TOPIC 7
ATOMIC and NUCLEAR PHYSICS
NEW PARTCILE PHYSICS
Solutions
1. The Electromagnetic Spectrum and the Nature of Light
1
Zumdahl Text p. 293 ex. 7.2
Questions
1. What is electromagnetic radiation? At what speed does it travel?
Energy produced by electromagnetism
speed of light = 3 x 10 8 ms-1
2. How are the different types of electromagnetic radiation similar? How are
they different?
Speed,
wavelength and f
3. What is the relationship between wavelength and frequency?
Inversely proportional
2
2. H , the Bohr Model, Quantum Numbers and Orbital
Shapes
Questions
1. How does the energy of a photon of visible light emitted by an atom
compare to the energy change within the atom itself?
The energy of a photon is given by E = hv where h is Planck’s constant =
6.626x10-34 J
and v is the frequency
The energy contained in the photon is equal to the change in energy that the
atom experiences and is related to a specific color, frequency and wavelength
as we learned about in the electromagnetic spectrum.
2. An atom in the lowest possible energy state is said to be in the _______
state. ground
.
3
3. When an atom in an excited state returns to its ground state, what
happens to the excess energy of the atom?
This is where the photon comes in. When going from the excited state to the
ground state the energy absorbed is being released in the form of photons ( tiny
energy packets) and as these photons are released a certain color of light is
given off
4. How is the energy carried per photon of light related to the wavelength of
the light ?
The equation E = hc relates photons to
λ
energy changes , wavelength , the speed of light (c) and Plank’s constant (
h) . Different wavelengths ( and therefore frequencies) of light correspond to
different amounts of energy per photon and different colors of light.
5. Compare short wavelength light to high wavelength : which one carries
more energy? short
6. When an atom absorbs energy it goes into a (n)________________state.
excited
4
5. Isotopes , Atomic Mass and Atomic Number
Examples
1.Write the atomic symbol
A
X
Z
for Z = 8 and neutrons = 9.
Z is the atomic number , in this case 8 , and this can be used to identify the
element as oxygen:
Since A = Z + neutrons = 8+9 = 17 :
17
O
8
final answer
2 . Write the isotope of chlorine in which A = 37.
37
Cl
Z
Z= atomic number for Cl , doesn’t change and =17
3. Write symbol for number of protons = 26 and neutrons = 31
Z = number of protons = 26 and used to identify element immediately as Fe:
A
Fe A = neutrons + protons = 31 + 26 = 57
26
5
4. Which of the following gives the correct number of protons and neutrons in a
nucleus of carbon-14 ( 146 C )? B
Protons
Neutrons
A.
8
6
B.
6
8
C.
14
6
D.
6
14
AP 2008 # 24
Which of the following shows the correct number of protons, neutrons and
electrons in a neutral cesium – 134 atom?
a)
Protons
55
Neutrons
55
Electrons
55
b)
55
79
55
c)
55
79
79
d)
79
55
79
134
55
134
e)
134
Cs
55
answer B
6
43. Give the number of protons and neutrons in the nucleus of each of the
following atoms:
a)
238
Pu
94
e = p = 94
b)
n = 144
65
Cu
29
e = p = 29
n = 36
45. Write the atomic symbol for each of the following atom described below in
the form of
A
X
Z
a) Z = 5 , A = 12
12
B
5
b) The isotope with 7 protons and 8 neutrons in the nucleus
15
N
7
7
c) Atomic number = 17, number of neutrons = 18
35
Cl
17
6. Nuclear Forces : Strong Nuclear Force vs. Coulomb Force
IB Which nucleons in a nucleus are involved in the Coulomb interaction and the strong
short-range nuclear interaction? A
Coulomb interaction
Strong short-range interaction
A.
protons
protons, neutrons
B.
protons
neutrons
C.
protons
protons
D.
protons, neutrons
neutrons
IB Which of the following is true in respect of both the Coulomb interaction and the
strong interaction between nucleons in an atom? C
Coulomb interaction exists
between
Strong interaction exists
between
A.
protons only
neutrons only
B.
both protons and neutrons
neutrons only
C.
protons only
both protons and neutrons
D.
both protons and neutrons
both protons and neutrons
8
RADIOACTIVE DECAY Tsokos pp. 373-378
1.
Nuclear Stability and Radioactive Decay
2. Types of Radioactive Decay – Note A and Z are balanced
A) Beta particle (βparticle) decay - losing an electron
0
e
-1
Example:
234
Th
90

