Download UNIT 15: NUCLEUS

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is defined as the central
core of an atom that is
positively charged and
contains protons and
neutrons.
UNIT 15: NUCLEUS
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15.1 Nuclear Structure
{
A nucleus of an atom is made up of protons and neutrons that known
as nucleons (is defined as the particles found inside the nucleus)
nucleus
as shown in figure 15.1a.
Proton
{
Proton and neutron are characterised by
the following properties in table 15.1a.
Proton (p)
Neutron
Charge (C)
Mass (kg)
Electron
+e
(1.60 × 10 −19 )
Neutron (n)
0
(uncharged)
1.672 × 10 −27 1.675 × 10 −27
Table 15.1a
{
For a neutral atom :
The number of protons = the number of electrons orbiting the
inside the nucleus
nucleus
z
This is because the magnitude of an electron charge equals to the
magnitude of a proton charge but opposite sign.
z
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{
Nuclei are characterised by the number and type of nucleons they
contain as shown in table 15.1b.
Number
Symbol
Definition
Atomic number
Z
The number of protons in a nucleus
Neutron number
N
The number of neutrons in a nucleus
Mass (nucleon)
number
A
The number of nucleons in a nucleus
Table 15.1b
Relationship :
{
{
Any nucleus of elements in the periodic table called a nuclide is
characterised by its atomic number Z and its mass number A.
The nuclide of an element is represented as
Mass number
Element X
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Atomic number
{
The number of protons Z is not necessary equal to the number of
neutrons N.
23
e.g. : 11 Na
;
32
16
S ;
195
78
Pt
Z = 11
N = A − Z = 12
{
Since a nucleus can be modeled as tightly packed sphere where each
sphere is a nucleon, thus the average radius of the nucleus is given by
femtometre (fermi)
fermi)
1
(15.1a) 1 fm = 1 × 10 −15 m
3
R = R0 A
where
{
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R0 : constant = 1.2 × 10 -15 m ( 1.2 fm)
A : mass number
Example 1 :
Based on the periodic table of element, Write down the symbol of
nuclide for following cases:
a. Z=20 ; A=40
b. Z=17 ; A=35 (exercise)
c. 50 nucleons ; 24 electrons
d. 106 nucleons ; 48 protons (exercise)
e. 214 nucleons ; 131 protons (exercise)
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Solution:
a. Given Z=20
A
Z
; A=40
40
20
X
Ca
c. Given A=50 and Z=number of protons = number of electrons =24
A
Z
{
X
50
24
Cr
Example 2 :
What is meant by the symbols below :
1
0
n ; 11 p ; −10 e
State the mass number and sign of the charge for each entity above.
Solution:
1
Neutron ; A=1
0
Charge : neutral (uncharged)
n
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{
1
1
p
0
−1
e
Proton ; A=1
Charge : positively charged
Electron ; A=0
Charge : negatively charged
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Example 3 : (exercise)
Complete the table below :
Element
nuclide
1
1
9
4
14
7
16
8
23
11
59
27
31
16
133
55
238
92
Number of
protons
Number of
neutrons
Total charge
in nucleus
Number of
electrons
11
12
11e
11
H
Be
N
O
Na
Co
S
Cs
U
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15.2 Isotope
{
{
Definition – is defined as the nuclides/elements/atoms that have the
same atomic number Z but different in mass number A.
From the definition of isotope, thus the number of protons or electrons
are equal but different in the number of neutrons N for two isotopes
from the same element.
