Mini Progress Check – Self, Peer or Teacher Assess
Name:
Score/Grade:
Nuclear Fission and Fusion - Review Question 34 marks
I think/reply.....
1.
This question is about nuclear fission of uranium-235.
(i)
State what is meant by a thermal neutron.
........................................................................................................................
........................................................................................................................ [1]
(ii)
State the importance of thermal neutrons in relation to the fission of uranium-235.
........................................................................................................................
............................................................................................................... [1] [Total 2 marks]
2.
The figure below shows the variation with nucleon number (mass number) of the binding energy
per nucleon for various nuclides.
10
9
8
binding energy
per nucleon 7
/ MeV
6
5
4
3
2
1
0
0
Mr Powell 2013
50
100
150
200
300
250
mass number
1
Mini Progress Check – Self, Peer or Teacher Assess
2)
(a)
(i)
State the number of nucleons in the nucleus of
(ii)
State the number of protons in the nucleus of
(iii)
State the number of neutrons in the nucleus of
94
37 Rb .
........................
142
55 Cs .
.........................
235
92 U .
.........................
[2]
(b)
Use the figure above to calculate the energy released when a
fission, producing nuclei of
94
37 Rb
and
235
92 U
nucleus undergoes
142
55 Cs .
energy = .............................................. MeV
[4]
[Total 6 marks]
3.
A uranium-235 nucleus
and bromine-87
Mr Powell 2013
87
35 Br
235
92 U
undergoes fission, producing nuclei of lanthanum-146
146
57 La
. The binding energies per nucleon of these nuclides are shown below.
nuclide
binding energy per
nucleon / MeV
235
92 U
7.6
146
57 La
8.2
87
35 Br
8.6
2
Mini Progress Check – Self, Peer or Teacher Assess
(i)
Plot these values on the grid below.
10.0
8.0
binding energy
per nucleon
/ MeV
6.0
4.0
2.0
0
0
50
100
150
200
nucleon number
250
[1]
(ii)
Sketch a graph on the grid above, to show how the binding energy per nucleon varies with
nucleon number for all nuclei.
[2]
(iii)
Use information from the table to calculate how much energy in MeV is released when a
235
92 U nucleus undergoes fission.
energy = ................................................ MeV
[3]
[Total 6 marks]
4.
In nuclear fission, energy is released.
(a)
Explain what is meant by nuclear fission.
........................................................................................................................
........................................................................................................................
[1]
Mr Powell 2013
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Mini Progress Check – Self, Peer or Teacher Assess
235
92 U captures a neutron
92
Kr releasing three
and 36
4 (b) In a possible fission reaction
before splitting into
141
56 Ba
to become a compound nucleus
neutrons.
Write down the nuclear reaction equation for this event.
........................................................................................................................
[2]
(c) The total mass of the compound nucleus 236
92 U before fission is 236.053 u. The total mass
of the fission products is 235.867 u. Use these data to calculate the energy released in the fission
process.
energy = …………………..J
[3]
4 (d) Most of the energy released arises from the electrostatic repulsion of the two nuclei as they
move apart. Use the information in (b) to show that the force F between the two nuclei at
the instant after fission occurs is about 3000N.
Assume the nuclei act as point charges a distance r apart of 1.3 × 10–14 m.
[4]
[Total 10 marks]
Mr Powell 2013
4
Mini Progress Check – Self, Peer or Teacher Assess
5.
In the JET fusion experiment, a plasma consisting of a mixture of deuterium ( 21H ) and tritium (
3
1H )
is confined within a magnetic field of high flux density.
The plasma is heated using two methods.
method 1
A very large current is passed through the plasma.
Fig. 1 shows the variation with time of this current.
The average electromotive force driving this current is 1.2 V.
4
current /
106 A
3
2
1
0
0
5
10
15
20
25
30
time / s
Fig. 1
method 2
Fast-moving deuterium atoms are injected into the plasma. The nuclei of the
injected deuterium atoms collide with nuclei in the plasma and so transfer energy
to it.
When the plasma temperature is high enough, deuterium and tritium nuclei fuse, producing a
helium nucleus and a neutron. This reaction may be represented as follows.
2
1H

