Radioactivity

PhysicsNucleiFor JEE aspirants

Radioactivity is the spontaneous emission of $α$ -particles, $β$ -particles or $γ$ -rays by unstable nuclei. Each decay changes the nucleus in a fixed way ( $α$ : $A - 4$ , $Z - 2$ ; $β^{-}$ : $Z + 1$ ), and the number of undecayed nuclei falls exponentially, $N = N_{0} e^{- λ t}$ , halving every half-life $T_{1/2} = \frac{0.693}{λ}$ . Radioactivity, with its decay law, half-life, mean life and activity, is a core topic for JEE Advanced and also appears in JEE Main and NEET.

On this page1Discovery2α, β, γ rays3Displacement laws4α-decay energy5β-decay and neutrino6β⁺ and K-capture7γ-decay8Nuclear stability9Decay law10Half-life, mean life, activity11Special cases12Dating

Key Formulas - Quick Reference

★ Must learn $α$ : $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} He$ ; $β^{-}$ : $_{Z}^{A} X \to_{Z + 1}^{A} Y + e^{-} + \overset{ν}{ˉ}$ ; $β^{+}$ : $_{Z}^{A} X \to_{Z - 1}^{A} Y + e^{+} + ν$
Q values (atomic masses): $Q_{α} = (M_{X} - M_{Y} - M_{H e}) c^{2}$ , $Q_{β^{-}} = (M_{X} - M_{Y}) c^{2}$ , $Q_{β^{+}} = (M_{X} - M_{Y} - 2 m_{e}) c^{2}$ , $Q_{E C} = (M_{X} - M_{Y}) c^{2}$
★ Must learnKE of $α$ (parent at rest): $K_{α} = \frac{m _{Y}}{m _{Y} + m _{α}} Q \approx \frac{A - 4}{A} Q$
★ Must learnDecay law: $\frac{d N}{d t} = - λ N$ , $N = N_{0} e^{- λ t} = N_{0} (\frac{1}{2})^{t / T_{1/2}}$
★ Must learn $T_{1/2} = \frac{ln 2}{λ} = \frac{0.693}{λ}$ , mean life $τ = \frac{1}{λ} = 1.44 T_{1/2}$
★ Must learnActivity $A = λ N = A_{0} e^{- λ t}$ ; $1 Bq = 1$ decay per second, $1 Ci = 3.7 \times 1 0^{10} Bq$
Parallel decay: $λ = λ_{1} + λ_{2}$ , $\frac{1}{T} = \frac{1}{T _{1}} + \frac{1}{T _{2}}$ ; production at rate $R$ : $N = \frac{R}{λ} (1 - e^{- λ t})$
Series decay: $n_{α} = \frac{Δ A}{4}$ , $n_{β} = 2 n_{α} - Δ Z$ (with $Δ Z = Z_{parent} - Z_{end}$ )

1. Discovery and Nature of Radioactivity

In 1896 Henri Becquerel found that uranium salts fogged a wrapped photographic plate without any light. Marie and Pierre Curie then discovered the far more active elements polonium and radium. Rutherford showed that the radiation has three parts, which he named $α$ , $β$ and $γ$ .

Radioactivity is the spontaneous disintegration of an unstable nucleus with the emission of $α$ -particles, $β$ -particles (electrons or positrons) and $γ$ -rays. It is a purely nuclear process: it is not affected by temperature, pressure, chemical combination or electric and magnetic fields.

After emitting an $α$ or $β$ particle, the daughter nucleus is often left in an excited state and then emits a $γ$ -ray. All nuclei with $Z > 83$ are radioactive, and so are many lighter ones with too many or too few neutrons (Section 7).

Property	$α$ -particle	$β$ -particle	$γ$ -ray
Nature	Helium nucleus $_{2}^{4} He^{2 +}$ (2p + 2n)	Electron ( $β^{-}$ ) or positron ( $β^{+}$ )	Photon (electromagnetic wave)
Charge	$+ 2 e$	$- e$ or $+ e$	0
Rest mass	$6.64 \times 1 0^{- 27} kg$ ( $4.0015 u$ )	$9.1 \times 1 0^{- 31} kg$	0
Typical speed	$\approx 0.05 c$ ( $4$ to $9 MeV$ )	up to $0.99 c$	$c$
Energy spectrum	Discrete lines	Continuous up to a maximum	Discrete lines
Ionising power	Highest (about 100 × β)	Medium	Lowest
Penetrating power	Lowest: stopped by paper or $\approx 5 cm$ of air	Medium: a few mm of Al	Highest: several cm of Pb
Deflection in E and B fields	Small (heavy)	Large, opposite to $α$ for $β^{-}$	None

Figure 1: Left:

F = q v \times B

bends

α

and

β^{-}

opposite ways;

β

bends far more because

r = \frac{m v}{q B}

is small for the light electron. Right: ionising power falls and penetrating power rises from

α

β

γ

The $α$ -particle's mass is less than $4 m_{p}$ : it equals the mass of a helium atom minus two electrons, $4.002603 - 2 (0.000549) = 4.001506 u$ , which is $0.0304 u$ less than $2 m_{p} + 2 m_{n}$ because of its $28.3 MeV$ binding energy.

Antiparticles. Every particle has an antiparticle of the same mass and opposite charge: the positron $e^{+}$ is the antiparticle of the electron, and the antineutrino $\overset{ν}{ˉ}$ that of the neutrino. A particle and its antiparticle can annihilate: $e^{-} + e^{+} \to 2 γ$ , each photon carrying $0.511 MeV$ .