For example :
z
Hydrogen isotopes :
1
1
2
1
3
1
z
H : Z=1, A=2, N=1
H : Z=1, A=3, N=2
Oxygen isotopes :
16
8
17
8
18
8
proton (11p)
H : Z=1, A=1, N=0
equal
deuterium (12D)
tritium ( 31T )
not equal
O : Z=8, A=16, N=8
O : Z=8, A=17, N=9
O : Z=8, A=18, N=10
not equal
equal
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15.3 Mass-Energy equivalent
{
{
From the theory of relativity leads to the idea that mass is a form of
energy.
energy
Mass and energy can be related by following relation :
where
(15.3a)
E = mc 2
E : amount of energy
m : rest mass
c : speed of light in vacuum (3.00 × 10 8 m s -1 )
e.g. : The energy for 1 kg of substance is
E = mc 2
E = (1)(3.00 × 10 8 ) 2
E = 9.00 × 10 16 J
{
Unit conversion of mass and energy :
z
The electron-volt (eV)
{
is a unit of energy.
{
is defined as the kinetic energy gained by an electron in
being accelerated by a potential difference (voltage) of 1
volt.
volt
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1 eV = 1.60 × 10 −19 J
1 MeV = 106 eV = 1.60 × 10 -13 J
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z
The atomic mass unit (u)
{
is a unit of mass.
{
is defined as exactly
atom.
{
1
the mass of a neutral carboncarbon-12
12
mass of 126C
1u =
12
1 u = 1.66 × 10 - 27 kg
1 atomic mass unit (u) can be converted into the unit of
energy by using mass-energy equivalent equation (eq. 15.3a) :
(j):
E = mc 2
E = (1.66 × 10 −27 )(3.00 × 10 8 ) 2
E = 1.49 × 10 −10 J
z
in joule
z
in eV or MeV:
1 u = 1.49 × 10 −10 J
1.49 × 10 −10
= 931.5 × 10 6 eV
−19
1.60 × 10
1 u = 931.5 × 106 eV
E=
or
1 u = 931.5 MeV
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15.4 Mass Defect and Binding Energy
15.4.1 Mass Defect
{
The mass of a nucleus (MA) is always less than the total mass of
its constituent nucleons (Zmp+Nmn) where
M A < (Zm p + Nmn ) with m p : mass of a proton
mn : mass of a neutron
{
{
{
{
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Hence the difference in this mass is given by
∆m = (Zm p + Nmn ) − M A
(15.4a)
where ∆m is called mass defect.
defect
The mass defect is defined as the mass difference between the
total mass of the constituent nucleons and the mass of a
nucleus.
nucleus
The reduction in mass arises because the act of combining the
nucleons to form the nucleus causes some of their mass to be
released as energy.
Any attempt to separate the nucleons would involve them being
given this same amount of energy. This energy is called the
binding energy of the nucleus.
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15.4.2 Binding Energy
{
Definition - The binding energy of a nucleus is defined as the
energy required to separate completely all the
nucleons in the nucleus.
nucleus
{
The binding energy of the nucleus is equal to the energy
equivalent of the mass defect. Hence
Binding energy
in joule
EB = (∆m )c 2
(15.4b)
Speed of light in vacuum
Mass defect in kg
{
Example 4 :
Calculate the binding energy of a lithium nucleus 37 Li in MeV.
(Given mass of neutron, mn=1.00867 u ; mass of proton,
mp=1.00782 u ; speed of light in vacuum, c=3.00 x 108 m s-1and
mass of lithium nucleus, MLi=7.01600 u)
( )
Solution:
7
3
Li
Z = 3 and N = A − Z
N =4
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The mass defect of lithium nucleus is
∆m = (Zm p + Nmn ) − M Li
∆m = [(3 × 1.00782) + (4 × 1.00867 )] − 7.01600
∆m = 0.04214 u
The binding energy of lithium nucleus can be calculated by using
two method :
2
1 u = 1.66 × 10 - 27 kg
z
Use equation 15.4b : E B = ∆m c
(
(
)
in kg
∆m = (0.04214 ) 1.66 × 10 −27
∆m = 6.995 × 10 −29 kg
(
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)(
)
)
2
EB = 6.995 × 10 −29 3.00 × 10 8
EB = 6.296 × 10 −12 J
1 MeV = 1.60 × 10 -13 J
in MeV :
6.296 × 10 −12
EB =
1.60 × 10 −13
EB = 39.4 MeV
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z
Using unit conversion ( u⇒ MeV )
1 u = 931.5 MeV
EB = ∆m × 931.5 MeV
in u
{
EB = (0.04214 )× 931.5
EB = 39.3 MeV
Example 5 :
35
If the mass of chlorine-35 nucleus 17 Cl is 34.97 u, calculate its
binding energy in joule.
