Radioactive series — uranium, thorium, actinium: Question bank

True or false — justify

Every member of one radioactive series has the same

A mod 4

True. An

α

decay changes

A

by exactly

- 4

(remainder unchanged) and a

β

decay changes

A

0

, so the remainder is frozen for the whole chain — that is what defines the family.

β^{-}

decay changes the mass number

A

of the nucleus.

False. The emitted electron comes from a neutron turning into a proton; the total count of nucleons stays the same, so

A

is unchanged (only

Z

rises by 1).

Since there are four possible values of

A mod 4

, all four series are found abundantly in nature.

False. There are four mathematically possible families, but the

4 n + 1

(neptunium) series has no long-lived member, so it has fully decayed away and is essentially absent naturally.

The uranium and actinium series both start from a uranium isotope.

True. Uranium series starts at

_{92}^{238} U

(

4 n + 2

) and actinium series starts at

_{92}^{235} U

(

4 n + 3

); same element

Z = 92

, different mass numbers put them in different families.

All natural radioactive series end on an isotope of lead.

False. Three end on lead (

^{206},^{207},^{208} Pb

), but the neptunium series ends on bismuth,

_{83}^{209} Bi

— a different element entirely.

If two nuclides have the same

A mod 4

, they must belong to the same series.

False. Sharing

A mod 4

is necessary but not sufficient; they also need to lie on the same actual decay chain. A nuclide from an unrelated chain can share the remainder without being connected.

^{238} U

survives on Earth today because its half-life is comparable to the planet's age.

True. With

t_{1/2} \approx 4.5 \times 1 0^{9}

yr (about the age of the Earth), only roughly half of the original

^{238} U

has decayed, so plenty remains to be found.

Why questions

Why is

A mod 4

— and not

Z mod 4

A mod 2

— the conserved label?

Because the two decays' effects on

A

are "

- 4

0

", both multiples of 4, so remainder mod 4 is untouched.

Z

changes by

- 2

+ 1

(no clean invariant), and

A mod 2

would merge families that alphas can never mix.

Why must a heavy nucleus like

^{238} U

decay through a chain rather than one step?

It must shed roughly 32 mass units and rebalance its proton count; a single decay changes

A

by at most 4, so many sequential steps are the only way down to a stable configuration (Nuclear stability and band of stability).

Why does emitting an alpha lower

Z

while emitting a beta raises it?

α

carries away 2 protons (

Z

drops by 2); a

β^{-}

converts a neutron into a proton inside the nucleus and ejects an electron, so the proton count rises by 1 (Alpha decay, Beta decay).

Why do we compute the alpha count

x

before the beta count

y

The beta formula

y = Z_{2} - Z_{1} + 2 x

needs

x

as an input to cancel the alphas' effect on

Z

; without knowing

x

first, the

+ 2 x

term can't be evaluated.

Why can two of the four series be distinguished only by their endpoint's mass number?

Because their parents share element

Z = 92

(both uranium) but differ in

A

; the family is then read off from the stable endpoint —

^{206} Pb

(

4 n + 2

) versus

^{207} Pb

(

4 n + 3

) — where

A mod 4

shows the difference.

Why does the survival of a series in nature depend on its longest-lived member, not its parent alone?

If any bottleneck nuclide in the chain is long-lived it can hold the series "in stock"; but for these series the parent is the longest-lived member, so its half-life alone decides whether the whole chain persists (Half-life and decay constant).

Edge cases

Can a decay chain ever increase the mass number

A

at some step?

No. The only two moves are

α

(

A

drops by 4) and

β^{-}

(

A

unchanged); neither can raise

A

, so a chain marches monotonically downward or holds steady.

What is the smallest possible number of alpha decays in a chain, and when does it happen?

Zero — if parent and product have the same mass number (

A_{1} = A_{2}

), then

x = (A_{1} - A_{2}) /4 = 0

and the change is achieved by beta decay(s) alone, since only betas leave

A

fixed.

If a chain produces a fractional alpha count from

x = (A_{1} - A_{2}) /4

, what does that tell you?

That the two nuclides do not lie on the same series:

A_{1} - A_{2}

must be a multiple of 4 for a real chain, so a non-integer means the parent and product have different

A mod 4

and cannot be connected.

Is it possible for the beta count

y

to come out negative?

In principle

y = Z_{2} - Z_{1} + 2 x < 0

would signal

β^{+}

/ electron-capture behaviour instead of

β^{-}

; for the natural heavy series it never happens, because the alphas always overshoot

Z

downward and betas must climb it back up (

y > 0

What would it mean for the "

n

" in

4 n + k

to be zero or negative?

For real nuclei

A

is a large positive integer, so

n

is large and positive;

n = 0

or negative has no physical nucleus — the label only ever uses

n

big enough to make

A = 4 n + k

match an actual mass number.

Could a stable endpoint ever be reached that is not lead or bismuth?

For these four specific series, no — the valley of stability at high

A

funnels each family to

^{206, 207, 208} Pb

^{209} Bi

. The remainder

k

pins which endpoint, and only those four nuclides are simultaneously stable and reachable (Radioactive dating relies on exactly this fixed endpoint).

Connections

Alpha decay — the single move that changes $A$ (by $- 4$ ); the reason $A mod 4$ is an invariant.
Beta decay — raises $Z$ by 1 at fixed $A$ ; the source of the $+ 2 x$ correction term.
Half-life and decay constant — decides which series survive on Earth and which vanish.
Group displacement law (Soddy–Fajans) — the per-step $Z$ -shift bookkeeping behind every trap here.
Nuclear stability and band of stability — why heavy nuclei must shed mass through a chain at all.

Question bank — Radioactive series — uranium, thorium, actinium

True or false — justify

Spot the error

Why questions

Edge cases

Connections