2. Displacement Laws

Soddy and Fajans found simple rules for the product of each decay. They follow from conservation of charge ( $Z$ ) and of nucleon number ( $A$ ):

Figure 2: Displacement laws.

α

Z - 2

A - 4

β^{-}

turns a neutron into a proton (

Z + 1

β^{+}

and electron capture turn a proton into a neutron (

Z - 1

β

decays keep

A

, so they move along a line of isobars.

$α$ -decay: $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} He$ . Example: $_{92}^{238} U \to_{90}^{234} Th + α$ .
$β^{-}$ -decay: $_{Z}^{A} X \to_{Z + 1}^{A} Y + e^{-} + \overset{ν}{ˉ}$ . Example: $_{90}^{234} Th \to_{91}^{234} Pa + e^{-} + \overset{ν}{ˉ}$ .
$β^{+}$ -decay: $_{Z}^{A} X \to_{Z - 1}^{A} Y + e^{+} + ν$ . Example: $_{11}^{22} Na \to_{10}^{22} Ne + e^{+} + ν$ .
$γ$ -decay: $_{Z}^{A} X^{*} \to_{Z}^{A} X + γ$ : no change in $A$ or $Z$ .

Exam Trick

Count decays in a series in two lines. Only $α$ changes $A$ , so $n_{α} = \frac{A _{start} - A _{end}}{4}$ . Each $α$ lowers $Z$ by 2 and each $β^{-}$ raises it by 1: $n_{β} = 2 n_{α} - (Z_{start} - Z_{end})$ . For $_{92}^{238} U \to_{82}^{206} Pb$ : $n_{α} = 8$ , $n_{β} = 16 - 10 = 6$ .

Figure 3: The uranium series (

4 n + 2

family). From

^{238} U

to stable

^{206} Pb

Δ A = 32

needs

8 α

, which lower

Z

16

; since

Z

falls only by

10

, there must be

6 β^{-}

decays.

Key idea

α

moves a nucleus 4 down in

A

and 2 down in

Z

;

β

keeps

A

and shifts

Z

by one;

γ

changes neither.

3. Energy in Alpha Decay

The $Q$ value is the rest-mass energy lost, shared as kinetic energy of the products. The nuclear masses are $M_{X} - Z m_{e}$ for the parent, $M_{Y} - (Z - 2) m_{e}$ for the daughter and $M_{H e} - 2 m_{e}$ for the $α$ , written with atomic masses $M$ . The electron masses cancel:

Q_{α} = (M_{X} - M_{Y} - M_{H e}) c^{2}

3.1 How the energy is shared

The parent is at rest and no external force acts, so the total momentum stays zero: $p_{α} = p_{Y} = p$ .
Kinetic energy $K = \frac{p ^{2}}{2 m}$ , so $m_{α} K_{α} = m_{Y} K_{Y}$ , and $K_{α} + K_{Y} = Q$ .
Solve:
$K_{α} = \frac{m _{Y}}{m _{α} + m _{Y}} Q \approx \frac{A - 4}{A} Q, K_{Y} = \frac{m _{α}}{m _{α} + m _{Y}} Q \approx \frac{4}{A} Q$

Figure 4:

α

-decay of

^{226} Ra

(

Q = 4.871 MeV

). The

α

takes

\frac{A - 4}{A}

of the available energy. Two groups of

α

-particles with sharp energies appear because the daughter can be left in an excited state, which then emits a

γ

-ray.

So the $α$ -particle carries about $98%$ of $Q$ for a heavy nucleus, and every $α$ from the same transition has the same energy. In a magnetic field perpendicular to its velocity it moves in a circle of radius $r = \frac{m v}{q B} = \frac{2 m K}{2 e B}$ , the same for all of them. Experiment shows a few groups of sharp radii, because the daughter may be left in an excited state $Y^{*}$ of energy $E^{*}$ which then emits a $γ$ -ray; those $α$ 's share only $Q - E^{*}$ :

K_{α} + K_{Y} = Q - E_{γ}, K_{α} = \frac{m _{Y}}{m _{α} + m _{Y}} (Q - E_{γ})

JEE Advanced

Why $α$ and not a single proton or neutron? $^{4} He$ is so tightly bound ( $28.3 MeV$ ) that emitting it gives heavy nuclei a positive $Q$ , while emitting one nucleon would cost about $6$ to $8 MeV$ . The $α$ is formed inside the nucleus but trapped by the Coulomb barrier ( $\approx 25$ to $30 MeV$ for uranium), far above its $4$ to $9 MeV$ ; it escapes by quantum tunnelling (Gamow, 1928). The tunnelling probability is extremely sensitive to energy, which gives the Geiger-Nuttall law: $lo g T_{1/2}$ falls roughly linearly with $\frac{Z}{K _{α}}$ . $^{238} U$ ( $4.27 MeV$ ) has $T_{1/2} = 4.5 \times 1 0^{9} y$ , while $^{212} Po$ ( $8.95 MeV$ ) has $0.3 μ s$ .

4. Beta Decay and the Neutrino

In $β^{-}$ decay a neutron inside the nucleus turns into a proton; the electron is created at that moment (there are no electrons inside nuclei):

n \to p + e^{-} + \overset{ν}{ˉ}

If only the electron and the daughter were emitted, momentum conservation would give the electron a fixed energy (almost all of $Q$ , since $m_{e} ≪ m_{Y}$ ), and every $β$ would move on the same circle in a magnetic field. Instead the $β$ -particles have every energy from zero up to a maximum $K_{ma x} \approx Q$ :

Figure 5:

α

-particles come in sharp lines;

β

-particles form a continuous spectrum up to

T_{ma x} = Q

(1.71 MeV for

^{32} P

; shape from the allowed statistical formula without the Coulomb correction). The missing energy is carried off by the antineutrino.

