# Physics Heinemann 11 Worked Solutions

September 12, 2017 | Author: Grief Egg | Category: Radioactive Decay, Neutron, Atomic Nucleus, Nuclear Physics, Isotope

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Worked solutions Chapter 1 Nuclear physics and radioactivity 1.1 Atoms, isotopes and radioisotopes 1

In each case the mass number A is given. A = the number of nucleons = the number of (protons + neutrons) in the nucleus. The number of protons in each nucleus can be established from the periodic table. a 20 protons, 25 neutrons, 45 nucleons b 79 protons, 118 neutrons, 197 nucleons c 92 protons, 143 neutrons, 235 nucleons d 90 protons, 140 neutrons, 230 nucleons

2

a b c

3

A radioisotope is an unstable isotope. At some time, it will spontaneously eject radiation in the form of alpha particles, beta particles or gamma rays from the nucleus. Three isotopes that are not radioisotopes could be any three stable isotopes, e.g. carbon-12, lead-206 and bismuth-209.

4

Yes, a natural isotope can be radioactive. For example, every isotope of uranium is radioactive.

5

Polonium-210 and uranium-238. These have atomic numbers of 84 and 92 respectively; and every isotope beyond bismuth (Z = 83) in the periodic table is radioactive.

6

a

27 protons, 33 neutrons 94 protons, 145 neutrons 6 protons, 8 neutrons

4 π(1.07 3

× 10–15)3 = 5.13 × 10–45 m3

i

V=

ii

Density = mass/volume

1.67 "10 !27 = 3.25 × 1017 kg m–3 ! 45 5.13 "10 Mass = density × volume = (3.25 × 1017)(1.0 × 10–6) = 3.25 × 1011 kg (325 million tonnes) =

b c d

3.25 ! 1011 = 3.25 × 108 (325 million cars) 3 1.0 ! 10

The density of normal matter is many orders of magnitude lower than that of an atomic nucleus.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

7

Chapter 1: Nuclear physics and radioactivity

Volume of nucleus = Volume of atom = Required ratio =

4 π(6.2 3

4 π(1.3 3

× 10–15)3 = 1.0 × 10–42 m3

× 10–10)3 = 9.2 × 10–30 m3

1.0 " 10!42 = 1.1 × 10–13 ! 30 9.2 " 10

volume of nucleus 1.1 " 10!13 !2 3 (1.0 "10 ) Then (volume of sphere occupied by electrons)3 = 1.1"10 !13 and radius of sphere occupied by electrons ≈ 210 m.

8

Volume of the sphere to be occupied by the electrons =

9

a b

Radioactive and nonradioactive isotopes of the same element will have identical chemical properties. There are no differences. 89 Kr has five more neutrons in its nucleus.

10 Because the radioisotope has an extra neutron in its nucleus, it will have a mass number of 28. The radioisotope produced will be 28Al.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

Chapter 1: Nuclear physics and radioactivity

1.2 Radioactivity and how it is detected 1

a b c

2

An alpha particle consists of two protons and two neutrons, and is identical to a helium nucleus. A beta particle is an electron that has been ejected from the nucleus. A gamma ray is very high frequency electromagnetic radiation that originates from the nucleus.

3

a b c d

beta particle proton alpha particle neutron

4

a

218 = A + 4 + 0, then A = 214 84 = Z + 2 = 0, and Z = 82, lead 235 – 4 – 0 = A, then A = 231 92 – 2 – 0 = Z, and Z = 90, thorium 228 = A + 0 + 0, then A = 228 88 = Z + (–1) + 0, and Z = 89, actinium 198 – 0 – 0 = A, and A = 198 79 – (–1) – 0 = Z, and Z = 80, mercury

b c d

nucleus nucleus nucleus

5

The identity of the unknown particle, or radiation, is determined by equating the mass numbers and atomic numbers on either side of the nuclear equation. a X = alpha particle b X = beta particle c X = beta particle d X = alpha particle e X = gamma ray

6

Equating atomic and mass numbers on either side of the nuclear equation shows that the new element X has an atomic number of 3, and mass number 7. Then X = lithium-7.

7

The identity of the unknown particle is determined in each case by balancing the nuclear equation. a X = proton b X = neutron c X = neutron d X = alpha particle

8

a b

7 protons, 7 neutrons, 1 electron, 1 antineutrino A neutron has decayed.

c

1 0n

d

! 11p + "01e + í (neutron → proton + beta particle + antineutrino) Kinetic energy carried by the beta particle, antineutrino and nitrogen-14 nucleus.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

9

a b c d e f g

10 a b

Chapter 1: Nuclear physics and radioactivity

40

Ca, 42Ca, 43Ca, 44Ca, 46Ca, 48Ca Niobidium has only one stable nuclide. Potassium-48 is a beta emitter. 48 K → 48Ca + beta particle. The daughter nucleus is unstable. Ratio of neutrons to protons for K = 29/19 = 1.53, and for Ca = 28/20 = 1.40. alpha emitter francium-217 → actinium-213 + alpha particle actinium-213 → bismuth-209 + alpha particle The stable nuclide is bismuth-209. 197

Au + neutron → 198Au 198 Au → 198Hg + beta particle

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

Chapter 1: Nuclear physics and radioactivity

1.3 Properties of alpha, beta and gamma radiation 1

a b

2

B. This type of radiation will lose less energy as it travels through matter.

3

The only suitable type of radiation for this application would be gamma radiation, as it is the only type of radiation with enough penetrating power to reach the site of the tumour.

