Topic 7, 12 - Selected problems Solutions
1. Determine missing element X :
235
U +
92
1
n
0

137
Te +
52
X
235
U +
92
1
n
0

137
Te +
52
97
Kr +
40
+
1
2 n
0
Answer :
1
2 n
0
2.
answer
3. An isotope of radium has a half-life of 4 days. A freshly prepared sample of this isotope contains N
7N
atoms. The time taken for
of the atoms of this isotope to decay is
8
answer : C ( i.e decay to 1/8 = 3 half lives = 3x4 = 12 days)
A.
32 days.
B.
16 days.
C.
12 days.
D.
8 days.
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4.
Which one of the following provides direct evidence for the existence of discrete energy levels in
an atom? B
A.
The continuous spectrum of the light emitted by a white-hot metal.
B.
The line emission spectrum of a gas at low pressure.
C.
The emission of gamma radiation from radioactive atoms.
D.
The ionization of gas atoms when bombarded by alpha particles.
Topic 7 notes
Line emission spectrum
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5.
In a fission chain reaction, C
A.
energy from one fission reaction causes further fission reactions.
B.
nuclei produced in one fission reaction cause further fission reactions.
C.
neutrons from one fission reaction cause further fission reactions.
D.
gamma radiation produced in one fission reaction causes further fission reactions.
B is close but key word is neutron not nuclei
235
U +
92
6.
1
n
0