234
Pa
91
+
131
I
53

131
Xe
54
+
0
e
-1
0
e
-1
9
B) Positron production - a positron is like an electron but with +1
charge
0
e
mass
1
don’t confuse with proton because protons have much more
IB example This question is about radioactive decay.
Iodine-124 (I-124) is an unstable radioisotope with proton number 53. It undergoes
beta plus decay to form an isotope of tellurium (Te).
(a)
answer :
State the reaction for the decay of the I-124 nuclide.
124
124
0 
53 I  52 Te  1 β
NOTE + sign
Example:
22
Na
11

22
Ne
10
+
0
e
+1
C) Alpha particle (α particle ) decay – losing a helium nucleus
4
He
2
Examples:
230
Th
90

226
Ra
88
+
4
He
2
10
D) Gamma ray decay ( γ ray) - nucleus is excited to a higher energy
level, loses photon and release light.
0
γ
0
Example :
238
U

92
238
U
92
+
0
γ
0
NOTE : no change in A or Z
Matter and Antimatter : a positron collides with an electron and mass is
changed into energy in the form of high energy photons or gamma rays :
0
e
1
+
0
e 
-1
0
γ
0
E) Electron capture - an electron is captured by the nucleus
0
e
-1
Example :
Note the electron is on the reactant side of the equation
201
Hg +
80
0
e 
-1
201
Au
79
+
0
γ
0
11
F) Neutron capture - a neutron is captured by the nucleus
1
n
0
Example :
Note the neutron is on the reactant side of the equation
238
U +
1
n 
0
239
Pu
Review table 18.2
12
sample exercise 18. 1 18.2
13
Zumdahl Text p. 906 : 9 – 12
a) electron capture = gaining electron
0
e
-1
b) Beta particle (βparticle) decay - losing an electron
0
e
-1
a) Beta particle (βparticle) decay - losing an electron
0
e
-1
b) Alpha particle (α particle ) decay – losing a helium nucleus
4
He
2
14
a) electron capture = gains electron
decay
b) positron
c) (α particle )
d) (βparticle) decay - losing an electron
a) d) (βparticle) decay - losing an electron
b) (α particle ) decay
c) positron
d) electron capture = gains electron
IB The nuclear equation below is an example of the transmutation of mercury into gold.
2
1H

199
80 Hg

197
79 Au
The particle X is a
+X
B
A.
gamma-ray photon.
B.
helium nucleus.
C.
proton.
D.
neutron.
(Total 1 mark)
15
IB A nucleus of the isotope potassium-40 decays to a nucleus of the isotope argon-40.
The reaction equation for this decay may be written as
40
19 K

40
Z Ar
Xν
Which of the following correctly identifies the proton number of argon-40 and the
particle X? B
Z
X
A.
18
β–
B.
18
β+
C.
19
β+
D.
19
β–
IB
In 1919, Rutherford produced the first artificial nuclear transmutation by
bombarding nitrogen with α-particles. The reaction is represented by the
following equation.
α+
14
7
N 178 O + X
Identify X.
answer : 1
p
1
solution:
4
He +
2
14
17
N  O
7
8
1
+ p
1
16
Nuclear Fusion- when two atoms combine and fuse to form 1 atom.
IB The nuclear reaction
2
1H
 31 H  42 He  01 n
is an example of C
A.
nuclear fission.
B.
radioactive decay.
C.
nuclear fusion.
D.
artificial transmutation.
17
3. Nuclear Fission – splitting the atom
fig. 18. 11 p. 898
Nuclear reactors work on the principle of nuclear fission and produce 26 million
times more energy than an equivalent mass of gasoline. The nuclear equation of
a fission reaction is where an atom , Uranium for example, is bombarded with
neutrons and the Uranium atom splits:
235
U +
92
1
n
0
Another example:
235
U +
92
1
n
0

