TUTOR COPY
Section A: Review of concepts
JC2 H2 Physics Tutorial 2011 Chapter 19: Nuclear Physics
This section helps you to consolidate your learning for this chapter. Complete the following summary of concepts before attempting the tutorial questions; all answers can be found in your lecture notes. (I) Atomic Structure a) In the space below, draw a well labeled sketch for the set up of the Experiment.
-particle Scattering
Detector
Vacuum
Source Slit
Gold foil
State and explain what the expected results of the experiment, based on Thomson’s model. •
-particles carry a charge of +2e and are about 7300 times more massive than electrons.
• If Thomson’s model was correct, maximum deflecting force on the -particle as it passes near a positive charge will be far too small to deflect the particle by even 1º. • Electrons in the atom would also have very little effect on the massive, energetic -particle. b) Describe the actual results of the -particle Scattering Experiment and infer from the results, the possible structure of the nucleus. 1. Large deflection of -particles • To produce such a large deflection, there must be a large force • Such a large force is only possible if we assume that the atom consists of a positively charged nucleus of very small dimensions compared with the ordinarily accepted magnitude of the diameter of the atom. When an -particle gets close to the centre of the concentration of positive charges (nucleus), it cannot penetrate the nucleus but gets deflected instead due to the large repulsive force between it and the positively charged nucleus. 2. Most particles passed through undeflected or with a small deflection
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• Since very few of the particles were scattered through large angles, the probability of the particle getting close to the centre of the positive charge is small. This shows that the nucleus occupies only a small proportion of the available space. 3. Few particles are reflected backwards, through an angle close to 180o • The nucleus is small and very massive. c) Define Isotope Isotopes are atoms that have the same number of protons but different number of neutrons. II) Nuclear Physics d) State Einstein’s mass-energy relation. E = mc2 e) Define one electron-volt. One electron-volt is the energy gained by a charge equal to that on an electron in moving through a potential difference of one volt. f)
Define mass defect and list the two equations for mass defect for a nucleus and for a neutral atom. The mass defect of a nucleus is defined as the difference between the mass of the separated nucleons and the combined mass of the nucleus. M = Zmp + Nmn - Mn M = Zmp + Nmn + Zme – Ma
g) Explain the relation of mass defect to binding energy. From Einstein’s mass energy relationship, we can infer from this increase in mass, that the total energy of the separated components is greater than the total energy of the bound system (i.e. the nucleus). The energy is released when the nucleons are separated in a nucleus is known as nuclear binding energy. h) Define nuclear binding energy of a nucleus. The nuclear binding energy of a nucleus is defined as the work done on the nucleus to separate it into its constituent neutrons and protons. i)
Sketch the graph of binding energy per nucleon against mass number.
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j)
Explain the relevance of binding energy per nucleon to nuclear fusion and to nuclear fission. i. Except for the lighter nuclei, the average binding energy per nucleon is about 8 MeV. ii. The peak, i.e. the maximum B.E. per nucleon occurs at around mass number A = 50, and corresponds to the most stable nuclei. From the graph, it can be seen that the iron nucleus 56 26 Fe is located close to the peak with a B.E. per nucleon value of approximately 8.8 MeV. It is one of the most stable nuclides that exist. iii. Nuclei with very low or very high mass numbers have lesser binding energy per nucleon and are less stable because the less the B.E. per nucleon, the easier it is to separate the nucleus into its constituent nucleons. iv. Nuclei with low mass numbers (located to the left of the peak), may undergo nuclear fusion, where light nuclei are joined together under certain conditions so that the final product may have a greater binding energy per nucleon v. Nuclei with high mass numbers may undergo nuclear fission --- the nucleus may split to give two daughter nuclei with the release of neutrons. The daughter nuclei will possess a greater binding energy per nucleon.
k) State the quantities conserved in a nuclear reaction. • • • • l)
nucleon number proton number (charge) mass-energy momentum
Explain the spontaneous and random nature of nuclear decay. Radioactive decay occurs spontaneously; the process cannot be speeded up or slowed down by physical means such as changes in pressure or temperature. The decay of a radioactive atom is not affected by any chemical condition or the chemical compound that it exists in and is independent of physical conditions such as temperature, pressure and most importantly the decay of other atoms. •
Radiation is emitted at random. By random, we mean that it is impossible to predict which nucleus and when any particular nucleus will disintegrate.
m) Define Nuclear fission Nuclear fission is the disintegration of a heavy nucleus into two lighter nuclei of approximately equal masses. n) Define Nuclear fusion Nuclear fusion is the combining of two light nuclei to produce a heavier nucleus.