To save energy, momentum and angular momentum conservation, Wolfgang Pauli (1930) proposed that a third, undetected particle shares the energy: the neutrino $ν$ (named by Fermi; its antiparticle $\overset{ν}{ˉ}$ is emitted in $β^{-}$ decay). It was detected directly in 1956 by Cowan and Reines.

Charge zero; spin $\frac{1}{2}$ (like the electron, proton and neutron, so spin is balanced in $n \to p + e^{-} + \overset{ν}{ˉ}$ ).
Extremely small but non-zero rest mass (neutrino oscillations, 1998); in decay problems treat it as massless, moving at nearly $c$ with $E = p c$ .
Interacts only through the weak force, so it passes through the whole Earth almost unaffected.

Since $\overset{ν}{ˉ}$ carries away a random share of $Q$ , the electron energy varies: $K_{e} + K_{\overset{ν}{ˉ}} + K_{Y} = Q$ . With nuclear masses $M_{X} - Z m_{e}$ and $M_{Y} - (Z + 1) m_{e}$ plus the emitted electron $m_{e}$ , the electrons balance exactly:

Q_{β^{-}} = (M_{X} - M_{Y}) c^{2}

Quick Recall: tap to check

Why do

α

-particles from one transition all have the same energy?

Only two bodies share

Q

; momentum conservation fixes the split,

K_{α} = \frac{A - 4}{A} Q

Why is the

β

spectrum continuous?

Three bodies share

Q

: the antineutrino takes a variable part.

Where does the electron in

β^{-}

decay come from?

It is created when a neutron turns into a proton:

n \to p + e^{-} + \overset{ν}{ˉ}

5. Positron Emission and Electron Capture

In a nucleus with too many protons, a proton can turn into a neutron in two ways.

β⁺ decay (positron emission)

$p \to n + e^{+} + ν$ . A free proton cannot do this (the neutron is heavier); inside a nucleus the energy comes from the binding. With atomic masses:

Q_{β^{+}} = (M_{X} - M_{Y} - 2 m_{e}) c^{2}

so it needs $M_{X} - M_{Y} > 2 m_{e}$ ( $1.022 MeV$ ).

Electron capture (K-capture)

$p + e^{-} \to n + ν$ : the nucleus absorbs an inner (K-shell) electron.

_{Z}^{A} X + e^{-} \to_{Z - 1}^{A} Y + ν, Q_{E C} = (M_{X} - M_{Y}) c^{2}

Possible even when $β^{+}$ is not; it competes with $β^{+}$ .

Figure 6: K-capture (electron capture).

_{Z}^{A} X +_{- 1}^{0} e \to_{Z - 1}^{A} Y + ν

. The only particle leaving the nucleus is a neutrino, so the process is detected by the X-rays emitted as the daughter atom refills its K shell.

Electron capture leaves a vacancy in the K shell. An outer electron drops into it and the atom emits the characteristic X-ray of the daughter element, which is how K-capture was discovered. The rules to remember: every $p \to n$ change (in $β^{+}$ or EC) releases a neutrino, and every $n \to p$ change ( $β^{-}$ ) releases an antineutrino. Other processes, such as fission, release neutrons with no neutrino.

Exam Trick

Only $β^{+}$ needs the $- 2 m_{e}$ . With atomic masses: $α$ , $β^{-}$ and EC use $(M_{X} - M_{Y} - \dots) c^{2}$ directly; for $β^{+}$ subtract $2 m_{e} c^{2} = 1.022 MeV$ . Example: $^{22} Na$ ( $21.994438 u$ ) $\to^{22} Ne$ ( $21.991385 u$ ) gives $Q_{E C} = 2.84 MeV$ but $Q_{β^{+}} = 1.82 MeV$ .

6. Gamma Decay

Like an atom, a nucleus has discrete energy levels, but they are spaced by keV to MeV instead of eV. A nucleus left in an excited state after $α$ or $β$ decay drops to lower levels by emitting $γ$ -ray photons: $_{Z}^{A} X^{*} \to_{Z}^{A} X + γ$ .

Figure 7:

γ

-decay. After

β^{-}

decay the daughter

^{60} Ni

is left excited and drops to its ground state by emitting two

γ

-rays. Their energies equal the gaps between nuclear levels, just as spectral lines equal gaps between atomic levels, but are a million times larger.

$γ$ -rays have wavelengths below about $0.01 nm$ . Their energies are sharp and identify the nucleus; the $1.17$ and $1.33 MeV$ rays of $^{60} Co$ are used in cancer therapy and to sterilise medical equipment.

7. Nuclear Stability and the N/Z Ratio

Figure 8: The belt of stability (dots: all naturally occurring nuclides). It follows

N = Z

for light nuclei and bends to

\frac{N}{Z} \approx 1.5

for the heaviest, because extra neutrons are needed to offset Coulomb repulsion. Radioactive decay moves a nucleus towards the belt.