141
Ba
56
+
92
Kr +
36

137
Te +
52
97
Kr +
40
1
3 n
0
1
2 n
0
Note that A and Z are balanced as in all other nuclear reactions.
18
AP 2002 # 23
235
1
141
1
92 U + 0 n  55 Cs + 3 0 n + X
23. Neutron bombardment of uranium can induce the reaction represented above. Nuclide X is
which of the following?
92
(A) Br
35
(B)
94
Br
35
(C)
91
Rb
37
92
(D) Rb
37
Ap 2008 # 27
answer : D
19
Zumdahl text p. 907 : 18
4. Half- Life
The half-life ( t1/2) of a radioactive sample is defined as the time it takes for the
amount of the substance to decay to ½ its original value . Original value is
sometimes called N0 and half of this value is called N0 / 2.
fig. 18.3 p. 885
20
Many times you are asked questions requiring multiple ½ lives. For example
how much is left after four half lives or if it takes so many seconds for a
material to reach 1/8 of its original amount ( 1/8 = ½ x ½ x ½ = three half lives)
AP 2008 # 32
Gaseous cyclobutene undergoes a first order reaction to form gaseous
butadiene. at a particular temperature, the partial pressure of cyclobutene in the
reaction vessel drops to one- eighth its original value in 124 seconds. What is
the half-life for this reaction at this temperature?
a) 15.5 s
b) 31.0 s c) 41.3 s d) 62.0 s
e) 124 s
Ans. c)
1/8 = ½ x ½ x ½ = three half lives
124/ 3 = 41.3 = ½ life
example 18.4 p. 884 + fig. 18.4
21
AP 2002 # 55
Time(days)
0
1
2
3
4
5
6
7
% Reactant
remaining
10
0
79
63
50
40
31
25
20
...
10
10
55. A reaction was observed for 20 days and the percentage of the reactant remaining
after each day was recorded in the table above. Which of the following best
describes the order and the half-life of the reaction?
Reaction Order
Half-life (days)
(A) First
3
(B) First
10
(C) Second
3
(D) Second
6
(E) Second
10
Ans. A
½ Life = time it takes for half of the original amount to decay. So 50 % = 3 days
thus, ½ life = 3 days
Order : Basically the reactant will be reduced at a constant rate based on the ½
life. Every 3 days it will be reduced by 50% . Checking data at day 3 and 6
went from 100%  50%  25 % therefore 1st order.
22
…
20
1
IB example
In 1919, Rutherford produced the first artificial nuclear transmutation by
bombarding nitrogen with α-particles. The reaction is represented by the following
equation.
α+
14
7
N 178 O + X
(a)Identify X : ………………………
(b) The reaction in (a) produces oxygen (O-17). Other isotopes of oxygen include O19 which is radioactive with a half-life of 30 s.
(i)
State what is meant by the term isotopes.
...................................................................................................................
...................................................................................................................
(ii)
Define the term radioactive half-life.
...................................................................................................................
...................................................................................................................
(c)
A nucleus of the isotope O-19 decays to a stable nucleus of fluorine. The halflife of O-19 is 30 s. At time t = 0, a sample of O-19 contains a large number
N0 nuclei of O-19.
On the grid below, draw a graph to show the variation with time t of the number N of
O-19 nuclei remaining in the sample. You should consider a time of t = 0 tot = 120 s.
23
Answer:
(a) In 1919, Rutherford produced the first artificial nuclear transmutation by
bombarding nitrogen with α-particles. The reaction is represented by the
following equation.
α+
14
7
N 178 O + X
Identify X.
answer : 1
p
1
solution:
4
He +
2
14
17
N  O
7
8
1
+ p
1
b) (i) (nuclei of same element with) same proton number,
different number of neutrons / OWTTE;
(ii)
1
the time for the activity of a sample to reduce by half / time
for the number of the radioactive nuclei to halve from original value;
1
24
(c)
scale drawn on t axis; (allow 10
N 0grid squares
N 0= 30 s or 40
N 0s)
smooth curve passes through 2 at 30 s, 4 at 60 s, 8
