 238 ^{238}_{92} 92 238 U →
82 206 ^{206}_{82} 82 206 Pb (
4 n + 2 4n+2 4 n + 2 ).
Parent and end-product of the thorium series? 90 232 ^{232}_{90} 90 232 Th →
82 208 ^{208}_{82} 82 208 Pb (
4 n 4n 4 n ).
Parent and end-product of the actinium series? 92 235 ^{235}_{92} 92 235 U →
82 207 ^{207}_{82} 82 207 Pb (
4 n + 3 4n+3 4 n + 3 ).
Parent and end-product of the neptunium series? 93 237 ^{237}_{93} 93 237 Np →
83 209 ^{209}_{83} 83 209 Bi (
4 n + 1 4n+1 4 n + 1 ).
Why is the neptunium series missing in nature? Its longest-lived member
237 ^{237} 237 Np (
t 1 / 2 ∼ 2.1 × 10 6 t_{1/2}\sim2.1\times10^6 t 1/2 ∼ 2.1 × 1 0 6 yr) is far shorter than Earth's age, so it has fully decayed.
Formula for number of alpha particles emitted? x = ( A 1 − A 2 ) / 4 x = (A_1 - A_2)/4 x = ( A 1 − A 2 ) /4 .
Formula for number of beta particles emitted? y = Z 2 − Z 1 + 2 x y = Z_2 - Z_1 + 2x y = Z 2 − Z 1 + 2 x .
How many α \alpha α β \beta β 238 ^{238} 238 206 ^{206} 206 8 α 8\alpha 8 α and
6 β − 6\beta^- 6 β − .
How many α \alpha α β \beta β 232 ^{232} 232 208 ^{208} 208 6 α 6\alpha 6 α and
4 β − 4\beta^- 4 β − .
Cannot reach stable in one jump
Intuition Hinglish mein samjhoDekho, bhaari nuclei jaise 238 U ^{238}\text{U} 238 U chain me decay karte hain — kabhi alpha particle nikaalte hain (jisse mass number A A A Z Z Z A A A Z Z Z stable lead (ya bismuth) isotope na mil jaaye.
Ab sabse zabardast trick: kyunki alpha sirf 4 ka multiple hata sakta hai aur beta A A A A m o d 4 A \bmod 4 A mod 4 chaar series possible hain: 4 n 4n 4 n 4 n + 1 4n+1 4 n + 1 4 n + 2 4n+2 4 n + 2 4 n + 3 4n+3 4 n + 3 237 ^{237} 237
Number of alpha-beta nikaalne ke liye formula yaad rakho: pehle x = ( A 1 − A 2 ) / 4 x = (A_1 - A_2)/4 x = ( A 1 − A 2 ) /4 y = Z 2 − Z 1 + 2 x y = Z_2 - Z_1 + 2x y = Z 2 − Z 1 + 2 x 238 U → 206 Pb ^{238}\text{U} \to {}^{206}\text{Pb} 238 U → 206 Pb x = 32 / 4 = 8 x = 32/4 = 8 x = 32/4 = 8 y = 82 − 92 + 16 = 6 y = 82-92+16 = 6 y = 82 − 92 + 16 = 6
