Radioactivity, Physics tutorial


The decay of the nucleus of an atom can either be natural or artificial. This decay occurs in nuclide in order that they may attain stability. This disintegration or decay occurs with the emission of some particles or energy.

Stability of Nuclides:

The stability of nuclides is mainly determined by the atomic mass (A) and the (N⁄Z) ratio.  The condition for the stability of eight elements is N/Z =1, and for heavy elements, N/Z  ≈ 1.5 Nuclides that are not stable due to this ratio seek stability by undergoing inter-nuclear spontaneous transformation which shifts the N/Z  ratio to a more stable configuration.

During this transformation, the following could occur:

i) 11p→10n + 01β + V

ii) 10n→11p + 01β + V

iii) α particles may be emitted (that is 2 protons and 2 neutrons)

iv) Splitting of nucleus into two nearly equal fragments through nuclear fission.

Radioactivity can therefore be defined as the tendency of unstable nuclides, seeking to become stable through the emission of particles and energy. The emitted particles during radioactivity are referred to as nuclear radiation.

Nuclear radiations can well be referred as ionizing radiation because they have sufficient energy to cause the production of ion pairs in any medium which they pass through. The most common particles usually emitted are β+, β-, α and γ particle.

α Particles:

The nuclear transformation equation for an α- particle is given by:

AZX→A-4Z-2Y + 42HE(42α) + γ + Q

23892U→23490TH + 42HE(42α) + γ + Q


i) α- particles can be stopped by a thin sheet of paper

ii) They cause intense ionization in air

iii) A group of α- particle emitted from the same type of nuclides usually have definite velocity and energy

iv) α - particles cover a definite distance in a given material practically without any loss of intensity and suddenly in a small distance are absorbed completely

The definite distance traveled with a given material is called the RANGE

β - Particles:

 The nuclear transformation equation for an

 AZX→ AZ+1Y + 0-1β-1 + V ‾ + Q

146C→147N + 0-1β-1 + V ‾ + Q


i) They cause less ionization in air.

ii) 100 times more penetrating than α-particle and can penetrate asheet of aluminum, a few millimeters thick.

iii) A particular β - active element emits β- particles with energies varying between zero and certain maximum. This maximum energy is called the end point energy.

Kinematics of Radioactivity:

When a nucleus disintegrates by emitting particle α, β, γ or captures the electron from atomic shell (k-capture), this procedure is known as Radioactive decay. All nuclear decay follows single law known as a Decay law.

Number of nuclei of given radioactive sample disintegrating per second is known as Activity of sample i.e. A = dN/dt.

Activity (dN/dt) at any instant of time is proportional to number N of parents type present at that time.

i.e. dN/dt α N

A = dN/dt = -λN

Here λ is decay constant or disintegration constant that only depends on nature or characteristics of radioactive sample and not on amount of substance. Also λ- provides probability of decay per unit interval of time.

N0NdN/N = ∫0t - λdt

ln(N/N0) = -λt

N/N0 = e-λt

N = Noe-λt

Multiply both sides of equation by λ

λN = λNoet

A = Aoe-λt

Here A stands for activity. Activity is estimated in Becquerel (1dps) and 1 curie (1ci) = 3.7x1010 Bq.

Time interval during which half of given sample of radioactive substance decays is referred to as Half Life.

A = A0/2 = A0e-λt1/2

1/2 = e-λt1/2

Taking log on both sides

-ln2 = -λt1/2

T1/2 = ln2/λ = 0.693

As individual radioactive atoms may have life spans between o and ∞, we can then talk of average life or mean life.

Tmean = total life time of all nuclei in given sample/total number of nuclei in that sample

[t1dN1 + t2dN2------+tNdNN]/[dN1 + dN2 + -------dNN]

Tmean = ∫N00tdN/∫N00dN = -1/N0N00tdN

But N = N0e-λt

λN = λN0e-λt

Since dN/dt = -λN

dN/dt = -λN0e-λt

dN = -λN0e-λtdt

By substituting values

= -1/N0N00t(+λ)N0e-λtdt

Tmean = λ∫0∞te-λtdt

N = Noe-λt

For N = O = N0e-λt

O = e-λt

O = 1/eλt = t → ∞ and for N = N0 = N0e-λt

1 = et

1 = e/eλt = t→0

For relationships to hold, then integrate equation by parts.

∫udv = uv-∫vdu

Where u = t and dv = e-λt

du = 1dt and v = ∫e-λt = -e-λt/λ + C

Tmean = λ[t(-e-λt/λ) - ∫0 -e-λt/λdt]

Tmean = λ[(-te-λt(λ)-e-λt)/λ2]0

= λ/λ2[-te-λt(λ)-et]0∞ + c

= 1/λ[0+1] + c

= 1/λ

Radioactive Equilibrium:

Considering this decay process A→B→C (Stable). As number of nuclei entering B will be decay of A.