N0
at 90 s, 16 at 120 s (to within 1 square); (points not necessary)
2
[10]
25
NUCLEAR REACTIONS , FISSION and FUSION
Tsokos pp. 380 – 387
2. The Mass Defect , E = mc2 and Binding Energy
2. Find the mass defect of silver
108
Ag
47
p = 47
n = 61
In Data booklet
mass of 1 proton ( mp) = 1.007276 u
mass of 1 neutron ( mn) = 1.008665 u
mass of 1 electron ( me) = 0.0005486 u
ᵟ = ( total mp + total mn ) - Mnucleus
Mnucelus = Matom – Zme
1st step : ( total mp + total mn ) : REPRESENTS NUCLEAR MASS NOT
BONDED
( total mp + total mn ) = 47 x 1.007276 u + 61 x 1.008665 u = 108.870537 u
2nd step : Mnucelus = Matom – Zme
Matom (Ag) = 107.87 u
REPRESENTS NUCLEAR BONDED
(given from periodic table) , Z = 47 , me = 0.0005486 u
Mnucelus = Matom – Zme
= 107.87 u - 47 x 0.0005486 u
= 107.8442158
Final step : ᵟ = ( total mp + total mn ) - Mnucleus
ᵟ = 108.870537 u - 107.8442158 u = 1.0263212
26
3. Find the mass defect of copper
65
Cu
29
p = 29
n = 36
ᵟ = ( total mp + total mn ) - Mnucleus
Mnucelus = Matom – Zme
1st step : ( total mp + total mn ) :
( total mp + total mn ) = 29 x 1.007276 u + 36 x 1.008665 u = 65.522944 u
2nd step : Mnucelus = Matom – Zme
Matom (Cu) = 63.55 u
(given from periodic table) , Z = 29 , me = 0.0005486 u
Mnucelus = Matom – Zme
= 63.55 u - 29 x 0.0005486 u
= 63.5340906 u
Final step : ᵟ = ( total mp + total mn ) - Mnucleus
ᵟ = 65.522944 u - 63.5340906
= 1.9888534 u
27
4. Find the mass defect of chlorine
37
Cl
17
p = 17
n = 20
ᵟ = ( total mp + total mn ) - Mnucleus
Mnucelus = Matom – Zme
1st step : ( total mp + total mn ) :
( total mp + total mn ) = 17 x 1.007276 u + 20 x 1.008665 u = 37.296992 u
2nd step : Mnucelus = Matom – Zme
Matom (Cl) = 35.453 u
(given from periodic table) , Z = 17 , me = 0.0005486 u
Mnucelus = Matom – Zme
= 35.453 u - 17 x 0.0005486 u
= 35.4436738 u
Final step : ᵟ = ( total mp + total mn ) - Mnucleus
ᵟ = 37.296992 u - 35.4436738 u
= 1.8533182 u
28
5. Sometimes you will be given the mass of the nucleus directly.
You only need to use this formula:
ᵟ = ( total mp + total mn ) - Mnucleus
5. The mass of a nucleus of plutonium ( 239
94 Pu ) is 238.990396 u. Find the mass
defect:
ᵟ = ( total mp + total mn ) - Mnucleus
ᵟ
= (94 × 1.007276 + 145 × 1.008665) – 238.990396 = 1.95u
29
3. E = mc2 and Binding Energy
To calculate the binding energy get the mass defect (ᵟ ) and multiply by 931.5
Example:
Find the binding energy of nickel,
62
Ni
28
p = 28
n = 34
ᵟ = ( total mp + total mn ) - Mnucleus
Mnucelus = Matom – Zme
1st step : ( total mp + total mn ) :
( total mp + total mn ) = 28 x 1.007276 u + 34 x 1.008665 u = 62.498338 u
2nd step : Mnucelus = Matom – Zme
Matom (Ni) = 58.69 u
(given from periodic table) , Z = 28 , me = 0.0005486 u
Mnucelus = Matom – Zme
= 58.69 u - 28 x 0.0005486 u
= 58.6746392 u
Final step : ᵟ = ( total mp + total mn ) - Mnucleus
ᵟ = 62.498338 u - 58.6746392 u
= 3.8236988u
Binding energy = mass defect x 931.5 MeV
= 3.8236988u X 931.5 MeV = 3561.8 MeV ( don’t worry about sig. dig. )
30
Example 2 :
Use the following data to deduce the binding energy of the isotope
nuclear mass of 23 He
mass of proton
mass of neutron
3
He
2
= 3.01603 u
= 1.00728 u
= 1.00867 u
ᵟ = ( total mp + total mn ) - Mnucleus
Binding energy = mass defect x 931.5 MeV
1st step : ( total mp + total mn ) :
3
He
2
has 2 protons and 1 neutron
( total mp + total mn ) = 2 x 1.00728 u + 1 x 1.00867u = 3.02323
ᵟ = ( total mp + total mn ) - Mnucleus ( Nuclear mass)
= 3.02323 - 3.01603 = 0.0072 u
Binding energy = mass defect x 931.5 MeV
= (0.0072 u ) x 931.5 MeV = 6.7 MeV
31
1.
This question is about nuclear physics.
(a)
(i)
Define binding energy of a nucleus.
Answer :
the (minimum) energy required to completely separate the
nucleons of a nucleus / the energy released when a nucleus
is assembled;
1
(1
(ii)
The mass of a nucleus of plutonium ( 239
94 Pu ) is 238.990396 u. Deduce
that the binding energy per nucleon for plutonium is 7.6 MeV.
32
(b)
The graph shows the variation with nucleon number A of the binding energy
