Radioactivity and nuclear structure.

Radioactivity is the spontaneous disintegration of unstable atoms of an element

by giving new stable and fresh nuclides with emission of either alpha particles,
Beta or Gamma rays. The element which disintegrates is said to be a radioactive
element.
Properties of a radioactive elements
They are very unstable and disintegrate to give new nuclides.
They emit rays or particles on disintegration such particles are alpha and Beta
particles while the rays are Gamma.
Radioactive substances affect photographic materials
When they disintegrate they release a lot of heat energy.
Alpha particles
- These are Helium particles
- They are deflected by magnetic and electric field in direction showing that
they are positively charged

- They are identified as He particles ( ) having a mass number of 4 and

atomic no. 2
1
- They move with a very high velocity of about 10 of that of light.
Beta particles
- These are negatively charged particles
- They can also be deflected by magnetic and electric field
- They are identified as fast moving electrons similar to cathode rays
- They have a mass number of zero and a charge of -1
- They penetrate more than Alpha particles and have the high velocity.

- They are presented as

Gamma rays
- These are neutral particles and are never deflected by either magnetic or
electric field.
- They are electromagnetic radiation similar to x- rays
- They have no charge
- They do not have any mass
- They have a high penetrating power with greater velocity
Detection of radiation.
• Use of photographic film
The radiation destroy them
• They cause surfaces for some substances to glow e.g if Beta particles which
are fast moving particles strike surfaces they cause it to glow.
• They also cause ionization of air molecules e.g when the fast moving
electron of Beta particles collide with air molecule the knock off leaving
charged ions.

1
Radioactive decay equation/reactions
(a) Alpha decay.
• An alpha particles is a Helium nucleus with a mass of 4 and atomic
number 2.
• The loss of alpha particles results in reduction in mass number by 4 units
and atomic no by 2 units. When an element undergoes alpha decay it
produces a new element whose position is 2 places earlier.
• Most of the radioactive isotopes of element with atomic no. greater than
83 undergo alpha decay.
(b) Beta decay.
• Loss of beta particles results in increase of atomic number by one but the
mass number remains the same.
• A Beta particle is a fast moving electron with no mass but a charge of -1.
• Beta decays also occurs in isotopes having atomic number over 83.
• The overall effect of Beta decay is to produce an element one place later in
the periodic table.
(c) Emission of gamma rays.
• These are fast moving electromagnetic radiations.
• They have no mass when they are being emitted there is no loss in mass
or change in atomic number.
Difference radioactive reactions and ordinary
• Radioactivity reaction are affected by physical factors e.g pressure,
temperature etc while chemical reactions are not affected by them.
• Radioactive reactions involve the nucleus which splits to give fresh
nuclides while chemical reactions involve valence shell electrons.
• In nuclear reactions a lot of heat is given off while in chemical reactions
give less heat.

RADIOACTIVITY
Is the natural/spontaneous disintegration (decay) of unstable nuclei by
emission of alpha, beta and gamma rays to form stable nuclei.
Stable nuclei do not undergo radioactive decay while unstable nuclei
undergo radioactive decay.

EFFECTS OF ∝, 𝛃 AND 𝛄 EMISSIONS ON ATOMIC AND MASS NUMBERS

OF NUCLEI FORMED

2
 a)
• All disintegrations by emission/loss of an alpha particle
decrease the mass number of the new nucleus by 4 units and atomic
number by 2 units.
• This displaces the new nucleus formed by two places
backwards/ left in the periodic table e.g.

Be, Th

He + etc
b) Emission of Beta particles
Emission / loss of a beta particle increases the atomic number of the new
nuclei formed by 1 unit but has no change on the mass number.
The new nucleus formed is displaced 1 position ahead/ on the right in the
periodic table. e.g.

Fr +

c) Emission of a gamma ray

The loss of gamma rays is accompanied by release of energy but there is
no change in the identity of the new nucleus formed. i.e.

energy

NB: When the emitted particle is known, the new nucleus formed is
identified by its atomic number and NOT mass number.

