Download ATOMIC STRUCTURE

NUCLEAR PHYSICS (5 lectures)
Nuclear model of the atom

Geiger-Marsden experiment.

Nuclear structure.

Atomic mass unit

Binding energy and mass defect.

Nuclear stability and natural radioactivity.

Fission and fusion.
Radioactivity
Radioactive decay and its equation. Activity.
Radioactive dating.
Medical and other applications of radioactivity.
Xray production and spectrum.
Simple radioactive detectors.
1
ATOMIC STRUCTURE

Rutherford, having studied the phenomenon of
radioactivity in depth, realized that alpha ()
particles were highly energetic and penetrated
matter easily, knocking electrons from atoms in
their path. Armed with such good projectiles,
Rutherford had two of his students, Geiger and
Marsden, fire away at bits of materials. They fired
alpha particles at thin metal foils with the following
results

Most of the alpha particles passed through without
any change of direction. Clearly. These particles
were passing through empty or charge-free space.
A few particles were deflected through small
angles.
Still fewer particles (about 1 in 10 000) were
deflected through a right angle or more. These
particles behaved as if they had approached an
area of dense positive charge and been repelled.


Based on these observations, Rutherford proposed a
nuclear model of the atom in which:

The bulk of the atom is empty or charge-free
space;
 99.9% of the mass is concentrated in a small
region of positive charge at the centre of the atom.
This region is the nucleus of the atom:
 Electrons are found in the space outside the
nucleus.
Rutherford pictured the atom as a tiny solar system in
which the nucleus plays the role of the Sun, and the
electrons the role of orbiting planets