Light stable nuclei have $N \approx Z$ ( $^{4} He$ , $^{12} C$ , $^{16} O$ , $^{40} Ca$ ). Heavy stable nuclei need extra neutrons to dilute Coulomb repulsion, so $\frac{N}{Z}$ rises to about $1.5$ for $^{208} Pb$ .
Above the belt (neutron-rich, $\frac{N}{Z}$ too large): $β^{-}$ decay turns $n \to p$ , moving the nucleus down and to the right towards the belt.
Below the belt (proton-rich): $β^{+}$ decay or electron capture turns $p \to n$ .
Very heavy ( $Z > 83$ ): $α$ decay reduces both $Z$ and $N$ . No element beyond bismuth is stable; all of the elements up to $Z = 118$ that have been made are radioactive. Technetium ( $Z = 43$ ) and promethium ( $Z = 61$ ) also have no stable isotope.
Even-even nuclei are especially stable, and nuclei with $2, 8, 20, 28, 50, 82, 126$ protons or neutrons (magic numbers) are extra stable.

Key idea

Radioactive decay always moves a nucleus towards the belt of stability:

β^{-}

if it has too many neutrons,

β^{+}

or EC if too many protons,

α

if it is too heavy.

8. The Law of Radioactive Decay

Rutherford and Soddy (1902) found that the rate of decay of a sample is proportional to the number of undecayed nuclei present. Each nucleus has the same probability $λ d t$ of decaying in the next short time $d t$ , whatever its age.

- \frac{d N}{d t} = λ N

$λ$ is the decay constant (SI unit $s^{- 1}$ ). It depends only on the nuclide: not on the amount, the time, temperature, pressure or chemical state. A larger $λ$ means a more unstable nucleus.

Separate the variables: $\frac{d N}{N} = - λ d t$ .
Integrate from $N_{0}$ at $t = 0$ to $N$ at $t$ :
$\int_{N_{0}}^{N} \frac{d N}{N} = - λ \int_{0}^{t} d t \Rightarrow ln \frac{N}{N _{0}} = - λ t$
So
$N = N_{0} e^{- λ t}$
and the number that have decayed is $N_{0} - N = N_{0} (1 - e^{- λ t})$ , which equals the number of daughter nuclei formed if the daughter is stable.

Figure 9:

N = N_{0} e^{- λ t} = N_{0} (\frac{1}{2})^{t / T_{1/2}}

. Each half-life halves what is left, whatever the starting amount. If decay continued at its initial rate

λ N_{0}

, everything would be gone at

t = τ

(the tangent). Plotting

ln N

(or

ln A

) against

t

gives a straight line of slope

- λ

, the standard way to measure

λ

8.1 Half-life

The half-life $T_{1/2}$ is the time in which half the nuclei present decay. Putting $N = \frac{N _{0}}{2}$ : $\frac{1}{2} = e^{- λ T_{1/2}}$ , so

T_{1/2} = \frac{ln 2}{λ} = \frac{0.693}{λ}

After $n$ half-lives ( $t = n T_{1/2}$ , $n$ need not be a whole number): $N = N_{0} (\frac{1}{2})^{n}$ .

Half-lives range from $1 0^{- 22} s$ to more than $1 0^{24}$ years: $^{214} Po$ $164 μ s$ , $^{131} I$ $8.0$ days, $^{60} Co$ $5.27$ years, $^{14} C$ $5730$ years, $^{226} Ra$ $1600$ years, $^{238} U$ $4.47 \times 1 0^{9}$ years. Decay is a statistical law: it predicts the behaviour of a large number of nuclei, not when a particular nucleus will decay.

8.2 Mean life

Individual nuclei live for different times. The mean (average) life is the sum of the lives of all nuclei divided by their number. The $λ N d t$ nuclei that decay between $t$ and $t + d t$ each lived for a time $t$ :

τ = \frac{1}{N _{0}} \int_{0}^{\infty} t λ N_{0} e^{- λ t} d t = \frac{1}{λ}, τ = \frac{T _{1/2}}{0.693} = 1.44 T_{1/2}

In one mean life $N$ falls to $\frac{N _{0}}{e} = 0.37 N_{0}$ , so $63%$ of the nuclei decay.

8.3 Activity

The activity $A$ (also written $R$ ) of a sample is its rate of decay, the number of disintegrations per second:

A = - \frac{d N}{d t} = λ N = λ N_{0} e^{- λ t} = A_{0} e^{- λ t}

Activity falls with the same half-life as $N$ . Units: $1 becquerel (Bq) = 1$ decay per second (SI); $1 curie (Ci) = 3.7 \times 1 0^{10} Bq$ , roughly the activity of $1 g$ of $^{226} Ra$ ; $1 rutherford = 1 0^{6} Bq$ .

Specific activity is the activity per unit mass, $\frac{A}{m} = \frac{λ N _{A}}{M}$ for a pure sample of molar mass $M$ . Since activity is what a Geiger counter measures, most numericals give $A$ rather than $N$ : $\frac{A}{A _{0}} = \frac{N}{N _{0}}$ .

Half-life

T_{1/2}

Time for half of the nuclei to decay. $T_{1/2} = \frac{0.693}{λ}$ . After it, $50%$ remain.

Mean life

τ

Average lifetime of a nucleus. $τ = \frac{1}{λ} = 1.44 T_{1/2}$ . After it, $37%$ remain.

Exam Trick

Use powers of $\frac{1}{2}$ whenever $t$ is a neat multiple of $T_{1/2}$ . $1% \approx \frac{1}{128}$ means about $7$ half-lives; $\frac{1}{1000} \approx \frac{1}{1024}$ means about $10$ . For other times use $t = \frac{T _{1/2}}{0.693} ln \frac{N _{0}}{N} = 3.32 T_{1/2} lo g_{10} \frac{N _{0}}{N}$ .

Quick Recall: tap to check

What fraction remains after

2.5

half-lives?