BALANCING RADIOACTIVE EQUATIONS:

In balancing radioactive equations, the
• sum of the atomic numbers on the left should equal to
the sum of atomic numbers on the right hand side.
• sum of the mass numbers on the left hand side should be
equal to that on the right hand side. e.g.
1. Complete the following reactions

i
ii. Ag iii.
232 228 228 228
Th α X β Y β Z
90 88 89 90
Identify X, Y and Z
Solution:

X
X: Radium

3
 Y: Actinium

Z: Thorium
a) Name the three main types of radioactive emissions
b) State two properties of any 1 of the emissions named in (a) above.
c) ) Complete the equation of the decay of Bismuth

Po
a) Complete the following equations for the nuclear reactions

i.
ii.
iii. Rn

a) Th

b)

UNEB 2006
Complete the following equations for the nuclear reaction.

1. S

Pb

2. Ne

Cd

Stability of the nucleus

This is a measure or an extent by which it remains undissociated and depends on
neutron to proton ratio (n/p ratio) together with bonding energy. The nucleus
contains protons and neutrons forming a nucleon. The protons are positively
charged and continuously repelling each other with in the nucleus but they never
split the nucleus and the nuclide remain stable. In forming the nucleus there is
a loss in mass such that the sum of neutrons and protons mass is greater than
the mass of the nucleus. The loss in mass is called mass defect which is
converted into energy according to the following equation.
E = mc2 where M is mass, C is velocity of light and E is energy. This energy is
called binding energy and it keeps the protons and neutron together hence
preventing the nucleus from splitting.
The nucleus is said to be stable if,
• It does not undergo radioactivity and so does not emit radiation.

4
 • The nucleus has a very long half time if it is to decay stable nuclei have
low atomic numbers usually less than 20
𝑛
• It has equal number of neutrons and protons such that the ratio 𝑝 = 1.
The nucleus is said to be unstable if it is radioactive and emits radiation and
particles forming a stable nuclei.
• For unstable nucleus the half time is very short ranging from fractions,
seconds or minutes to a few hours.
• They have a very high atomic number and the number of neutrons is not
equal to the number of protons i.e
𝑛
≠1
𝑝
Heavy nuclei with atomic number between 21- 83 are unstable and undergo beta
particle emission to form stable nuclei. Heavy nuclei with atomic number greater
that 83 are equally unstable and they undergo alpha particle emission and to
give stable nuclei.
A plot of neutron-proton number for stable and unstable nuclides.

Stability belt

Y
Neutrons X
(n)

Protons (p)
From the graph above all stable nuclides lie in the region which is shaded
Isotopes X, Y and X are unstable and they can become stable by approaching
the stability bell. X has no excess neutrons with few protons. To approach the
stability belt it reduces on the number of neutrons and excess on the no of
proton.
The neutron splits into an electron and a proton

This reduces on the number of neutrons but increases the number of protons.
The electron produced as a beta particle in a process called beta particle
emission. Nuclide Y has an excess of protons and neutrons which combine to
form alpha particles

5
This reduces both protons and neutrons in alpha particles emitted. Z has an
excess of neutrons and protons. To attain stability it reduces on the number of
protons in its nucleus but increasing the number of neutrons. It does this
through capturing an electron and liberating a neutron.

Laws of radioactivity.
1. Radioactive nuclides disintegrate spontaneously giving new and fresh
element(s)
2. The disintegration of an element is as a result of either emission of an
alpha particle or Beta or Gamma rays.
3. The rate of disintegration of a nuclide is not affected by physical factors
e.g. concentration, temperature, pressure etc
4. The rate of disintegration of an element depend on the amount of it present
at time (t).
Decay law
It states that the rate of decay is directly proportional to the number of un
decayed atom or un decayed mass of a given sample of an element.
Ie Rate of decay ∝ un decayed atoms
−𝑑𝑁
∝ Nt -------------(1)
𝑑𝑡
−𝑑𝑁
∝ Nt -------------(2)
𝑑𝑡
Where x is decay constant
𝑑𝑁
-∫ 𝑁𝑡 =  dt
Integration
𝑑𝑁
-∫ 𝑁𝑡 =  ∫ 𝑑𝑡 Note InN can also be written as log 𝑒 𝑁
In Nt = λt + c--------------------(3)
If t = 0, Nt = No = C
Substituting in equation (3) we get
In Nt = λt – InNo
𝑁𝑜
Therefore In( 𝑁𝑡 ) = λt----------------(3)
In term of logarithms to base ten(i.e. change of base from e to 10)
𝑁𝑜
From In ( 𝑁𝑡 ) = λt
Taking log10 on either sides
𝑁𝑜
log10 ( ) 1
𝑁𝑡
= λt but log10 𝑒 = log
log10 𝑒 𝑒 10
𝑁𝑜
log10 ( ) 𝑁𝑜
∴ 𝑁𝑡
= λt becomes 2.303 𝑙𝑜𝑔10 ( 𝑁𝑡 ) = 𝜆𝑡.Dividing through by 2.303,we obtain;
log10 𝑒