2
Particle
Symbol
Charge
mass / u,
Relative mass
Location in the atom
electron
e
-e
-
proton
p
+e
1836me
in orbitals or shells
around the nucleus
nucleons, i.e.
neutron
n
neutral
0.00055
me
1.0073
mp
1.0087
mn
1839me
Particles found in the
nucleus
Properties of Particles in the Atom
Particle
Electric Charge (C)
Mass (kg)
Mass (amu)
Electron
-1.60 x 10-19
9.109390 x 10-31
5.485799 x 10-4
Proton
+1.60 x10-19
1.672623 x 10-27
1.007276
Neutron
0
1.674929 x 10-27
1.008665
The unified atomic mass unit u = 1.661 x 10 –27 kg.
The charge of one electron =  1.602 x 10 –19 Coulombs.
The protons and neutrons are called nucleons
The Proton has a positive is about 1836 times the mass of the electron.
The neutron is electrically neutral its mass is about 1889 times the mass of all electron.
The atomic number (Z) of an element is number of protons in the nucleus of an atom.
The mass number (A = Z + N) (nucleon number) is sum of the number of neutrons and
protons in its nucleus of an atom.
Isotopes these are atoms having the same atomic number but different number of neutrons.
These atoms are also chemical identical.
A
ZX
4
3Li
2
1H
3
1H
Nuclide is the species of atoms of which every atom has the same number of proton and the
same nucleon number.
3
Atomic mass = mass of atom………….
1/12 mass of 12 C atom
The atomic mass unit (u) is 1/12 of mass of a 12C
The atomic mass of 1 mole of 12C = 12g = 0.012kg
mass of atom of 12C
= 0.012 ………
6.022  1023
= 1.99  10-26 kg
1 a. m. u. = 1.99  10-26 kg / 12 = 1.66  10
Atomic mass of carbon = 12 u
= 12  1.66  1027
= 1.99  10 26 kg
4
27
kg
THE MASS SPECTROMETER
Positive rays are produced in a gas discharge tube and directed through slits S1 and S2 and
energy as a narrow beam with a range of speeds and specific charge (q/m) . Between S2 and
S3 crossed electric and magnetic fields are applied. Only charges with speed = E / B are
undeflected.
All ions leaving S3 have the same speed. Beyond S3 only a magnetic field acts and the ion
will describe circular path and strike the photographic plate
BQ v = EQ
v=E/B
BEYOND AND S3
B Q v= mv2
R
R=Mv
BQ
But v = E/B
R=ME
B2 Q
rM
Q
R = radius of curvature
M = mass of ion
5
WORKED EXAMPLE
Carbon of atomic mass 12u is found to be mixed with another unknown element. In a mass
spectrometer the carbon traverse a path of radius 22.4 cm and the unknown path has r
=26.2 cm . What is the unknown element? Assume they have the same charge.
rM
Q
for carbon 24.4 = 12u unknown element
Q
26.2 = M1
Q
22.4 = 12u
26.2 M
M = 12u  26.2
22.4
M= 14.03 u
Questions
1.
In a mass spectrometer, germanium atoms have radii of curvature equal to 21.0, 21.6,
21.9 and 22.8 cm, the largest radii correspond to an atom of 76u. What are the masses
of other isotopes?
2.
Suppose the electric field between the plates in the mass spectrometer is
2.18  10 4 V/m and the magnetic field is 0.68 T. The source contains boron isotopes
of mass number 10 and 11. How far apart are the lines formed by single charged ion
of each type on the photographic film?
6
Einstein’s Mass Energy Equivalence
The law of conservation of mass
The nucleon number, proton number, and the energy and mass are all conserved in nuclear
processes.
Einstein showed that mass and energy could be changed from one to another. The energy
E produced by a change of mass m is given by
E = mc2
c = 3  108 ms-1
Question
Find the energy produced when a mass of 1 g is converted
m = 0.001 kg
E = 0.001 (3  108)2
= 9  1013 J
1 kg of water absorbs 4.2  10³ J / kg K To raise it temperature by 1K
m=?
E = 4.2  10³ J
m = E = 4.2  103
= 4.67  10 -14 kg
c2
(3.0  10 8)2
1u= ? eV
1u = 1.66  10 -27 kg
E = mc²
= 1.66  10 -27  (299,792,458)2
= 1.5  10 -10 J or = 9.4  108 eV
or = 932 MeV
BINDING ENERGY
This is the energy required to just take all the nucleons (Protons and Neutrons) apart so that
they are completely separated. The total mass of the separated nucleons is greater than the
total mass of the nucleus. The difference in mass is called the mass defect of the nucleus, it
is a measure of the binding energy.
mass defect of the nucleus = mass of the separated nucleons  mass of the nucleus
7
Example
Consider Helium 42He Nucleus
Atomic mass
= 4.00260 u
Mass of proton = 1.00728 u
Mass of neutron = 1.00867 u
Total mass of proton = 2  1.00728 u = 2.01456 u
Total mass of neutron = 2  1.00867 u = 2.01734 u
Total mass of electrons = 2  0.00055 = 0.0011
Total mass of nucleons
= 4.033 u
Mass defect = 4.0330 u  4.00260 u
m
= 0.0304 u
0.0305 u represent binding energy
1u