(\frac{1}{2})^{2.5} = 0.177

, about

18%

Does the half-life of a sample change as it gets older?

No.

λ

is constant, so every half-life is the same length.

How is activity related to the number of nuclei?

A = λ N

What fraction decays in one mean life?

1 - \frac{1}{e} = 63%

9. Special Cases of Decay

9.1 Parallel (branching) decay

Some nuclei can decay by two routes, for example $^{64} Cu$ by $β^{-}$ and by $β^{+}$ /EC. The probabilities add: $λ d t = λ_{1} d t + λ_{2} d t$ .

Figure 10: (a) Parallel decay (

λ_{1} = 0.6

λ_{2} = 0.4

per unit time): the decay constants add, so the effective half-life is shorter than either, and the products form in the fixed ratio

N_{B} : N_{C} = λ_{1} : λ_{2}

. (b) Production at rate

R

N = \frac{R}{λ} (1 - e^{- λ t})

approaches

\frac{R}{λ}

, when decay balances production, but never reaches it; this is how radioisotopes are made in a reactor.

λ = λ_{1} + λ_{2}, \frac{1}{T} = \frac{1}{T _{1}} + \frac{1}{T _{2}}, \frac{N _{B}}{N _{C}} = \frac{λ _{1}}{λ _{2}}

The effective half-life is shorter than either partial half-life.

9.2 Production at a constant rate

If a nuclide is produced at a constant rate $R$ (for example by neutron bombardment in a reactor) while decaying, $\frac{d N}{d t} = R - λ N$ (Figure 10, panel b). The number rises towards $\frac{R}{λ}$ , at which the activity $λ N$ equals the production rate. Irradiating for more than about $4$ half-lives gains little.

9.3 Successive (chain) decay

When the daughter is itself radioactive, $A λ_{1} B λ_{2} C$ , the rate of change of $B$ is gain from $A$ minus its own decay: $\frac{d N _{B}}{d t} = λ_{1} N_{A} - λ_{2} N_{B}$ .

Figure 11: Successive decay

A λ_{1} B λ_{2} C

N_{B}

is maximum when

λ_{1} N_{A} = λ_{2} N_{B}

, at

t = \frac{ln ( λ _{2} / λ _{1} )}{λ _{2} - λ _{1}}

. If the parent is longer-lived (

λ_{1} < λ_{2}

N_{B}

settles at

\frac{λ _{1}}{λ _{2} - λ _{1}} N_{A}

, which is

\approx \frac{λ _{1}}{λ _{2}} N_{A}

when

λ_{1} ≪ λ_{2}

: small, but not zero.

JEE Advanced

Radioactive equilibrium. If the parent lives much longer than the daughter ( $λ_{1} ≪ λ_{2}$ ), after a few daughter half-lives $\frac{d N _{B}}{d t} \approx 0$ and

λ_{1} N_{A} = λ_{2} N_{B} \Rightarrow \frac{N _{B}}{N _{A}} = \frac{λ _{1}}{λ _{2}} = \frac{T _{B}}{T _{A}}

(secular equilibrium): parent and daughter have equal activities. If the parent is only moderately longer-lived, the exact limit is $\frac{N _{B}}{N _{A}} = \frac{λ _{1}}{λ _{2} - λ _{1}}$ (transient equilibrium, Figure 11, panel b), which reduces to $\frac{λ _{1}}{λ _{2}}$ when $λ_{1} ≪ λ_{2}$ . This is how the half-life of $^{238} U$ is found from the ratio of radium to uranium in old ores, and why every member of the uranium series in an old rock has the same activity.

Key idea

Rates add for parallel routes; production balances decay at

N = \frac{R}{λ}

; in a chain the daughter peaks when

λ_{1} N_{A} = λ_{2} N_{B}

10. Radioactive Dating and Uses

Carbon dating. Cosmic rays keep the ratio $^{14} C :^{12} C$ in the atmosphere nearly constant ( $\approx 1.3 \times 1 0^{- 12}$ ). Living things take in carbon and keep the same ratio, giving about $15.3$ decays per minute per gram of carbon. After death no new $^{14} C$ enters, and it decays with $T_{1/2} = 5730$ years. Measuring the present activity $A$ gives the age:

t = \frac{T _{1/2}}{0.693} ln \frac{A _{0}}{A}

Figure 12: Carbon dating. A living sample shows

15.3

decays per minute per gram of carbon; after death the activity halves every

5730

years. A sample with

3.8

has age

t = \frac{T _{1/2}}{ln 2} ln \frac{A _{0}}{A} = 5730 \times lo g_{2} \frac{15.3}{3.8} \approx 11500

years, just over two half-lives.

Carbon dating works up to about $50000$ years. Older rocks are dated with longer-lived pairs such as $^{238} U \to^{206} Pb$ or $^{40} K \to^{40} Ar$ , from which the age of the Earth, about $4.5 \times 1 0^{9}$ years, was found. Other uses: $^{131} I$ for thyroid diagnosis and treatment, $^{60} Co$ in radiotherapy, radioactive tracers in medicine, agriculture and industry, and smoke detectors ( $^{241} Am$ ).

Safety. Radiation damages living cells and DNA. Work behind shielding, keep your distance (intensity falls as $\frac{1}{r ^{2}}$ for a point source), limit exposure time, and never ingest $α$ -emitters, which are harmless outside the body but very damaging inside it.

11. Problem-Solving Map and Revision

Use the flowchart to pick the formula, then the mind map to revise the whole concept.