(Given mass of neutron, mn=1.009 u ; mass of proton, mp=1.007 u
and speed of light , c=3.00 x 108 m s-1)
Solution:
(
35
17
Cl
)
Z = 17and N = A − Z
N = 18
The mass defect of chlorince-35 nucleus is
∆m = (Zm p + Nmn ) − M Cl
∆m = [(17 × 1.007 ) + (18 × 1.009 )] − 34.97
∆m = (0.311 u )× (1.66 × 10 −27 )kg
13
∆m = 5.16 × 10 −28 kg
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The binding energy of chlorince-35 nucleus is
EB = (∆m )c 2
(
)(
EB = 5.16 × 10 −28 3.00 × 10 8
EB = 4.65 × 10 −11 J
or
(
)
)
2
1 u = 1.49 × 10 −10 J
EB = ∆m × 1.49 × 10 −10 J
EB = (0.311 u )× 1.49 × 10 −10 J
EB = 4.63 × 10 −11 J
{
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(
)
Example 6 : (exercise)
Calculate the binding energy in joule of a deuterium nucleus. The
mass of a deuterium nucleus is 3.344275 x 10-27 kg.
(Given mass of neutron, mn=1.674954 x 10-27 kg ; mass of proton,
mp=1.672648 x 10-27 kg and speed of light , c=3.00 x 108 m s-1)
Ans. : 2.99 x 10-13 J
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15.5 Nucleus Stability
{
{
{
{
Since the nucleus is viewed as a closed packed of nucleons, thus
its stability depends only on the forces exist inside it.
The forces involve inside the nucleus are
z
repulsive electrostatic (Coulomb) forces between protons and
z
attractive forces that bind all nucleons together in the nucleus.
These attractive force is called nuclear force and is responsible
for nucleus stability.
The general properties of the nuclear force are summarized as
follow :
z
The nuclear force is attractive and is the strongest force in
nature.
z
It is a short range force . It means that a nucleon is attracted
only to its nearest neighbours in the nucleus.
z
It does not depend on charge; neutrons as well as protons are
bound and the binding is same for both.
e.g. :
protonproton-proton (p
(p-p)
The magnitude of
neutronneutron-neutron (n
(n-n)
nuclear forces are
protonproton-neutron (p
(p-n)
same.
same.
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The nuclear force depends on the binding energy per nucleon.
Note that a nucleus is stable if the nuclear force greater than
the Coulomb force and vice versa.
The binding energy per nucleon of a nucleus is a measure of
the nucleus stability where
z
{
{
Binding energy per nucleon =
{
Example 7 :
Why is the uranium-238 nucleus 238
less stable than carbon92 U
12 nucleus 12
?
Give
an
explanation
by
referring to the
C
6
repulsive coulomb force and the binding energy per nucleon.
(Given mass of neutron, mn=1.009 u ; mass of proton, mp=1.008 u
; mass of carbon-12 nucleus, MC=12.000 u ; mass of uranium-238
nucleus, MU=238.051 u and 1 u =931.5 MeV)
( )
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Binding energy ( EB )
Nucleon number( A)
(
)
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Solution:
z
From the aspect of repulsive coulomb force :
{ Uranium-238 nucleus has 92 protons but the carbon-12
nucleus has only 6 protons.