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o) Sketch the path of alpha, beta and gamma rays in a strong magnetic field pointing into the paper in the space below.
×
×
×
×
× × - rays × × ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
- particles
×
×
- particles
p) Define activity and decay constant. The activity of a radioactive substance is defined as the average number of atoms disintegrating per unit time. The decay constant of a nucleus is defined as its probability of decay per unit time. q) Define Half-life Half-life is defined as the time taken for half the original number of radioactive nuclei to decay. r) Discuss qualitatively the effects, both direct and indirect, of ionising radiation on living tissues and cells. Radiation can cause immediate severe damage to body tissue such as radiation burns. Delayed effects such as cancer and eye cataracts may appear many years later. Hereditary defects may also occur in succeeding generations due to genetic damage. When radiation passes through living tissues, it can damage the structure of molecules leading to the malfunction or death of living cells. Some cells can recover, others cannot and the effects on tissues are cumulative. Chromosomes are particularly sensitive to ionising radiation at the moment of cell division resulting in genetic mutations which are probably harmful. Such genetic mutations can lead to birth defects if an unborn child and / or the mother is exposed to radiation.
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Section B: Question-Solving 1 Two alpha-particles with equal energies are fired towards the nucleus of a gold atom. Which diagram could represent their paths (in the plane of the paper)?
[N96P1Q29] Ans: A 2
a) Find the sum of masses of the constituents of the 126 C atom, giving your answer in the unified atomic mass unit. You are given that mp = 1.00728u, mn = 1.00867u and me = 0.00055u. Total mass of nucleons and electrons = 6mp + 6mn + 6me = 12.099 u b) By definition, the mass of the exactly?
12 6
C atom is 12 u exactly. Why is your answer to a) not 12 u
Loss of mass in the form of mass defect when the protons, neutrons and electrons come together to form the atom.
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3
a) When an 168 O nucleus is bombarded by a neutron, a deuteron 21H is emitted. Write down the reaction equation. What is the resulting nucleus? 16 8
O + 01n → 12 H + 157 N
Use conservation of nucleon number and proton number. Based on proton number, can identify from the periodic table that the element is nitrogen. b) Why are neutrons such good projectiles for produing nuclear reactions? Neutrons are good projectiles for producing nuclear reactions because they are neutral and they are massive. If you want a particle to hit the nucleus with a lot of energy, a more massive particle is the better choice. A light electron would not be as effective. Using a positively charged projectile like an alpha or a proton means that the projectile will have to overcome the large electrical repulsion from the positively charged nucleus. Neutrons can penetrate directly to the nucleus and cause nuclear reactions.
4
In a nuclear reaction, uranium-235 ( 235 92 U) nuclei are transformed into unstable uranium-236 nuclei ( 236 92 U) through bombardment by slow-moving neutrons. The unstable uranium-236 nucleus undergoes nuclear fission to form stable products of a lathium-139 nucleus ( 139 57 La) and a nuclide of bromine ( ZA Br). 235 92 U
+ 01 n →
236 92
U →
139 57
La +
A Z
Br + 3 01 n + γ
Use the following masses in answering this question: mass of 235 92 U nucleus = 235.044 u mass of
139 57
La nucleus = 138.906 u
mass of 11 p proton mass of
1 0n
neutron
= 1.00728 u = 1.00866 u
(i)
Explain the meaning of the term nuclide.
(ii)
Determine the mass number A and proton number Z of the bromine nuclide.
(iii)
Calculate the binding energy of
(iv)
Given that the nuclear binding energy per nucleon of lathium-139 nucleus is 8.18930 MeV and that the energy released in the above reaction is 197 MeV, determine the mass of the bromine nuclide in terms of the unified atomic mass constant u to 4 significant figures.