6
 𝑁𝑜 𝜆
𝑙𝑜𝑔10 ( 𝑁𝑡 ) = 2.303 𝑡
𝑁𝑜
If a graph of log10 ( 𝑁𝑡 ) is plotted against time, a straight line through the origin is
λ
obtained with a positive gradient equal to as shown below.
2.303

𝑁𝑜
Log10( 𝑁𝑡 )
𝜆
Slope =2.303

Time (t)
𝑁𝑜
However when a graph of ( 𝑁𝑡 ) is plotted against time, a straight line graph from
the origin but with a negative gradient is obtained as shown below

Time(s)

𝑁𝑡
Log10 (𝑁0)
𝜆
Slope = - 2.303

Deriving an expression for half-life(t½).

Half-life is the time taken for half of the initial amount of a substance to
disintegrate and remains constant regardless of the initial amount.
𝑁𝑜
From In ( 𝑁𝑡 ) = λt
If t = t½, No = ½No
𝑁𝑜
In (1 ) = λt½
𝑁𝑜
2
2 𝑁𝑜
In ( 𝑁𝑜 ) = λt½
0.693
In 2 = λt½ , t½ = λ
If a graph of initial mass/ concentration is plotted against time, a curve which
does not pass through the origin is obtained and the half life of the given sample
can be determined.

7
 Initial
concentration
(mol/l)
𝑁𝑜
2

𝑵𝒐
𝟒

𝒕 𝒕
Time(s)
𝟐 𝟒

Examples.
1. A compound W was found to contain a radioactive X which emits alpha
particles and has half life of 5720 years. Calculate the % of X that would remain
w after 228800 years.
Solution.
t½ =5720 years.
t=22880 years.
No =100%
Nt= ?
𝑁𝑜 𝑡
= 2( 𝑡1 )
𝑁𝑡
2
100 = 16
𝑁𝑡 1
Nt = 6.25%

2. (a)Define the term radio activity.

• Radio activity refers to the spontaneous disintegration of an unstable
nucleus to form a stable nucleus with emission of radiations like alpha, beta
gamma rays.
(b)Name three types of radiation emitted during radioactivity. State how they
affected by nucleus of the radio isotope.
• Alpha particles
• Effects; when the nucleus emits an alpha particle, it loses 2 electrons
and 4 neutrons hence decreases the atomic number by 2 units and the
mass number by 4units and it is converted to another element of a lower
atomic number by 2 units.
• Beta particles.
• Effects; the atomic number increases by 1unit but the mass number
remains the same; the atom is converted to another with a higher atomic
number by 1 unit.
• Gamma rays.

8
 • Effects; the emission of gamma rays by the atom does not affect the
atomic number and mass number but only energy is given out.