Figure 13: Choosing the formula for a decay problem. Always check the units of

λ

and

t

match.

Figure 14: Mind map of this concept. Read the left column, then the right; cover a branch, recall its three points, then check.

12. Solved Examples

Solved Example 1

A radioactive nucleus can decay by two different processes. The half-life for the first process is

t_{1}

and that for the second is

t_{2}

. Show that the effective half-life

t

of the nucleus is given by

\frac{1}{t} = \frac{1}{t _{1}} + \frac{1}{t _{2}}

Solution:

The decay constants are $λ_{1} = \frac{ln 2}{t _{1}}$ and $λ_{2} = \frac{ln 2}{t _{2}}$ .

The probability that an undecayed nucleus decays by the first process in time $d t$ is $λ_{1} d t$ , and by the second $λ_{2} d t$ . The probability that it decays by either is $λ_{1} d t + λ_{2} d t$ . If the effective decay constant is $λ$ , this is also $λ d t$ :

λ = λ_{1} + λ_{2} \Rightarrow \frac{ln 2}{t} = \frac{ln 2}{t _{1}} + \frac{ln 2}{t _{2}}

Answer: $\frac{1}{t} = \frac{1}{t _{1}} + \frac{1}{t _{2}}$ . For example $t_{1} = 3 h$ and $t_{2} = 6 h$ give $t = 2 h$ .

Solved Example 2

A factory produces a radioactive substance A at a constant rate

R

, which decays with decay constant

λ

to form a stable substance B. Production starts at

t = 0

. Find (i) the number of nuclei of A and (ii) the number of nuclei of B at time

t

, and (iii) the maximum number of nuclei of A present at any time.

Solution:

(i) Net rate of change of A: $\frac{d N}{d t} = R - λ N$ .

\int_{0}^{N} \frac{d N}{R - λ N} = \int_{0}^{t} d t \Rightarrow - \frac{1}{λ} ln \frac{R - λ N}{R} = t \Rightarrow N_{A} = \frac{R}{λ} (1 - e^{- λ t})

(ii) Every nucleus made is either still A or has become B: $N_{B} = R t - N_{A} = \frac{R}{λ} (λ t - 1 + e^{- λ t})$ .

(iii) $N_{A}$ increases all the time and approaches $\frac{R}{λ}$ as $t \to \infty$ .

Answer: $N_{A} = \frac{R}{λ} (1 - e^{- λ t})$ , $N_{B} = \frac{R}{λ} (λ t - 1 + e^{- λ t})$ ; the maximum (limiting) value of $N_{A}$ is $\frac{R}{λ}$ , approached but never exactly reached.

Solved Example 3

A radioactive substance A with

N_{0}

active nuclei at

t = 0

decays to a radioactive substance B with decay constant

λ_{1}

. B decays to a stable substance C with decay constant

λ_{2}

. (a) Find the numbers of nuclei of A, B and C at time

t

. (b) What does the answer for B become if

λ_{1} ≫ λ_{2}

, and if

λ_{1} ≪ λ_{2}

Solution:

(a) For A: $N_{1} = N_{0} e^{- λ_{1} t}$ . For B: $\frac{d N _{2}}{d t} = λ_{1} N_{1} - λ_{2} N_{2}$ , i.e. $\frac{d N _{2}}{d t} + λ_{2} N_{2} = λ_{1} N_{0} e^{- λ_{1} t}$ .

Multiply by the integrating factor $e^{λ_{2} t}$ : $\frac{d}{d t} (N_{2} e^{λ_{2} t}) = λ_{1} N_{0} e^{(λ_{2} - λ_{1}) t}$ .

Integrate: $N_{2} e^{λ_{2} t} = \frac{λ _{1} N _{0}}{λ _{2} - λ _{1}} e^{(λ_{2} - λ_{1}) t} + C$ . With $N_{2} = 0$ at $t = 0$ , $C = - \frac{λ _{1} N _{0}}{λ _{2} - λ _{1}}$ .

N_{2} = \frac{λ _{1} N _{0}}{λ _{2} - λ _{1}} (e^{- λ_{1} t} - e^{- λ_{2} t}), N_{3} = N_{0} - N_{1} - N_{2}

(b) If $λ_{1} ≫ λ_{2}$ : $e^{- λ_{1} t}$ dies out almost at once and $λ_{2} - λ_{1} \approx - λ_{1}$ , so $N_{2} \approx N_{0} e^{- λ_{2} t}$ : A turns into B almost instantly and B then decays on its own.

If $λ_{1} ≪ λ_{2}$ : after a short time $e^{- λ_{2} t}$ is negligible and $λ_{2} - λ_{1} \approx λ_{2}$ , so $N_{2} \approx \frac{λ _{1}}{λ _{2}} N_{0} e^{- λ_{1} t} = \frac{λ _{1}}{λ _{2}} N_{1}$ .

Answer: $N_{2} \approx N_{0} e^{- λ_{2} t}$ for $λ_{1} ≫ λ_{2}$ ; $N_{2} \approx \frac{λ _{1}}{λ _{2}} N_{0} e^{- λ_{1} t}$ (small, not zero: secular equilibrium) for $λ_{1} ≪ λ_{2}$ .

Solved Example 4

_{88}^{226} Ra

decays by

α

-emission to

_{86}^{222} Rn

. Atomic masses:

^{226} Ra = 226.025408 u

^{222} Rn = 222.017576 u

^{4} He = 4.002603 u

. Find

Q

and the kinetic energy of the

α

-particle.