(v)
A similar reaction is used in nuclear power plants to generate nuclear energy. Explain the subsequent behaviour of the neutrons produced in such reactions and a potential problem they present.
235 92 U
in MeV to 4 significant figures.
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4
(i) A nuclide is a species of nucleus with a particular nuclear structure. (ii)
94 35
Br :
A=94 ,Z=35
(both must be correct)
(iii) Binding energy of 235 92 U = (mass of neutrons + mass of protons – mass of nucleus) c2 = [((235-92) x 1.00866) + (92 x 1.00728) – (235.044)] u c2 = 1.86414 u c2 = (1.86414 ) (1.66x10-27) (3.00 x 108)2 = 2.7850 x10-10 J = 1740.641 MeV = 1741 MeV (4 sf)
(iv)
235 92 U
+ 01 n →
236 92
U →
139 57
La +
A Z
Br + 3 01 n + γ
Energy released = BELa + BEBr – BeU 197 MeV = (8.18930 x139) + BEBr – 1740.64073 BEBr
= Energy released + BeU – BELa = 197 + 1740.64073 – (8.18930 x139) = 799.3 MeV (1.279x10-10 J)
Since BEBr = (Mass of neutrons + Mass of protons – Mass of Br) c2 Mass of Br = Mass of neutrons + Mass of protons – BEBr / c2 =(59x1.00866u)+(35x1.00728u)–(799.3x106)(1.6x10-19)/[(1.66x10-27)(3.00x108)2] = 93.91 u (4 sf) Alternatively: Energy released = (minitial – mfinal) c2 = (mU + mn – mLa – mBr – 3mn) c2 197 MeV = (235.044 + 1.00866 – 138.906 – mBr – 3x1.00866) u c2 mBr = 93.91 u (v) The neutrons produced can trigger a chain reaction. If the reaction is not controlled, it will produce an enormous amount of energy and possibly lead to an explosion.
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5.
A nitrogen nucleus an oxygen nuclues
14 7 17 8
N is bombarded with an α-particle of a certain energy and transmutes to O and a proton.
a) Write an equation for this nuclear reaction. 14 7
N + 24 He → 178 O + 11H
b) Find the minimum energy of the α-particle to make this reaction occur. Given that: m(N) = 14.003074u m(O) = 16.998800u m(He) = 4.002602u m(p) = 1.007285u EK,N + EK,α + mN c2 + m α c2 = EK,O + EK,p + mO c2 + mp c2 Assume nitrogen nucleus is originally at rest
If α particle has minimum energy, will only be able to create product particles but not any excess energy for them
Hence, the kinetic energy of the α particle can be found by making it the subject of the equation: EK,α = [(mO + mp ) – (mN + m α)] c2 = (0.000409 u) c2 = 6.11 × 10-14 J
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6. The uranium isotope
238 92 U decays
naturally by a series of stages to the lead isotope Two schemes for the decay are suggested: 238 92 U
238 92 U
206 82 Pb
+ 10p + 22n 206 82 Pb
206 82 Pb .
---------- (I)
+ x 42 He + ye
---------- (II)
(i) Define the term nuclear binding energy and explain how it is related to the mass-defect of a nucleus. Nuclear binding energy is the amount of energy released when individual protons and neutrons come together to form a nucleus. From the principle of conservation of mass-energy, it follows that the release in energy results in a corresponding decrease in mass of the nucleus as compared to the total mass of the individual protons and neutrons that make up the nucleus. This decrease in mass is known as the mass-defect. (ii) Given that the atomic mass of uranium-238 is 238.05080 u, show that the binding energy of uranium-238 is 1807 MeV given the following data: rest mass of electron: rest mass of proton: rest mass of neutron: (ii)
5.48580 × 10-4 u 1.00728 u 1.00867 u
Mass defect of U-238 atom = 92(1.00728) + 92(5.48580 × 10-4) + (238 - 92)(1.00867) - 238.05080 = 1.935249 u Binding energy of U-238 = 1.935249 × (1.66×10-27) x (3.0×108)2 ÷ (1.60×10-19) = 1.807 × 109 eV = 1807 MeV (4 s.f.)