932 MeV
or
Binding energy = 932  0.0304 MeV = 28.3 MeV
The higher the binding energy the more stable the nucleus.
Question 1
Calculate:
(a) the mass defect,
(b)
the binding energy per nucleon for
238
92
U
Given that
238
Atomic mass of 92 U = 238.050 76 u,
mass of neutron = 1.008 67 u,
mass of proton = 1.007 28 u,
mass of electron = 0.000 55 u,
1 u = 932 MeV.
Ans: m=1.93542u and B.E=1802.84373MeV
8
Question 2
c)
Carbon
14
6
C (atomic mass = 14.0033241 u) is converted into nitrogen
14
7
N (atomic
mass = 14.003074 u) via  decay.
-
i)
Write this process in symbolic form, giving Z and A for the parent and daughter
nuclei and the - particle.
ii)
Determine the energy released (in MeV). Ans.
0.27935685 MeV
Question 3
22
11
0 
Na22
10 Ne 1  m
m = 21.994435u  ( 21.991383u + 0.000549u ) = 0.002503 u
E = 932  0.002503 u = 2.332796 MeV
9
Binding Energy Per Nucleon
Binding Energy Per Nucleon =
Binding Energy
Mass Number
It is a measure of the stability of a nucleus
Plot a graph of binding energy per nucleon against mass number make comment on
the shape of the graph in term of the stability of the element.
The nuclides of intermediate mass number have the largest values for the binding energy per
nucleon and the most stable is 56
26 Fe has a value 8.8 MeV it’s the most stable since it need
the most energy to disintegrate.
The smaller values of binding energy per nucleon for higher and lower mass numbers imply
that potential source of nuclear energy.
The figures show that nuclides with low mass numbers can produce energy by fusion, when
two light nuclei have fused to produce a heavier nucleus.
In contrast heavier nuclides (i.e. higher mass numbers) can produce energy by fission
(disintegration) into their nuclei into two lighter nuclei.
In both cases, nuclei are produced having great binding energy per nucleon and therefore
1.
more stable nuclei
2
there is consequently a mass transfer during their formation
10
The potential energy (PE) of each nucleon in a nucleus is also negative and equal in
magnitude to the binding energy per nucleon for that nucleus. If nucleon potential energies
are plotted against nucleon number we get a graph, which is simply the inverse of the one
above. Nucleons in iron-56 have the lowest potential energy of all and so are most difficult
to prise out of their nucleus.
This graph helps explain the energy released in two very important nuclear processes.
NUCLEAR FISSION. This is disintegration of heavy nuclei into two lighter
nuclei. This decreases nucleon potential energy and
releases it in other forms.
235
92
92
1
U  01n 141
56 Ba  36 Kr 30 n  energy
This type of reaction takes place in nuclear reactors.
Use the Data given and calculate the energy released in from this reaction.
235
92
U =235.0439299u
1
0
n = 1.0086649u
141
56
Ba = 140.88589u
11
92
36
Kr = 91.926156u
NUCLEAR FUSION: is combining of two light nuclei to form heavy nucleus.
This decreases the average nucleon potential energy
(and increases the binding energy per nucleon), and the
PE lost is emitted in other forms such as gamma-ray
photons or kinetic energy of particles.
2
1
H  21H23 He  01n  energy
This type of reaction takes place on the Sun. Very high temperatures are required to make
the reaction
Use the Data given and calculate the energy released in from this reaction.
2H
= 2.01412u,
3He
= 3.01605u
&
1n
12
= 1.00867u
The Figure shows A slowly moving neutron causes the uranium nucleus
141
barium 56 Ba , krypton
92
36
235
92
U
to fission into
Kr and three neutrons.
The Figure shows Deuterium and tritium are fused together to form a helium
4
nucleus ( 2 He). The result is the release of an enormous amount of energy,
1
mainly carried by a single high-energy neutron ( 0 n)
13
RADIOACTIVITY
This is the spontaneous and random decomposition of unstable atoms. Three types of
radiation are emitted.
 particle
-particles
- Radiation
Contain:
He²+ nuclei
Contain:
stream of high-energy
electrons similar to
cathode ray.
Range
Range
of several millimetres in They are able to
air. Most are stopped: penetrate several mm
by ordinary thickness of thickness of aluminium .
paper or by a very thin
sheet of aluminium.
Speed
Speed
vary up to 7% of light.
vary up to 99% that of
light.
Least penetrating.
Most ionising.
E.M radiation of very
short wavelength
Range:
Infinite. And are
absorbed by a few mm
of lead.
Speed of light
More penetrating than - Most penetrating
particles.
More ionising than  rays Least ionising
Deflected by electric and Deflected by electric and Not deflected by electric
magnetic field
magnetic field
or magnetic field.
14
RADIOACTIVE DECAY
 - Decay
The Parent nucleus has lost two (2) protons and two (2) neutrons or moves two (2) up the
periodic table.
A
Z
XAZ42Y 42 He
Parent
nucleus
226
88
Daughter
nucleus
 particle
nucleus
4
Ra 222
Rn