{ Therefore the coulomb force inside uranium-238 nucleus is
 92 
  or 15.3 times the coulomb force inside carbon-12
6 
nucleus.
z
From the aspect of binding energy per nucleon:
12
Z = 6 and N = 6
{ Carbon-12 :
6C
The mass defect : ∆m = Zm p + Nmn − M C
(
∆m = 0.102 u
)
The binding energy per nucleon:
(∆m × 931.5 ) MeV
 EB 

 =
12
 A C
 EB 

 = 7.92 MeV / nucleon
 A C
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{
238
Uranium-238 : 92 U
The mass defect : ∆m
Z = 92 and N = 146
= (Zm p + Nmn ) − M U
∆m = 1.999 u
The binding energy per nucleon:
(∆m × 931.5 ) MeV
 EB 

 =
238
 A U
 EB 

 = 7.82 MeV / nucleon
 A U
{
z
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From the value of binding energy per nucleon for both
nuclei, we obtain that
 EB 
E 

 < B 
 A U  A C
Since the binding energy of uranium-238 nucleus less than
the binding energy of carbon-12 and the coulomb force
inside uranium-238 nucleus greater than the coulomb force
inside carbon-12 nucleus therefore uranium-238 nucleus
less stable than carbon-12 nucleus .
18
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{
Figure 15.5a shows a graph of the binding energy per nucleon as
a function of mass number A.
Binding energy per
nucleon ((MeV
MeV/nucleon)
/nucleon)
Greatest stability
Fig. 15.5a
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Mass number A
From the graph :
{ The value of EB/A rises rapidly from 1 MeV/nucleon to 8
MeV/nucleon with increasing mass number A for light
nuclei.
{ For the nuclei with A between 50 and 80, the value of EB/A
ranges between 8.0 and 8.9 Mev/nucleon. The nuclei in
these range are very stable. The maximum value of the
curve occurs in the vicinity of nickel, which has the most
stable nucleus.
{ For A > 62, the values of EB/A decreases slowly,
indicating that the nucleons are on average less tightly
bound.
{ For heavy nuclei with A between 200 to 240, the binding
energy is between 7.5 and 8.0 MeV/nucleon. These nuclei
are unstable and radioactive.
Example 8 : (exercise)
20
The mass of neon-20 nucleus 10
Ne is 19.9924 u. Calculate the
binding energy per nucleon of neon-20 nucleus in joule per
nucleon. (Given mass of neutron, mn=1.009 u ; mass of proton,
mp=1.008 u ; 1 u = 1.66 x 10-27 kg
20
Ans. : 1.33 x 10-12 J/nucleon
z
{
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(
)
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15.6 Liquid Drop Model
{
{
{
{
The liquid drop model is one of the successful models that permits
us to correlate many facts about nuclear masses and binding
energies.
The model is proposed by Niels Bohr and later expanded on by
C.F. Von Weiszacker in 1935.
It is used to estimate the total binding energy of a nucleus by
driving semisemi-empirical mass (binding energy) formula.
formula
The liquid drop model based on assumptions below :
z
A nucleus likes a drop of incompressible liquid with uniform
density and spherical shaped.
z
The force between the nucleons does not depend on the spin
and charge of the nucleon. Thus the force between two
nucleons either (n-n), (n-p) or (p-p) is the same.
z
The attractive nuclear force which binds the nucleons is a
short range force. It means a nucleon can attract only a few of
its nearest neighbours.
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{
Figure 15.6a shows the diagram of liquid drop model.
nucleus
proton
Fig. 15.6a
{
nuclear
force
There are five major effects (factors) influence the binding energy
of the nucleus in the liquid drop model :
z
The volume effect (factor), V.
{ The binding energy of a nucleus is proportional to mass
number A and therefore proportional to the nucleus
volume.
{ The contribution of this effect to the binding energy of the
entire nucleus is given by
where
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neutron
EB1 = C1 A
(15.6a)
C1 : constant of volume effect (factor)
A : mass number
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z
The surface effect (factor).
{ The nucleons on the surface of a nucleus have fewer near
neighbours than those in the interior of the nucleus (Figure
15.6a), surface nucleons reduce the binding energy by an
amount to their number.