(iii) For reaction (II), state the values of x and y. (iii)
Equating mass number on both sides of the reaction: x=8 Equating proton number on both sides of the reaction: y=6
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238 = 206 + 4x 92 = 82 + 8(2) - y
4 (iv) The total binding energy of 206 82 Pb is 1613 MeV and that of 2 He is 28.3 MeV. Based on energy considerations, discuss the feasibilities of the two schemes. 238 92 U
206 82 Pb
238 92 U
+ 10p + 22n
206 82
---------- (I)
Pb + 8 24 He + 6e
---------- (II)
(iv) For reaction (I): Energy released = 1613 – 1807 = -194 MeV For reaction (II): Energy released = 1613 + 8(28.3) – 1807 = 32.4 MeV From the calculation of the energy changes involved, it appears that 194 MeV per nucleus of U-238 must be supplied in order for reaction (I) to proceed. Reaction (II) however, results in the release of 32.4 MeV of energy. The total binding energy of the products is greater than that of the parent nucleus. This implies that the products might be more stable than the parent. Hence, this reaction might proceed with greater ease than the first reaction.
7.
A radioactive isotope has a decay constant λ, and a molar mass M. Taking the Avogadro constant to be NA, the activity of a sample mass m of this isotope is
A
mN A λM
Ans: C
A = λN → A = λ
B
mλ MN A
C
λmN A M
D
λmN A m
m NA M
8. (a) Explain the meaning of the term half life. Half life means the time taken for half the number of nuclei of a radioactive element to decay
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(b) Carbon-14 has a half life of 5600 years. The activity of each gram of living wood due to C14 is 15 decays min-1. (i)
Calculate the number of C-14 atoms in one gram of living wood. = ln 2 / t1/2 = ln 2 / 5600 Since A = N No. of C-14 atoms, N = 15 / [ln 2 / (5600 x 365 x 24 x 60)] = 6.37 x 1010 = 6.4 x 1010
(ii) Determine the age of an ancient ship from which a sample of 5.0 g wood has an activity of 6.5 decays min-1. Activity of 5.0 g of living wood, A0 = 15 x 5 = 75 decays min-1 Since A = A0 e- t 6.5 = 75 e-(ln 2 / 5600) t t = -(5600 / ln 2) [ln (6.5/75)] = 1.98 x 104 yrs = 2.0 x 104 yrs
9
Nuclide X of decay constant = 7.85 x 10-10 s-1 is a beta emitter which decays to a stable nuclide Y. A sample initially contains N0 number of nuclei of X. (a)
Show that the number of nuclei of Y in the sample, N, after time t is given by N = N 0 1 − e − λt .
(
N X = N0 e − λt
)
(
N = N0 − N X = N0 − N0 e − λ t = N0 1 − e − λ t
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)
(b)
Sketch labeled graphs showing the variation of the number of nuclei of X and Y with time on the same axes.
T1 = 2
ln 2
λ
=
ln 2 = 8.83 × 108 s 7.85 × 10−10
N N0
Y
N0/2
X 0
(c)
8
8.83x10
t/s
It was found that a fresh sample of X emits 5.25 x 1013 particles in (i)
10 s.
Calculate the time taken by a 50 year-old sample of X to emit the same number of particles. A0 =
5.25 × 1013 = 5.25 × 1012 Bq 10
A = 5.25 × 1012 e
(
)
− 7.85×10−10 ( 50×365× 24× 60×60 )
= 1.52 × 1012 Bq t= (ii)
5.25 × 1013 = 34.5s 1.52 × 1012
Calculate the age of a sample of X if it takes 1 week to emit the same number of particles. 5.25 × 1013 = 8.68 × 107 Bq 7 × 24 × 60 × 60 −10 8.68 × 107 = 5.25 × 1012 e −7.85×10 t
A=
t = 1.40 × 1010 s
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10. A power source with an output of 1.00 kW is required for use in a research laboratory. This power is derived from the 6.35 x 10-12 J of energy released in the decay of a Given that the half-life of radium is 1622 years, find a.