141
Ba
56
+
92
Kr +
36
1
3 n
0
Which one of the following correctly identifies the atomic (proton) number and mass (nucleon)
number of a nucleus that has neutrons n and protons p? B
Atomic number
Mass number
A.
p
n
B.
p
n+p
C.
n
p
D.
n+p
p
3
7.
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The binding energy per nucleon of the nucleus 3 Li is approximately 5 MeV. The total energy
required to completely separate the nucleons of this nucleus is approximately C
A.
15 MeV.
B.
20 MeV.
C.
35 MeV.
D.
50 MeV.
Total number of nucleons = 7 : 7 x 5 MeV = 35 MeV
(1)
8.
The source of the Sun’s energy is C
A.
fission.
B.
radioactivity.
C.
fusion.
D.
ionization.
9. This question is about nuclear binding energy.
(a)
(i)
Define nucleon.
Subatomic particles that make up the nucleus. Either proton or neutron
(ii)
Define nuclear binding energy of a nucleus.
the energy required to separatea nucleus into its component nucleons
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The axes below show values of nucleon number A (horizontal axis) and average binding energy per
nucleon E (vertical axis). (Binding energy is taken to be a positive quantity).
E / MeV
10
9
F
8
U
7
6
5
4
3
2
1
H
0
0
(b)
(c)
25 50 75 100 125 150 175 200 225 250
A
Mark on the E axis above, the approximate position of
(i)
the isotope
56
Fe
26
(ii)
the isotope
2
H
1
(iii)
the isotope
238
U
92
(label this F).
(label this H).
(label this U).
Explain the difference between U and H interms of binding energy and nuclear reactions
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 are split (nuclear fission) = radioactive decay, the binding energy per nucleon
increases, again releasing energy and the nuclei produced are more stable.
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(d)
Use the following data to deduce that the binding energy per nucleon of the isotope
2.2 MeV.
nuclear mass of 23 He
mass of proton
mass of neutron
3
He
2
is
= 3.01603 u
= 1.00728 u
= 1.00867 u
answer: 2.2 MeV
mass of nucleons = (2 × 1.00728) + 1.00867 = 3.02323 u
mass difference = 0.0072 u = 6.7 MeV
binding energy per nucleon = 6.7 / 3 = 2.2 MeV
In the nuclear reaction
(e)
2
2
3
1 H  1 H  2 He
 01 n energy is released.
State the name of this type of reaction.
Answer : fusion
[13]
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10. Two different models have been developed to explain the existence of atomic energy levels. The
Bohr model and the Schrödinger model are both able to predict the principal wavelengths
present in the spectrum of atomic hydrogen.
Outline
(i)
the Bohr model, and
(ii)
the Schrödinger model.
BOHR MODEL:
H , the Bohr Model, Quantum Numbers and Orbital Shapes
Energy Levels , Bohr Model, electrons, excited state and color . Text p. 299
Electrons move around the nucleus in very complex ways that are still not fully understood.
There are several models of this complex movement and we will start with the ideas of
energy levels or orbits and the Bohr model. Niehls Bohr ( nobel prize winner from
Denmark) stated that the electron moved in concentric circles around the nucleus at
different energy levels , called n = 1, n = 2 , n=3, etc.
ELECTRON ABSORPTION – EMISSION
As atoms absorb energy the electrons can get further from the nucleus with this gained
energy. n= 1is called the ground state and represents the atom and its electrons in the
lowest possible energy state. The excited state or states n= 2, n= 3 etc. are different
energy states that the atom can exist when absorbing energy. The atom can also release
( emission) this energy going back to the ground state or lower energy state. This is
where the photon comes in. When going from the excited state to the ground state the
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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. 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.
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.
Because only certain photons are emitted , we know only certain energy changes are
occurring. This means that the atom has certain energy levels . For example hydrogen in
the excited state always emits photons with the same colors ( wavelengths-frequencies) .
They never emit photons in between those shown. So all hydrogen atoms have the same
set of discrete energy levels and these levels are quantized. This set of energy levels can
also be called a line spectra . Atoms can be identified by the spectrum they give when
excited.
The quantized nature of the energy levels in atoms was a surprise when scientists
discovered it. It had been assumed that the atom could exist at any energy level. This
discovery radically changed the view of the atom.
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9
Schrödinger Theory
In 1926 Erwin Schrödinger, an Austrian physicist, took the Bohr atom model one step
further. Schrödinger used mathematical equations to describe the likelihood of finding an
electron in a certain position. This atomic model is known as the quantum mechanical model
of the atom. Unlike the Bohr model, the quantum mechanical model does not define
the exact path of an electron, but rather, predicts the odds of the location of the
electron. This model can be portrayed as a nucleus surrounded by an electron cloud.
Where the cloud is most dense, the probability of finding the electron is greatest, and
conversely, the electron is less likely to be in a less dense area of the cloud. Thus, this
model introduced the concept of sub-energy levels.
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11. This question is about radioactive decay.
(a) The decay constant for a particular isotope is λ = 0.048 s–1. A sample of the isotope initially contains 2.0 x
1012 nuclei of this isotope.
(i) Define decay constant.
The Law of Radioactive decay states that the number of nuclei that will decay per second (i.e. the rate of
decay) is proportional to the number of atoms present that have not yet decayed.
dN
 N
dt
Where: N = number of nuclei that have not yet decayed
λ = decay constant [s-1] (Represents the probability of decay per unit time)
(ii) Estimate the number of nuclei that will decay in the first second.
nuclei that will decay per second = λ N
= ( 0.048 s-1 ) (2.0 x 1012) = 9.6 x 1010
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12. This question is about the wave nature of electrons.
(a)
Describe the de Broglie hypothesis.
Since an electromagnetic wave can behave like a particle, can a particle of matter behave
like a wave? In 1923, the French physicist Louis de Broglie said yes. His hypothesis, which
has been supported by experiment, is that a particle of mass m and speed v and thus linear
momentum p = mv, has an associated wavelength. This wavelength is called the de Broglie
wavelength.:
λ=h
p
p = momentum = mv
An experiment is carried out in which a beam of electrons is scattered from a single nickel crystal.
A schematic diagram of the apparatus is shown below.
Vacuum
Nickel
crystal
Incident
electron beam
Electron “gun”
Scattered electron beam
The electrons are accelerated in the electron “gun” by a potential difference of 75 V.
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(b)
Determine the wavelength associated with the electrons as predicted by the de Broglie
hypothesis.
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13. This question is about atomic spectra.
An electron undergoes a transition from an atomic energy level of 3.20  10
–15
0.32  10
λ
–15
J to an energy level of
J. Determine the wavelength of the emitted photon.
hc
;
E 
–15
E = 3.20  10
–15
J - 0.32  10
15
J = 2.88  10
J
 = ( 6.63 x 10-34 ) ( 3x 108)
2.88  10
11
6.9  10
14.
15
J
m
This question is about nuclear reactions.
(a)
State the meaning of the terms
(i)
nuclide
an atom or nucleus that is characterized by the constituents of
its nucleus – proton and neutron.
(ii)
isotope
an atom or nucleus that has the same number of protons but different number of neutrons
(b)
The isotope sodium-24 undergoes radioactive decay to the stable isotope magnesium-24.
(i)
Complete the nuclear reaction equation for this decay.
24
24
11 Na  12 Mg
+
0
1 e
14
(ii)
One of the particles emitted in the decay has zero rest-mass. Use the data below to
estimate the rest mass, in atomic mass units, of the missing particle emitted in the
decay of the reaction above :
rest mass of
24
11 NA
24
12 Mg
= 23.99096u
rest mass of
= 23.98504u
energy released in decay = 5.002160 MeV
Answer :
0
1 e
= 0.00055 u
Need to understand that the overall equation has energy released :
24
24
11 Na  12 Mg
0
1 e
+
mass of reactants
rest mass of
24
11 NA
23.99096u
+ energy
=
=
mass of products
rest mass of
24
12 Mg
23.98504u +
0
1 e
+ energy 5.002160 MeV
convert energy 5.002160 MeV to u : 1 u = 931.5 MeV ( in data booklet)
5.002160 MeV = 0.00537 u
rest mass of
24
11 NA
23.99096u
=
rest mass of
24
12 Mg
23.98504u +
0
1 e
+ energy 0.00537 u
SO:
0
1 e
=
24
11 NA
0
1 e
23.99096u
- (rest mass of
24
12 Mg
23.98504u + energy 0.00537 u)
= 0.00055 u
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(c)
The isotope sodium-24 is radioactive but the isotope sodium-23 is stable. Suggest which of
these isotopes has the greater nuclear binding energy.
Answer :
sodium-24 has more nucleons;
and more nucleons (usually) means greater (magnitude of) binding energy;
or
sodium-23 has less nucleons;
and less nucleons (usually) means less (magnitude of) binding energy;
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