3
1H

4
2 He

1
0n
+
energy
The energy released is shared between the helium nucleus and the neutron, which move off in
opposite directions.
4
2 He
1
0n
Fig. 2
Mr Powell 2013
5
Mini Progress Check – Self, Peer or Teacher Assess
(a)
For method 1, calculate the total energy input provided by the current source.
energy = ....................................... J
[4]
(b)
Explain why in method 2 a beam of neutral deuterium atoms is injected, rather than a
beam of deuterium nuclei.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
[2]
(c)
Show that the helium nucleus gains 20% of the total energy released in the fusion reaction,
and the neutron gains 80% of the energy released.
You may assume that the initial momentum of the helium-neutron system is zero.
[4]
[Total 10 marks]
Mr Powell 2013
6
Mini Progress Check – Self, Peer or Teacher Assess
Answers
1.
(i)
(ii)
(neutrons) having energies comparable with thermal energies /
slow moving / low kinetic energy / energy in range 6 - 100 eV /
energy similar to (energy of ) atoms of surroundings ;
1
either thermal neutrons will be captured / absorbed (by U-235 nuclei)
or
higher energy neutrons do not get absorbed;
1
[2]
2.
(a)
(b)
Rb 94
Cs 55
U143
–1 for each error
B2
Values from graph: U 7.4 MeV allow 7.3 to 7.4
Rb 8.6 MeV allow 8.5 to 8.6
Cs 8.4 MeV
C1
Total binding energies: U 235 × 7.4 (1739)
Rb 94 × 8.6 (808)
Cs 142 × 8.4 (1193)
B2
Total energy released = 808 + 1193 – 1739
= 262 MeV
allow 8.6 + 8.4 – 7.4 = 9.4 MeV for 1 mark only
A1
[6]
3.
(i)
3 points plotted;
(ii)
curve through 3 points and heads down towards zero; (1)
line peaks between Br and origin; (1)
2
BE per nucleus of 23592U = 7.60 × 235 (= 1786 MeV)
BE of products
= 8.20 × 146 + 8.60 × 87
both lines (1)
(= 1197 + 748 MeV)
so energy released
= (1197 + 748) – 1786 (1)
= 159 MeV (1)
omits multiplication by nucleon number to get 9.2 MeV gets 0,1,0 = 1
3
(iii)
any point incorrect loses this mark
1
[6]
4.
(a)
the splitting of a nucleus into two (or more) smaller nuclei/particles/
fragments (spontaneously/after absorption of a neutron)
Mr Powell 2013
1
7
Mini Progress Check – Self, Peer or Teacher Assess
(b)
(c)
(d)
235
92U
+ 10n → 14156Ba + 9236Kr + 310n
–1 mark per error
2
∆E = c2∆m; ∆m = 0.186 u (= 3.09 × 10–28 kg); (2)
∆E = 9 × 1016 × 0.186 × 1.66 × 10–27 = 2.78 × 10–11 (J) (1)
3
F = kQ1Q2/r2; Q1 = 56e, Q2 = 36e; (2)
F = 9 × 109 × 56 × 36 × (1.6 × 10–19)2/(1.3 × 10–14)2; = 2.7(4) × 103 (N) (2)
4
[10]
5.
(a)
energy
= V I t (1)
= V × (area under I-t graph) (1)
= 1.2 × 4 × 106 × (20 + 5) (1)
= 1.2 × 108 J (1)
4
no V gets 0/4 except if stated ‘area under graph = charge’ which gets ¼
area calculation errors eg wrong triangle areas can get ¾
omits 106 can get 3/4
(b)
(c)
nuclei have (net) charge but atoms don’t; (1)
nuclei would be deflected by B field / atoms are not; (1)
2
(momentum conservation: mH vH = mn vn mH = 4 mn so) vn = 4 vH (1)
ke = ½ m v2 (1)
ke of 10n = ½ m (4 vH)2 = 8 m vH2 }
ke of 42He = ½ × 4 m vH2 = 2 m vH2 } subs. (1)
so (ke of 10n) = 4 × (ke of He)
(so 10n has 80%, 42He has 20% of total ke) (1)
4
[10]
Mr Powell 2013
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