86
2 He
 - Decay
a neutron get converted into a proton and -particle is emitted
A
Z
XZA1Y01e
Parent
nucleus
24
11
Daughter
nucleus
 particle
nucleus
0
Na24
Mg

12
1
- Decay
The nucleus, like the orbital electrons, exists only in discrete energy states or levels. When a
nucleus changes from an excited energy state (denoted by an asterisk *) to a lower energy
state, a photon is emitted.
A
Z
X* AZ X  
60
27
60
Co* 27
Co  
15
RADIOACTIVE DECAY
The Activity (A) (or rate of disintegration) of a given unstable element is proportional to the
number of atoms N present at the time
N
N
t
N
A
 N
t
A
A  N
or
 is called the decay constant
The S.I. unit of Activity (A) is the Becquerel (Bq)
1 Bq = 1 disintegration per sec
Activity is also measured in Currie1 C = 3.7  1010 s-1
If the N0 is the number of radioactive atom present at time t=0 and N the number at the time
t, then it can be shown by using calculus that
N = No et
Since the activity A  N a similar expression can be written
A = A0 et
The activity is proportional to the count rate per second, which can be denoted as C or R
R = R0 e 
A general expression then would be
The decay
X = X0 e

t
t
State that a radioactive substance service decay exponentially
 - particle and -ray are emitted with a single energy or with a small number of discrete
values of energy.
16
Half Life (T1/2)
This is time taken for the Number atoms present decrease to half of it initial value.
This is time taken for the Activity to decrease to half of its initial value.
This is time taken for the Mass to decrease to half of its initial value.
This is time taken for the Count Rate to decrease to half of its initial value.
This is time taken for the Activity to decrease to half of its initial value.
N = N0 e -t
Starting with
At
time
t = 0,
N = N0
At
time
t = T½ (half-life),
N = ½N0
1
T
N 0  N 0e 1 2
2
1
 T
 e 12
2
 T
1
ln    ln e 1 2
2
 ln 2  T1 2

T1 2 

ln 2

17
Question1
A laboratory has 1.49 µg of pure N, which has a half-life of 10.0 min (600 s).
(a)
How many nuclei are present initially?
(b)
What is the activity initially`?
(c)
What is the activity after 1.0 h?
(d) After approximately how long will the activity drop to less than one per second?
18
19
If the stable end product of the complete uranium series is 206Po how many alpha particle are
emitted between 226Ra and the end of the series.
binding energy (B.E.) = Energy Released during nucleus formation
(Energy Released )= 931x (decrease in mass )
(
MeV
)
( mass defect
)
Energy released in (J) = mass defect x c2
210
209
Po
84

4
Pb +
82
He
2
P = 209.98 x u
Pb = 205.974u
He = 4.003 u
20
1.
238U
92

234Tn
90

234Pa

91
234U
92

230Th
90

226Ra
88
What particle is emitted at each decay?
1
How many are of isotopes are these?
ANS. Two pairs of isotopes
2
A radio - isotopes of silver has a half life of 20 mins.
(a)
How many half lives is has in 1 hr?
`ANS. # of half lives in 1 hour = 60/20 = 3 half lives
(b) What fractions of the original mass would remain after 1 hr
ANS.
1 / 23 x N = 1/8 of the mass.
(c)
What fraction would have delay after 2 hrs.
ANS.1 / 26 = 1/64 would have remained
1 - 1/64 = 63/64 disintegrated
3
Taking the half life of Radium - 226 to be 1600 years. What fraction of a given sample
remain after 4800 years
ANS. Number of half life = 4800 = 3
1600
1/23 = 1/8
i
r = m assuming that Q remains the same  r = m
and that 76u  22.8
x
21.9
x1 = 21.9 x 76 u =
22.8
ii
For a radius of 21.6
m= 21.6 x 76u =
22.8
iii
For a radius of 21.0
m = 21.0 x 76 =
22.8
2
Electric field intensity = 2.18  104 Vm-1
Magnetic field = 0.68 T
Isotopes of Boron mass N 10 & 11 charge +e.
photographic plate
r=m/ Q
21
Find how far apart they are the