-dNA/dt = λANA

Number of nuclei leaving B will be λBNB

Thus, net charge in number of nuclei per second of B is

dNB/dt = λANA - λBNB

But this is first order (linear d.e.) and NA = Noe-λAt

dNB/dt + λBNB = λANA

dNB/dt + λBNB = λAN0e-λAt

Multiply both sides of equation by integrating factor = eλBt

d/dt(λBtNB) = λAN0e-λAt(eλBt)

Now integrating both sides with t


Using integration by parts

du = -λAe-λAt and v = eλBtB

∫udv = uv - ∫vdu

eλAt.eλBtB + λAB∫eλBt.e-λAtdt

Then equation is

I = (e-λAt.eλBtB + λAB)I

I(I - λA/λB).I = (e-λAt.eλBt)/λB

(e-λAt.eλBt)/λB ÷ (λB - λA)/λB + C

(e-λAt.eλBt)/(λB - λA) + C

Then the equation will now be

e-λBtNB = λAN0[e-λAt.eλBt/(λB - λA)] + C

NB = (N0λAB - λA)[e-λAt - e-λBt]


1. At t maximum, dN/dt = 0 i.e λANA = λBNB and activity of parent and daughter are said to be at equilibrium. This is known as Ideal Equilibrium

2. Considering the case whereby daughter is short lived than parent i.e. TA>TB from equation above, activity of B is

λBNB = (N0λAB - λA).λB[e-λAt - e-λBt]

N0λA = λANA/e-λAt

λBNBANA = λB/(λB - λA)[1 - e-(λBt - λAt)]

Since T = 1/λ hence λB = 1/T

λB = 1/TB, λA = 1/TA

Then equation becomes

λBNBANA = (TA/(TA - TB))[1 - e-(λBt - λBt)]

As (λB - λA)t = (TA - TB/TBTA).t

At large time, t


When this equation holds, this means that transient equilibrium exists between parent and daughter, and its correlation is that

TA/(TA - TB>1 For(TA>TB)

For case in which daughter is long-curved, then parent i.e. (TA < TB). It follows from equation that ratio λBNBANA increases as t increases. Thus, after sufficient time, activity of daughter becomes independent of that of residual activity of parents.

4. If TA>>TB, λA<<λB then λBNB = λANA[1 - e-λBt]

For t = TB = e-λBt = 0


i.e. λBA = TA/TB = NA/NB

All these can be known as serial transformation.

Radioactive Series:

Radio nuclides which are related comprise a decay chain or series. Successive daughter products are formed through emission of β and α particles leading to stable end-product. There are four known series in nature but that of neptunium is artificial. They are:

i) Thorium 4n

ii) Neptunium 4n+1

iii) Uranium 4n+2

iv) Actinium 4n+3

There are also some that aren't of large atomic number e.g. 40K and all these comprise source of radioactivity in earth crust.


Normally, particular radio nuclide is assumed to decay through either α and β decay. In few cases, some will decay through α and β or both. This phenomenon is called as branching e.g.


Age Determination Using Radioisotopes:

Radioactivity is the best clock in determining or usually applied to estimate the absolute age of geological materials because it is totally not affected by environmental changes of natural processes like earthquakes, storms etc. Some of the radioisotopes that are useful for geological age dating include 87Sr / 87Rb, 14C/ 12C, Pb / U, Pb / Th etc.

The half life of these radioisotopes is usually used for determining the age ranges of interest. The determination of geological ages is done very often by the lead method which involves

AXX → A-4Z-2Y + 4zα + γ + Q

23892u → emission of 8(4zx)→20682pb

23592u → emission of 7(4zx)→ 20682pb

23290Th → emission of 6(4zx)→ 20682pb

These are natural decay series in which after enough time e.g. billion years only uranium and lead are elements left in appreciable amounts. This is for the reason that all elements in uranium series are in secular equilibrium with parents except 20682pb which is not in secular equilibrium.

NB = N0λA/(λB - λA)[e-λAt - e-λBt]

where NB = NPb

λB = λPb - 0(stable)

λA = λu

N0 = Nv

Then after substituting

NPb = Nv[1 - e-λut]

Thus, number of uranium atoms originally present = present number of lead atoms + present number of uranium i.e.

Nv = NPb + Nu

Solving equations simultaneously

NPb = Nv - Nve-λut

NPb = Nv - Nu

Then we have

0 = -Nve-λut + Nu

e-λut = Nu/Nv

ut = logNu/Nv

t = -1/λu[Nu/(Nv + NPb)]

t = -1/λu[logNu - log(Nu + NPb)]

= 1/λu[log(Nu + Npb) - logNu]

t = (1/λu)log[Nu + NPb/Nu]

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