Radioactivity

6 methods · 20 questions

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Keep two independent ledgers for a decay or decay chain. An alpha emission lowers the mass number by $4$ and the proton number by $2$ ; a beta $^{-}$ emission leaves the mass number unchanged and raises the proton number by $1$ (a neutron turns into a proton).

Trigger: a decay described by counts of alpha and beta particles, or by a net change in mass number or proton number.

Instances:
(i) net proton change $Δ Z = - 2 n_{α} + n_{β}$ , net mass change $Δ A = - 4 n_{α}$ .
(ii) since only alpha changes the mass number, read $n_{α} = Δ A /4$ straight off the mass drop, decoupled from the betas, then recover $n_{β}$ from $Δ Z$ .
(iii) with a single alpha in the chain, the mass number can only ever be the start value or that value minus $4$ , and beta steps only raise $Z$ , so any nuclide off those two mass numbers (or below the starting $Z$ while the mass is unchanged) is unreachable.

Linked questions (3)

NSAA 2016 Section 1 Part B (ESAT Physics) · Q20

A nuclide

_{82}^{214} Pb

changes by radioactive decay into the nuclide

_{82}^{210} Pb

.

Which combination of emissions produces this change?

A3 alpha
B2 alpha and 1 beta
C2 alpha and 2 beta
D1 alpha and 2 beta
E3 beta

Worked solution

Answer:D

An alpha particle emission reduces the mass number by 4 and the proton number by 2; a beta

^{-}

particle emission leaves the mass number unchanged and increases the proton number by 1.

The mass number falls from

214

210

, a decrease of

4

, so exactly one alpha particle must be emitted (any other number of alphas would change the mass number by a different multiple of 4).

The proton number is unchanged (

82

82

). One alpha emission alone would reduce the proton number by 2, down to 80. To bring it back up to 82, two beta

^{-}

particles must also be emitted (each raises the proton number by 1):

80 + 2 = 82

.

So the combination is 1 alpha and 2 beta.

The answer is D.

NSAA 2017 Section 1 Part B (ESAT Physics) · Q27

The nuclide

_{Q}^{P} X

decays to the stable nuclide Y. During this process four particles are emitted: an

α

-particle and three

β^{-}

particles.

Which of the following is not a nuclide that could be formed at any stage during this process?

Aatomic mass $P$ , atomic number $Q - 1$
Batomic mass $P$ , atomic number $Q + 1$
Catomic mass $P$ , atomic number $Q + 2$
Datomic mass $P$ , atomic number $Q + 3$
Eatomic mass $P - 4$ , atomic number $Q - 2$
Fatomic mass $P - 4$ , atomic number $Q - 1$
Gatomic mass $P - 4$ , atomic number $Q$
Hatomic mass $P - 4$ , atomic number $Q + 1$

Worked solution

Answer:A

Only one alpha particle is emitted in total, so at any stage of the process the mass number is either still

P

(before the alpha has been emitted) or already

P - 4

(after it). Beta

^{-}

emission never changes the mass number, so it cannot move a nuclide between these two mass numbers.

While the mass number is still

P

(options A to D), only beta

^{-}

emissions can have happened so far (0, 1, 2 or 3 of the three available), and each one only ever INCREASES the atomic number by 1. So the atomic number while mass

= P

can only be

Q

Q + 1

Q + 2

Q + 3

: it can never be

Q - 1

, since nothing in this decay chain ever decreases the atomic number while the mass number is still

P

. Option A (atomic mass

P

, atomic number

Q - 1

) is therefore impossible.

Options B, C and D (

Q + 1

Q + 2

Q + 3

at mass

P

) correspond to 1, 2 or 3 beta emissions having occurred before the alpha: all reachable. Once the mass number has dropped to

P - 4

(after the alpha, which itself lowers the atomic number by 2), the atomic number is

Q - 2

plus however many of the three betas have occurred so far (0, 1, 2 or 3):

Q - 2

Q - 1

Q

Q + 1

, matching options E, F, G and H exactly: all reachable (H is the final, stable nuclide Y itself, after all four emissions).

The answer is A.

NSAA 2018 Section 1 Part B (ESAT Physics) · Q19

An unstable nucleus X becomes a stable nucleus Y after a succession of decays, during which a total of 5 alpha particles and 2 beta (

β^{-}

) particles are emitted.

How many fewer protons does nucleus Y contain than nucleus X?

Worked solution

Answer:B

Each alpha particle emitted reduces the proton (atomic) number by 2; each beta

^{-}

particle emitted increases it by 1.

Change in proton number

= - 2 \times 5 + 1 \times 2 = - 10 + 2 = - 8

.

So Y has

8

fewer protons than X.

The answer is B.