Solution:

$Q = (226.025408 - 222.017576 - 4.002603) \times 931.5 = 0.005229 \times 931.5 = 4.871 MeV$ .

$K_{α} = \frac{A - 4}{A} Q = \frac{222}{226} \times 4.871$ .

Answer: $Q = 4.87 MeV$ , $K_{α} = 4.78 MeV$ ; the radon nucleus recoils with $0.086 MeV$ .

Solved Example 5

Find the Q value for the

β^{+}

decay

_{11}^{22} Na \to_{10}^{22} Ne + e^{+} + ν

. Atomic masses:

^{22} Na = 21.994438 u

^{22} Ne = 21.991385 u

m_{e} = 0.000549 u

Solution:

Nuclear masses: initial $M_{N a} - 11 m_{e}$ ; final $(M_{N e} - 10 m_{e}) + m_{e}$ (the positron). So $Q = (M_{N a} - M_{N e} - 2 m_{e}) c^{2}$ .

$= (21.994438 - 21.991385 - 0.001098) \times 931.5 = 0.001955 \times 931.5$ .

Answer: $Q \approx 1.82 MeV$ , shared between the positron and the neutrino.

Solved Example 6

The half-life of

^{131} I

8.0

days. (a) What fraction of a sample is left after

24

days? (b) How long does it take for the activity to fall to

1%

of its initial value?

Solution:

(a) $n = \frac{24}{8} = 3$ half-lives: $\frac{N}{N _{0}} = (\frac{1}{2})^{3} = \frac{1}{8}$ .

(b) $t = \frac{T _{1/2}}{ln 2} ln \frac{A _{0}}{A} = \frac{8.0}{0.693} ln 100 = 11.54 \times 4.605$ .

Answer: (a) $\frac{1}{8}$ ( $12.5%$ ); (b) $t \approx 53$ days (about $6.6$ half-lives).

Solved Example 7

Find the activity of

1 g

^{226} Ra

(

T_{1/2} = 1600

years) in becquerel and curie.

Solution:

$N = \frac{1}{226} \times 6.022 \times 1 0^{23} = 2.66 \times 1 0^{21}$ .

$λ = \frac{0.693}{1600 \times 3.156 \times 1 0 ^{7} s} = 1.37 \times 1 0^{- 11} s^{- 1}$ .

$A = λ N = 1.37 \times 1 0^{- 11} \times 2.66 \times 1 0^{21}$ .

Answer: $A \approx 3.66 \times 1 0^{10} Bq \approx 0.99 Ci$ , which is why the curie was defined from radium.

Solved Example 8

A piece of ancient wood shows a

^{14} C

activity of

3.8

decays per minute per gram of carbon, while living wood gives

15.3

. Find the age of the wood (

T_{1/2} = 5730

years).

Solution:

$\frac{A _{0}}{A} = \frac{15.3}{3.8} = 4.03$ , almost exactly $4 = 2^{2}$ , so about two half-lives.

Exactly: $t = \frac{5730}{0.693} ln 4.03 = 8268 \times 1.393$ .

Answer: $t \approx 1.15 \times 1 0^{4}$ years (about $11500$ years).

Solved Example 9

_{92}^{238} U

decays through a series of

α

and

β^{-}

decays to

_{82}^{206} Pb

. The numbers of

α

and

β^{-}

particles emitted are
(A)

8, 6

(B)

6, 8

(C)

8, 8

(D)

6, 6

Solution:

Answer: (A). $n_{α} = \frac{238 - 206}{4} = 8$ . The $8 α$ lower $Z$ by $16$ to $76$ ; reaching $82$ needs $6 β^{-}$ decays.

Solved Example 10

The fraction of a radioactive sample that decays during one mean life is
(A)

1/2

(B)

1/ e

(C)

1 - 1/ e

(D)

ln 2

Solution:

Answer: (C). At $t = τ = \frac{1}{λ}$ , $N = N_{0} e^{- 1}$ , so the fraction decayed is $1 - \frac{1}{e} \approx 0.63$ . Option (B) is the fraction remaining.

Solved Example 11

A source in a lead block sends

α

β^{-}

and

γ

rays vertically upward into a magnetic field directed into the page. Which statement is correct?
(A)

α

bends to the right
(B)

β^{-}

bends to the left
(C)

α

bends to the left, less sharply than

β^{-}

bends to the right
(D) all three bend the same way

Solution:

Answer: (C). $F = q v \times B$ with $v$ up and $B$ into the page points to the left for a positive charge, so $α$ bends left and $β^{-}$ right. The radius $r = \frac{m v}{q B}$ is far larger for the heavy $α$ ( $\frac{m}{q}$ about $3700$ times that of the electron, only partly offset by its lower speed), so the $β^{-}$ track curves much more sharply. $γ$ is undeflected (Figure 1).

Solved Example 12

The tangent to the

N

t

decay curve at

t = 0

meets the time axis at
(A)

T_{1/2}

(B)

1/ λ

(C)

2/ λ

(D)

T_{1/2} ln 2

Solution:

Answer: (B). The initial slope is $\frac{d N}{d t}_{0} = - λ N_{0}$ , so the tangent is $N = N_{0} - λ N_{0} t$ , which is zero at $t = \frac{1}{λ} = τ$ , the mean life ( $1.44 T_{1/2}$ ). If the sample kept decaying at its initial rate it would be gone in one mean life (Figure 9).