{ Since the number of surface nucleons is proportional to the
surface area 4πR2 of the nucleus and R2 ∝ A2/3, the
surface term can be expressed as
Negative sign means the
decreasing in binding
energy of the nucleus
z
2
EB 2 = −C2 A 3
where C2 : constant of surface effect (factor)
The Coulomb repulsion effect (factor).
{
{
Every one of the Z protons repels every one of the (Z-1)
other protons in the nucleus.
Therefore
The total repulsive
Z(ZZ(Z-1)
electric potential
1
1
or
energy
1
∝
∝
Nucleus radius
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{
R
A3
23
This repulsive energy decreases the binding energy, so
this term is negative and given by
EB 3 = −
C3 Z (Z − 1)
A
z
(15.6b)
1
3
(15.6c)
where C3 : constant of Coulomb repulsion effect (factor)
The symmetry effect (factor).
{ To be in a stable, low energy state, the nucleus must have
a balance between the energies associated with the
neutrons and with the protons.
z
{
For stable light nuclei (small A ) ⇒ N ≈ Z
Any large asymmetry between N and Z for light nuclei
reduces the binding energy and makes the nucleus
less stable.
z For stable heavy nuclei (larger A) ⇒ N >Z
This effect can be described by a binding energy term
2
below :
C4 ( A − 2 Z )
(15.6d)
A
where C4 : constant of symmetry effect (factor)
EB 4 = −
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z
The pairing effect (factor).
{ The nuclear force favours pairing of protons and of
neutrons.
{
{
The binding energy terms positive when both Z and N are
even and negative when Z and N are odd.
The pairing effect can be written as
C 
(15.6e)
EB 5 = ± 54 
 3
A 
where C5 : constant of pairing effect (factor)
C 
Z & N are even
+  54 
 3
A 
EB 5 =
0
C
−  54
 3
A
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{
A is odd




Z & N are odd
25
As a result, the total estimated binding energy EB is the sum of
these five term :
2
C3 Z (Z − 1) C4 ( A − 2 Z )  C5
EB = C1 A − C2 A −
−
± 4
1
A
 3
A3
A
2
3




(15.6e)
where the values of the constants are
C1 = 15.75 MeV
C2 = 17 .80 MeV
C3 = 0.710 MeV
C4 = 23.69 MeV
C5 = 39.00 MeV
{
determined from the
experimental results
The mass of a nucleus for any neutral atom, MA is given by
M A = (Zm p + Nmn ) −
EB
c2
(15.6f)
Mass defect (∆m)
SemiSemi-empirical mass formula
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{
Example 9 :
Consider the nuclide 37 Li in example 4. Calculate
a. the five terms in the binding energy and the total estimated
binding energy.
b. its neutral atomic mass using the semi-empirical mass formula.
(Given mn=1.00867 u ; mp=1.00782 u ; c=3.00 x 108 m s-1;
C1=15.75 MeV ; C2=17.80 MeV; C3=0.710 MeV; C4=23.69 MeV;
C5=39.00 MeV)
Solution: From the nuclide notation : Z=3, A=7 and N=4
a. The five terms are
( )
EB1 = C1 A
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2
3
EB1 = 110.25 MeV
EB 2 = −C2 A
EB 2 = −65.14 MeV
C Z (Z − 1)
EB 3 = −2.23 MeV
EB 3 = − 3 1
A3
2
C4 ( A − 2 Z )
EB 4 = −3.38 MeV
EB 4 = −
A
27
A is odd
EB 5 = 0
The total estimated binding energy is
EB = (110.25 − 65.14 − 2.23 − 3.38 ) MeV
EB = 39.5 MeV
This value is 0.51% greater than the value in example 4 which
is 39.3 MeV.
b. By applying the semi-empirical mass formula, thus
M A = (Zm p + Nmn ) −
EB
c2
 39.5 MeV 
M A = [(3)(1.00782) + (4 )(1.00867 )] − 

 931.5 MeV / u 
M A = 7.01574 u
This value is 3.77 x 10-3 % less than the value in example 4
which is 7.01600 u.