226
88 Ra
nucleus.
the activity A= dN/dt = dN/dE x dE/dt = P/(dE/dN) = 1.00 x 103 / 6.35 x 10-12 = 1.57 x 1014 s-1
b.
the mass of radium required to produce this power. (Take one year = 3.15 x 107 s) A=
N =( ln 2 /t1/2) (NA M/Mm)
1.57 x 1014 s-1 = (ln2/1622 x 3.15 x 107) (6.02 x 1023 x M/ 0.226) M = 4.34 kg Define the decay constant of a radioactive isotope.
11 (a)
The decay constant λ of a radioactive isotope is the fraction of the total number of atoms that decay per unit time. (b)
A freshly prepared sample of an unknown radioactive isotope X−80 has an initial activity of 3.6 × 10 8 Bq. After 4.0 h, the activity drops to 4.5 × 10 7 Bq. (i) Calculate the decay constant of the radioactive isotope. Using A = Ao e − λt , 4.5 × 10 7 = 3.6 × 10 8 × e − λ ×4×3600 λ ≈ 1.4 × 10 −4 s–1
(ii)
Deduce the number of radioactive nuclei present initially in the freshly prepared sample. Since A = λN , 3.6 × 10 8 = 1.4 × 10 −4 × N N ≈ 2.5 × 1012 nuclei
(iii)
Given that the density of the sample is 2.53 g cm–3, calculate the volume of the sample. M , V 2.5 × 10 12 × 80 × 1.66 × 10 −27 × 10 3 2.53 = V 3 −10 V ≈ 1.3 × 10 cm
Since ρ =
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12
(a)
Radioactive decay is a random and spontaneous process. Explain what is meant by random and spontaneous. 1. Random – do not know when or which particular nucleus will decay. 2. Spontaneous – not affected by external factors
(b)
Fig. 12.1 shows a simple experiment set up by a student to estimate the activity of a radioactive source. The source emits α and β -particles and is placed 10 cm from a detector that is connected to a counter. The detector is capable of detecting all types of radioactive emissions. counter detector
10 cm source
Fig. 12.1 (i) Explain why the detector does not detect the emission of α -particles in the above set-up. Alpha particles are highly ionizing and will not penetrate far in air. (ii) Without the source, the counter gives a count-rate of 600 min−1. When the source is placed 10 cm in front of the detector, the following count rates are observed at two different times t.
t / hour
Count-rate / min−1
0
7009
6.0
1401
Find the time t at which the observed count-rate falls to 650 min−1.
True count = observed count – background count t / hour 0 6.0
A ' C ' 801 1 = = = A0 C0 6409 2 lg
801 1 = n lg 6409 2 6.0 hrs = 3.0 t1/2
n
n = 3.0
True count-rate / min-1 6,409 801 n'
50 1 = 801 2 50 1 lg = n 'lg n '= 4.0 801 2 t = 4.0 × 2.0 + 6.0 = 14.0 hrs (Ans)
∴ t1/ 2 = 2.0hrs
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Extra Question: Radioactive Decay Data Analysis Question: TYS [N98/II/9] (a) (i) A photon is a ‘packet’ or quantum of electromagnetic radiation. (ii) A parallel beam is used to ensure that the radiation travels straight and incident normally on the absorber so that none will be deflected and reflected from other surfaces to bypass the absorber to get to the detector. (iii) As the thickness is increased, the count-rate approaches to a very small value but never reaches zero.