Solved Example 13

In an old uranium ore, secular equilibrium holds between

^{238} U

and its descendant

^{226} Ra

(

T_{1/2} = 1600

years). The ore contains

1

radium atom for every

2.8 \times 1 0^{6}

uranium atoms. Estimate the half-life of

^{238} U

Solution:

In secular equilibrium all members have equal activity: $λ_{U} N_{U} = λ_{R a} N_{R a}$ , so $\frac{T _{U}}{T _{R a}} = \frac{N _{U}}{N _{R a}}$ .

$T_{U} = 1600 \times 2.8 \times 1 0^{6} = 4.5 \times 1 0^{9}$ years.

Answer: $T_{1/2} (^{238} U) \approx 4.5 \times 1 0^{9}$ years, about the age of the Earth. This is how half-lives far too long to watch are measured.

Practice Questions

Write the products: (a) $_{92}^{238} U$ emits an $α$ ; (b) $_{90}^{234} Th$ emits a $β^{-}$ .Answer: (a) $_{90}^{234} Th$ ; (b) $_{91}^{234} Pa$
The half-life of a nuclide is $2 h$ . What fraction is left after $6 h$ ?Answer: $\frac{1}{8}$
Find the decay constant and mean life of a nuclide with $T_{1/2} = 1386 s$ .Answer: $λ = 5.0 \times 1 0^{- 4} s^{- 1}$ , $τ = 2000 s$
The activity of a sample falls from $800$ to $100$ counts per second in $30 min$ . Find the half-life.Answer: $10 min$
A nucleus decays by two routes with half-lives $3 h$ and $6 h$ . Find its effective half-life.Answer: $2 h$
How many $α$ and $β^{-}$ decays take $_{90}^{232} Th$ to $_{82}^{208} Pb$ ?Answer: $6 α$ and $4 β^{-}$
A nucleus with $A = 212$ at rest emits an $α$ -particle; $Q = 6.2 MeV$ . Find the kinetic energy of the $α$ .Answer: $\frac{208}{212} \times 6.2 = 6.08 MeV$

Common Mistakes to Avoid

Watch out

Thinking temperature, pressure or chemical combination changes the half-life. The decay constant is a property of the nucleus alone.
Assuming a sample is fully decayed after two half-lives. After $2 T_{1/2}$ one quarter is still left; the decay never quite finishes.
Confusing half-life and mean life: $τ = 1.44 T_{1/2}$ , not $T_{1/2}$ or $\frac{T _{1/2}}{2}$ .
Adding half-lives for parallel decay. It is the decay constants that add: $\frac{1}{T} = \frac{1}{T _{1}} + \frac{1}{T _{2}}$ .
Forgetting the $2 m_{e}$ in $β^{+}$ decay with atomic masses, or wrongly subtracting it in $β^{-}$ decay or electron capture.
Giving the whole $Q$ to the $α$ -particle. The daughter recoils: $K_{α} = \frac{A - 4}{A} Q$ .
Saying $β$ -particles are electrons from the atom's shells. They are created in the nucleus when $n \to p$ .
Writing that $γ$ -decay changes $Z$ or $A$ , or mixing up the directions: $α$ and $β^{-}$ are deflected in opposite directions, $γ$ not at all.

Frequently Asked Questions

What is radioactivity?

Radioactivity is the spontaneous emission of alpha particles, beta particles or gamma rays by unstable atomic nuclei. It was discovered by Henri Becquerel in 1896. It is a nuclear process, so it is not affected by temperature, pressure or chemical combination.

What are the differences between alpha, beta and gamma rays?

Alpha particles are helium nuclei with charge plus 2e, strongly ionising and stopped by paper. Beta particles are electrons or positrons, less ionising and stopped by a few millimetres of aluminium. Gamma rays are high-energy photons with no charge, the most penetrating, reduced only by thick lead.

What is the law of radioactive decay?

The rate of decay is proportional to the number of undecayed nuclei, so $\frac{d N}{d t} = - λ N$ . Integrating gives $N = N_{0} e^{- λ t}$ : the number of undecayed nuclei falls exponentially, halving every half-life. The decay constant depends only on the nuclide.

What is the relation between half-life, decay constant and mean life?

Half-life equals 0.693 divided by the decay constant, and mean life equals one divided by the decay constant. So the mean life is 1.44 times the half-life. After one half-life 50 percent of the nuclei remain; after one mean life about 37 percent remain.

Why is the beta spectrum continuous?

In beta decay three particles share the energy: the daughter nucleus, the electron and an antineutrino. The antineutrino takes a variable share, so the electron can have any energy from zero up to the Q value. Pauli proposed the neutrino in 1930 to explain this.

What is the activity of a radioactive sample?

Activity is the number of decays per second, equal to the decay constant times the number of undecayed nuclei. Its SI unit is the becquerel, one decay per second; the older unit curie equals 3.7 times ten to the ten becquerel. Activity decreases with the same half-life as the number of nuclei.

Is radioactivity in the NEET syllabus?

Radioactivity was dropped from the rationalised NCERT textbook, so recent NEET papers focus on nuclear binding energy, fission and fusion. Older papers and many test series still ask half-life, activity and the displacement laws, so learn the decay law and $T_{1/2} = \frac{0.693}{λ}$ for revision.

How is radioactivity tested in JEE Main and Advanced?

JEE Advanced lists radioactivity explicitly: alpha and beta Q values with atomic masses, kinetic energy sharing, the decay law, half-life, mean life, activity, parallel decay, production with decay and successive decay. JEE Main questions on half-life and activity also appear in older papers and mock tests.

Previous year questions on Radioactivity

14 questions from past papers, each with a step-by-step solution.

Show all 14 questions

Ready to master Nuclei?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test