(c)The table below shows the mass of protactinium varies with

time.
Mass of 60.0 38.5 26.0 17.2 11.1
Pa
Time in 0 40 80 120 160
Sec
(i) A graph of mass of Pa against time
(ii) Use the graph to determine half-life of Pa
(iii) Determine the time of 8g of Pa to decay up to 1 g
Solution.
𝑁𝑜
from in ( 𝑁𝑡 ) = λt
8
in(1) = 0.0108 t
(8) 0.0108𝑡
in =
0.0108 0.0108
2.0794
t = (0.0108)
t = 192.54s
(d) The data below shows the result of the disintegration of element y.
No Nt t(s)
0.01 0.01 0
0.00999 0.009 30
0.0985 0.008 60
0.0997 0.0073 90
0.01000 0.0066 120
0.00989 0.0053 180
0.01010 0.0044 240
0.0975 0.0028 360
0.01056 0.0020 480
0.0104 0.0013 600
(i) Plot a graph of Nt against time (t).
(ii)Use the graph to determine half life of the element.
(iii)Decay constant of radioactive element Y
𝑁𝑜
(iv) Plot a graph of log10 ( 𝑁𝑡 ) against t.
Use it to determine.
(v) Decay constant of Y
(vi) Half life of Y
Solution
No Nt 𝑁𝑜
( 𝑁𝑡 )
𝑁𝑜
Log10( 𝑁𝑡 )
𝑁𝑡
(𝑁0)
𝑁𝑡
Log10(𝑁0) t(sec)
0.01 0.01 1.000 0.000 1 0 0
0.00999 0.009 1.110 0.045 0.900 -0.046 30
0.0985 0.008 12.31 0.090 0.081 -1.092 60

9
 0.0997 0.0073 1.377 0.137 0.073 -1.137 90
0.01000 0.0066 1.515 0.180 0.660 -0.180 120
0.00989 0.0053 1.867 0.271 0.536 -0.271 180
0.01010 0.0044 2.295 0.360 0.436 -0.361 240
0.0975 0.0028 34.821 1.542 0.029 -1.538 360
0.01056 0.0020 5.280 0.723 0.189 -0.724 480
0.0104 0.0013 8.000 0.903 0.125 -0.903 600
(i) See on the graph.
(ii) Half life = 64 years
0.613
(iii) t½ = 𝜆
0.693
t½ = 𝜆
64 0.693
=
1 𝜆
64𝜆 0.693
64
= 64
λ = 0.0108
(iv) See on the graph.
(v) t½ = 168
1
but t½ = 𝜆
0.693
168 = 𝜆
168𝜆 0.693
= 168
168
λ = 0.00413
𝑁𝑜
(vi) from in ( ) = λt
𝑁𝑡
8
in(1) = 0.0413 t
(8) 0.0108𝑡
in =
0.0108 0.0108
2.0794
t = (0.0108)
t = 192.54s.

Exercise.
1. 1000g of a related was allowed to decay to 125g after 6 min. calculate the
decay constant and hence the half life of radioactive element.
Solution.
• No= 1000g, Nt= 125g, t= 6minutes.
𝑁𝑜
In ( 𝑁𝑡 ) = λt.
1000
In ( 125 ) =6λ.
1000
λ =in (125𝑥 6) =0.3466min-1
𝑖𝑛 2 𝑖𝑛 2
t½ = 𝜆 , t½ =𝑜.3466, t½ =2minutes.
2. If the decay constant of a radium is 1.356 x 10-11/sec, calculate the time
taken for 10% of the sample decay.
Solution.

10
 • λ = 1.356x10-11 s-1, 10% decayed and 90% remained.
𝑁0 100
In( 𝑁𝑡 ) = λt, in( 90 ) =1.356x10-11
100 1
t= in( 90 ) x 1.356𝑥1011 , t=7.7699x109s.
3. The half life of a radioactive element Y is 8days. Determine the time taken
for 32g of Y to decay to 2g.
Solution
• No= 32g , Nt = 2g, t½= 8days
𝑁𝑜 𝑡 32 𝑡
( 𝑁𝑡 ) =2(𝑡½) , 2 =2(8)
𝑡 8𝑙𝑜𝑔16
Log 16 = log2, t= , t=32days.
8 𝑙𝑜𝑔2
Uses of radioactivity;
• Carbon dating
The radioactivity of carbon 14 in archeological remains can be measured. The
radio activity of a sample of a similar object can that is living can be determined.
If the alf life of carbon 14 is known, then the values are used to calculate the age
of the archeological.
𝑁𝑜
3.303 log10 ( 𝑁𝑡 ) = λt
where N – is the number of radiation emitted per min.
λ = is the decay constant
t = is the time in years since the death of the object
N.B: the radioactivity of the 14C is measured by Geiger Muller which records the
radiation of the sample in 1 min by making pulses or clicking on an electronic
counter.
Examples.
1. Freshly killed piece of wood give 15 counts / min/ gram of carbon 14. An
Egyptian mummy gives 9.5 counts/g/min of 14C .How old is the mummy case
if the half of 14 C if the half life of carbon -14 is 5600years.
1𝑛2
λ=t½= 𝜆
1𝑛2
λ= t½
0.69
λ = 5600
λ = 0.00012378yr-1
𝑁𝑜
2.303 log10( 𝑁𝑡 ) =λt
𝑁𝑜
( )
2.303𝑙𝑜𝑔10 𝑁𝑡
t= 𝜆
15
( )
2.303𝑙𝑜𝑔10 9.5
t = 1.2378𝑥10−4
t = 3690.75years.
2. A sample of wood from an Egyptian tomb give disintegration of 8.25min-1g-1 if
the sample of the living gives 15.3 g/min/g of 14C. Determine the age of Egyptian
tomb.
Solution.
No= 8.25min-1g-1 , No = 15.3min-1g