(b) (i) let graph be ln
CX = m ' x (comparing to y = mx + c) ; y-intercept =0, C0 gradient = m’
C ln X = x C0 taking exp on both sides:
Cx = em ' x Co C x = Co e m ' x Comparing it with C x = Co e − µ x , gradient m’ = - µ
(ii) Gradient of the graph (Fig 4.4) =
Since gradient m’ = - µ =
µ = 0.444 cm -1
−4 − 0 −4 -1 = cm 9.0 − 0 9
−4 9
(c) (i) the unit for µ m = unit of µ / unit of ρ = (cm-1)/g cm-3) = cm2g-1 (ii) Complete the table for lead: µ of lead = 0.444 cm -1 (from bii)
µm of lead = 0.444/11.3 = 0.039 (d) (i) Average value = (0.035 + 0.037 + 0.039) /3 = 0.037 cm2g-1 Therefore approximate µ for concrete = µm ρ = (0.037)(2.4) = 0.0888 = 0.09cm-1
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(ii) For 4.0cm lead, from the graph Fig 43,
Given
Cx = 0.16 Co
Cx = e− µ x Co
For concrete; 0.16 = e− µ x where µ =0.09cm-1 ln 0.16 = - µ x ln 0.16 = (0.09)x x =20.4 = 20 cm (e) From answer in d(ii), the thickness of concrete required for shielding is 5 times more than lead for the same degree of shielding. This shows that lead , per unit thickness is more effective than concrete for radioactive shielding. However concrete is often used for shielding because it can be much more easily made into a fluid mixture to be molded and used to encase the radiaoactive source before solidifying (as in the Chernobyl disaster). In order to melt lead to encase the radioactive source, a very high temperature is required. Concrete is also a poor thermal conductor compared to lead. Hence the heat generated by the radioactive source can be contained within it more easily. Lead is a good thermal conductor and the heat transmitted through it by the radioactive source may result in fire hazard.
Assignment Solution 14. A nuclear explosion occurs as a result of the rapid release of energy from an uncontrolled nuclear chain reaction. The driving reaction may be nuclear fission, nuclear fusion or a multistage cascading combination of the two. The "Trinity" test was the outcome of the Manhattan Project and the first ever test of a nuclear weapon, conducted by the United States on July 16, 1945 at what is now White Sands Missile Range, headquartered near Alamogordo, New Mexico. It was a test of an implosion-design fission bomb with a 6.2 kg plutonium core, in which the plutonium core was rapidly compressed by high explosives to achieve super-criticality. This was the same type of weapon later dropped on Nagasaki, Japan. The picture below is obtained using high speed photography of the Trinity test shot 0.025 s after detonation. The "bubble" seen represents the wavefront of the advancing blast wave.
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280 m
High speed photograph of Trinity test shot after detonation The yield or total energy released by a nuclear explosion can be estimated from the blast radius at a known time using the formula:
cρR 5 yield, E = t2
equation (1)
where E is the energy (in joules), t is the time (in seconds), is the ambient density of air (in kg m-3) and R is the blast radius (in metres), and c is a constant of 1.033 in air. The yield is often expressed as the equivalent tonnage of conventional explosive TNT detonated instead of in joules. The majority of the energy released is in the form of the blast wave and thermal radiation. One of the delayed hazards of a nuclear explosion is the radioactive fallout, named from the fact that it "falls out" of the atmosphere into which it is spread during the explosion. It commonly refers to the radioactive dust created when a nuclear weapon explodes. Examples of these radioactive isotopes include relatively short-lived iodine-131 or long-lived strontium-90 and caesium-137. These can cause serious illnesses to humans if they are exposed to fallout through external radiation exposure as well as internal hazards due to inhalation and ingestion of radiocontaminants. Additional data is provided below: Density of air : 1.00 kg m-3 Energy equivalent of 1 kiloton of TNT detonated : 4.184 x 1012 J Mass of plutonium-239 atom, 239 94 Pu : 239.052 u Mass of strontium-90 atom, Half-life of strontium-90, Half-life of caesium-137, Half-life of iodine-131,
90 38
Sr
: 89.908 u
90 38
Sr : 28.90 years (decay mode : β-) 137 55 Cs : 30.07 years (decay mode : β -) 131 53
I
: 8.04 days (decay mode : β-)
(a) Write down the number of nucleons, neutrons and protons in plutonium-239. Page 17 of 5
[1]
239 nucleons, 145 neutrons, 94 protons [1]
(b) (i)
Estimate the yield of the Trinity nuclear explosion, in kiloton TNT equivalent, using equation (1) and other information from the passage. [2]
E=
(
(1.033)(1.00) 280 0.0252
)
5
2 = 8.8892 x 1013 = 8.89x1013 J (3s.f.)
[1]
In kiloton TNT equivalent = (8.8892x1013) ÷ (4.184x1012) = 21.2 kilotons TNT [1]
(ii) Calculate the amount of mass converted to energy. E = mc2 m=
[2] [1]
8.8892 × 10
13
(3.00 × 10 )
8 2
= 9.88 x 10-4 kg (3s.f.)