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15.7 Bainbridge Mass Spectrometer
{
{
Mass spectrometer is a device that detect the presence of
isotopes and determines the mass of the isotope from known
mass of the common or stable isotope.
Figure 15.7a shows a schematic diagram of a Bainbridge mass
spectrometer.
Ion source
S1
Plate P1
r
E
-
Evacuated
chamber
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r
B
Photographic plate
× × × × × × × × × × × ×m×1 ×m
×2× × × ×
r1 × ×r × × × × × × × × ×r× ×
××××××
× × × × × × × ×2× × × × × × × × ×B× ×
× × × × × × × × × × × × × × × × × ×2 ×
× × × × × × × × × × × × × × × × × ××
Fig. 15.7a
{
29
Working principle:
z
Ions from an ion source such as a discharge tube are
narrowed to a fine beam by the slits S1 and S2.
z
The ions beam then passes through a velocity selector (plates
P1 and P2) which uses a uniform magnetic field B1 and a
uniform electric field E that are perpendicular to each other.
z
The beam with selected velocity v passes through the velocity
selector without deflection and emerge from the slit S3. Hence,
the force on an ion due to the magnetic field B1 and the
electric field E are equal in magnitude but opposite in direction
(Figure 15.7b). The selected velocity v is given by
Plate P1
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Ions beam
- ××+ S2
+
Plate P2
- ××+
××
+
+
- ××+
1
- ××+
S3
−× ××
× ××
−r +
− F×E × ×
× ××
−× ××
r
v
Fig. 15.7b
Plate P2
× +
×
r+
×FB+
×
× +
FB = FE
B1qv sin 90 o = qE
v=
Using Fleming’
Fleming’s
left hand rule.
E
B1
(15.7a)
30
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z
The ions beam emerging from the slit S3 enter an evacuated
chamber of uniform magnetic field B2 which is perpendicular to
the selected velocity v. The force due to the magnetic field B2
causes an ion to move in a semicircle path of radius r given by
FB = Fc 2
mv
B2 qv sin 90 o =
r
E
mv
and v =
r=
B1
B2 q
mE
r=
(15.7b)
B1 B2 q
z
Since the magnetic fields B1 and B2 and the electric field E are
constants and every ion entering the spectrometer contains
the same amount of charge q, therefore
r = km
r∝m
with
k=
E
: constant
B1 B2 q
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31
z
If ions of masses m1 and m2 strike the photographic plate with
radii r1 and r2 respectively as shown in figure 15.7a then
m1 r1
=
m2 r2
{
(15.7c)
Example 10:
A beam of singly charged ions of isotopes Ne-20 and Ne-22
travels straight through the velocity selector of a Bainbridge mass
spectrometer. The mutually perpendicular electric and magnetic
fields in the velocity selector are 0.4 MV m-1 and 0.6 T
respectively. These ions then enter a chamber of uniform
magnetic flux density 0.8 T. Calculate
a. the selected velocity of the ions.
b. the separation between two isotopes on the photographic plate.
(Given the mass of Ne-20 = 3.32 x 10-26 kg; mass of Ne-22 =
3.65 x 10-26 kg and charge of the beam is 1.60 x 10-19 C)
Solution: m1=3.32x10-26 kg, m2=3.65x10-26 kg, B1=0.6 T
E=0.4x10-6 V m-1, q=1.60x10-19 C, B2=0.8 T
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32
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a. The selected velocity of the ions is
E
B1
v = 6.67 × 10 5 m s −1
v=
b. The radius of the circular path made by isotope Ne-20 is
Em1
B1 B2 q
r1 = 0.173 m
The radius of the circular path made by isotope Ne-22 is
Em2
r2 =
B1 B2 q
r2 = 0.190 m
r1 =
Therefore the separation between them is
∆r = r2 − r1
∆r = 0.190 − 0.173
∆r = 0.017 m
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33
THE END…
Next Unit…
UNIT 16 :
Nuclear Reaction
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34