per nucleon.
Plutonium ( 239
94 Pu ) undergoes nuclear fission according to the reaction given
below.
239
94 Pu
(i)
91
1
 01n  38
Sr  146
56 Ba  x 0 n
Calculate the number x of neutrons produced.
33
(ii)
(c)
Use the graph to estimate the energy released in this reaction.
Stable nuclei with a mass number greater than about 20, contain more
neutrons than protons. By reference to the properties of the nuclear force and
of the electrostatic force, suggest an explanation for this observation.
(c) Answer :
The nuclear force, electrostatic force and binding energy are all involved in holding the
nucleons ( protons and neutrons) together.
Typically, when nuclei of atoms smaller than Iron or Nickel fuse (nuclear fusion) to
form a single nucleus, the binding energy per nucleon increases , which tells us energy
is released in the process and the nucleus produced is more stable.
When nuclei larger than Iron or Nickel or , in this curve greater than 20 , are split
(nuclear fission) = radioactive decay, the binding energy per nucleon increases, again
releasing energy and the nuclei produced are more stable. Radioactive elements such as
plutonium ( Pu ) or uranium (U) are found in this category.
34
NEW 7.3 – PARTICLE PHYSICS : STRUCTURE OF
MATTER
Charge
1. The proton is composed of two up (u) quarks and one down (d) quark : uud
What is the charge of a proton? You should already know it is +1 and the chart
shows:
uud =
2. The neutron is composed of two down (d) quarks and one up (u) quark : ddu
What is the charge of a neutron? Explain using the chart:
3. The antiproton is composed of two anitup quarks and one antidown quark.
Write the 3 letter code and explain what the charge is using the chart:
4. The π + meson is composed of an up quark and an anti down quark. Write the
2 letter code and explain what the charge is using the chart:
35
5.
6.
7.
Since ccc = 1/3 + 1/3 + 1/3 = 1
Then
= antiquark = - 1
36
Example 1 : The π + meson is composed of an up quark and an anti down quark
Example 2 ; What is the code and charge of a pion π which is composed of an
anti up quark and a down quark?
-2/3 + -1/3 = -1
BARYON NUMBER Examples
1. What is the baryon number of the quark combination ccc?
Answer : 1/3 + 1/3 + 1/3 = +1
2.
Answer : It is – 1 since this is an antibaryon
3. What is the charge of the baryon Λ = uds?
Answer :
EXAMPLES of REACTIONS – USE CHARTS
37
Example 1: Determine whether a collision between two protons could produce
two protons and a neutron .
Step 1 write basic equation: p + p  p + p + n
Charge rule #1 Q : +1 + 1  +1 + 1 + 0
2=2
OK conservation of charge
Lepton rule # 2 L : 0 + 0  0 + 0 + 0
0 = 0 OK
conservation of lepton number
Baryon rule # 3 B : 1 + 1  1 + 1 + 1
2 = 3 NOT OK - conservation of baryon number
Answer : Reaction is NOT possible due to no conservation of baryon
number
Example 2 : Determine whether a collision between a proton and an electron
could produce a neutron .
Step 1 write basic equation: p + e  n
Charge rule #1 Q : +1 + - 1  0
0=0
OK conservation of charge
Lepton rule # 2 L : 0 + 1  0
1 = 0 NOT OK - conservation of lepton number
Baryon rule # 3 B : 1 + 0  1
0=0
OK conservation of baryon number
Answer : Reaction is NOT possible due to no conservation of lepton
number
38
Example 3: A proton collides with a negative pion π – and produces a neutron
and an uncharged pion. a) write the equation b) determine it the particle
interaction is possible
a)
b)
8. Feynman Diagrams
Try these: Remember does not have to be exact. Several versions are
acceptable.
1. A proton decays into a neutron and releases a positive boson. Which
fundamental force is responsible for this reaction?
Strong nuclear force
39
2. An electron absorbs a photon. Which fundamental force is responsible for this
reaction?
Electromagnetic force
3. A neutrino collides with an electron and releases a positive boson. Which
fundamental force is responsible for this reaction?
Weak nuclear force
40