[1]
(iii) The efficiency of a fission weapon is the fraction of the fissile material (plutonium) that actually fissions. Given that the efficiency of the weapon used in the “Trinity” test was 14%, calculate the mass of the plutonium core that underwent fission. [1] (6.2 kg) x (0.14) = 0.868 kg
[1]
(iv) Compare the answers to part (ii) and (iii) and comment on why they are different.
[1]
Part (ii) value > Part (iii) value. The value to Part (ii) is the mass equivalent of the total energy released, while the mass in Part (iii) is the mass of the Plutonium that fissions to release the energy, of which some is converted to energy and released while the rest constitute the mass of [1] the product nuclei/fission fragments (like 90Sr, 137Cs,131I)
Other comments Part (ii) also represents the INCREASE in total mass defect/binding energy of the nuclei that undergo fission.The main difference bet (iii) and (ii) value is due to the total mass of the product nuclei/fission fragments. i.e. (iii) - (ii) = mass of fission fragments (just saying "mass defect" or "the binding energy" insufficient, as it does not represent the total mass defect/binding energy of the Pu that fissions)
(v) Using your answer to (iii), determine how many plutonium-239 nuclei underwent fission. [2]
0.868 = 2.1873 x1024 = 2.19 x 1024 (3s.f.) (239.052) × 1.66 × 10− 27
(
)
[2]
(vi) Hence, calculate the energy in MeV released per fission of plutonium-239.
8.8892 × 1013 ÷ 1.60 × 10 −19 =2.5399x109 eV = 254 MeV (3s.f.) 24 2.1873 × 10
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[2]
(c) Assuming equal numbers of caesium-137, iodine-131 and strontium-90 nuclides, explain briefly which will have the highest activity. [2] As A = λN, and λ = (ln 2)/t½. Since N is assumed the same initially, then λ ∝ 1/t½. Thus the shortest half-life translates into the highest activity for the same number of radioactive nuclei present. Thus, assuming equal numbers of each nuclide present, iodine131 will have the highest activity. [2]
(d) Write an equation to represent the radioactive decay of strontium-90. You may use the symbol X to represent the daughter nucleus after the decay. [1] 90 38
(e) (i)
Sr →3990X + −10e
[1]
Calculate the decay constant λ of iodine-131.
[1]
λ = (ln 2)/t½ = (ln2)/(8.04) = 0.0086212 = 8.62x10-2 day-1 (3 s.f.)
(ii)
Iodine-131 is extracted from a sample of rock collected near a nuclear test explosion. If the activity is measured to be 6.7 x 1012 Bq, determine the number of iodine-131 atoms present in the rock sample. [2] N = A/λ
[1]
= (6.7 x 1012)/
(iii)
[1]
ln 2 ÷ [24 × 60 × 60] s-1 = 6.7146x1018 = 6.71 x 1018 s-1 (3s.f.) [1] 8.04
How long will it take for the activity of the above mentioned sample to fall to 4.3 x 1011 Bq? Give your answer in days. [2] A = Aoe-λt
ln t=
Ao
λ
A
ln =
6.7 × 1012 4.3 × 1011 = 31.8525 = 31.9 days 0.086212
[2]
(f) Suggest why exposure to radioactive fallout can be hazardous to one’s health.
[1]
The radiation/radioactive emissions given off by the fallout (beta particles, in the case of the 3 radionuclides mentioned in the passage) ionize matter it encounters. This will affect cells/tissue, which might lead to organ failure and death. If the radiation damages DNA, the mutations can result in cancers/tumours. [1] Other comments The harmful property of the radiation given off is ionization, not penetration (save for neutrons). This is especially so if it is inhaled or ingested. Alpha emitters ingested are extremely dangerous in the body even though they are not penetrating but highly ionizing. Beta emitters being both penetrating and ionizing and can cause significant damage from inside and outside the body. Gamma radiation on the other hand is so penetrating that a significant portion will pass through the human body without interaction. See http://www.fas.org/nuke/guide/usa/doctrine/dod/fm8-9/1ch5.htm
(HCI Prelim 2006, P2, Q8) Page 19 of 5