11
 𝑖𝑛2
From λ = 𝑡½
0.69
λ= = 1.2378x10-4.
5600
𝑁𝑜
2.303log10 ( 𝑁𝑡 ) = 𝜆t.
15.3
( )
2.303𝑙𝑜𝑔10 8.25
t= 1.2378𝑥10−4
t= 2672years.
• It is used to treat cancer. Cancerous tissues are destroyed by
radioactivity in preference for healthy carbon 60 which emits gamma rays
of 5 yrs is used.
• Sterilization of surgical instruments. Surgical instruments are
effectively sterilized by use of radioactivity by boiling.
• Detecting faults in metal sheet. The thickness of metal sheet is
continuously checked by the radio activity which is detected through the
meal sheets.
• Detecting underground leakages in water and oil pipes. The level of the
radioactivity on the surface can be monitored by the radioactive substance
with in water being moved within the pipe.
• Detect engine wear; they can also be detected by measuring the rate of
radioactivity I the engine oil. Normally piston rings that are made up of
radioactive are used.
• Investigating the mechanism of the reaction e.g in esterification i.e
RCOOH + ROH RCOOR + H2O
Acid alcohol Ester
In this reaction, the O2 in H2O comes from either an acid or an alcohol and to
prove where O2 come is labeled H2O and the reaction is given time to produce.
On test of radioactivity of H2O molecule given out it’s found to be not radioactive.
This can now lead to a conclusion that O2 comes from acid.
Investigation of Bio chemical reaction e.g. photosynthesis
CO2 + H2O CH2O + O2
The O2 of CO2 is labeled radioactive O2-18 and supplied to plants to carry out
photosynthesis. The O2 then given is then tested by radioactivity and its to be
non radioactive. This leads to a conclusion that the O2 evolved during
photosynthesis is comes from water.
NUCLEAR FUSION AND NUCLEAR FUSION.
Nuclear fusion.
This is a process by which light nuclides combine at a very high temperature to
form a heavy nuclide. This process is a companied by high amount of energy and
some radiations are emitted e.g protons and neutrons.
The hydrogen bond is obtained by fusing 2 hydrogen nuclei at very high
temperature of at about 107k and high amount of energy are release during
fusion and this can cause explosion. The sun obtains its energy from hydrogen
atoms and the sun’s temperature is about 10,000,0000C which is enough H2
atoms to fuse.
Nuclear fission.

12
This is a process by which unstable nuclide is bombarded with a nuclide to
disintegrate and form stable nuclides and other particles being emitted. This
process is accompanied by loss of large of amount of energy. For example
Uranium – 238 undergoes nuclear fission when combined with a neutron. This
process produces 3 more neutrons which can accelerate the reaction. The
nuclear fission of Uranium -238 is shown by the following chain reaction.
238 U + 1𝑛 98𝑆𝑟 + 138𝑋𝑒 + 3 1𝑛
92 0 38 54 0

Uses of uranium.
• Uranium – 238 is used in atomic bombs.
• It is also used in nuclear reaction to produce thermal electricity.

13