4

The wire should be a beta emitter, since the irradiation needs to be confined to a relatively small area. Alpha radiation does not have sufficient penetrating power, while gamma radiation would irradiate adjacent healthy cells.

5

The lead apron is only needed for protection against beta and gamma radiation. The alpha particles would be absorbed by the air.

6

a b c

E = (8.8 × 106)(1.6 × 10–19) = 1.4 × 10–12 J E = (0.42 × 106)(1.6 × 10–19) = 6.7 × 10–14 J E = (500 × 103)(1.6 × 10–19) = 8.0 × 10–14 J

7

a

E = 34(1.0 × 105) = 3.4 × 106 eV = 3.4 MeV or E = (3.4 × 106)(1.6 × 10–19) = 5.44 × 10–13 J Distance = energy of alpha particle/energy lost per collision = 5.6 MeV/3.4 MeV = 1.65 cm

b

α, β, γ γ, β, α

8

D. Gamma radiation has by far the highest penetrating ability of any radiation. The higher the energy of the gamma radiation, the greater its penetrating power.

9

A. Alpha radiation is the most damaging to human tissue because it has the greatest ionising ability.

10 D. Alpha particles only travel a few centimetres in air before losing their energy, whereas beta particles can travel a few metres and gamma rays can travel almost an unlimited distance.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

Chapter 1: Nuclear physics and radioactivity

1.4 Half-life and activity of radioisotopes 1

C. After one half-life, (100 mg)/2 = 50 mg will remain. After two half-lives, (50 mg)/2 = 25 mg will remain.

2

a b c d

3

a

b c d

(20 g)/2 = 10 g (20 g)/4 = 5.0 g (20 g)/8 = 2.5 g 1.5 hours = 6.0 half-lives, then amount remaining = (20 g)/26 = 0.31 g

235 Bq The half-life is found by determining the time which corresponds to a count rate of (400 Bq)/2 = 200 Bq. From the graph; time = 20 minutes. 60 minutes corresponds to three half-lives. Then activity = (400 Bq)/23 = 50 Bq.

4

375/6000 = 1/16; therefore, 1 hour (60 minutes) would correspond to four half-lives, and one half-life = 15 minutes.

5

Probability = 0.5 The fact that the nucleus has not decayed during the previous half-life is irrelevant, since the two events are independent. The probability of a nucleus decaying during any particular half-life is always 0.5.

6

The time interval involved is equal to four half-lives. During this time the amount of material remaining will be equal to 1/16 of the original amount. To satisfy the requirement of the hospital, the amount that must be produced = 16 × 12 µg = 192 µg.

7

a b

8

a b

0.1% = 0.001, and 1/2n = 1/1000 n = 10 or 10 half-lives 10 × 24 000 years = 240 000 years Uranium-235 would have the greater activity, because the shorter half-life indicates that more particles are being emitted per second. It has a much shorter half-life than uranium-238, and so has decayed more rapidly since the formation of the Earth.

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Worked solutions

9

a b

10 a b

Chapter 1: Nuclear physics and radioactivity

From the activity graph, the time that the activity takes to halve from 800 Bq to 400 Bq is 10 minutes. This is the half-life. 40 minutes is 4 half-lives. The activity will fall from 800 → 400 → 200 → 100→ 50 Bq. Over time, the radioactive elements contained in the ore transmute by a series of decays to form lead-206, which is stable. The half-life of polonium-214 is so small (160 µs) that there would be almost none of this isotope found in the sample.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

Chapter 1: Nuclear physics and radioactivity

1.5 Radiation dose and its effect on humans 1

a b

A, B, C, E D, E

2

a b c

Dose equivalent = absorbed dose × quality factor = 0.50 × 10–6 × 20 = 10 µSv Dose equivalent = absorbed dose × quality factor = 0.50 × 10–6 × 1 = 0.50 µSv Dose equivalent = absorbed dose × quality factor = 0.50 × 10–6 × 1 = 0.50 µSv

3

a b

Dose equivalent = absorbed dose × quality factor = 200 × 1 = 200 µSv Absorbed dose = energy/mass Energy = absorbed dose × mass = 200 × 10–6 × 80 = 1.6 × 10–2 J

4

a b

B, the dose equivalent of 400 µSv is the largest of the three. A, the dose equivalent directly gives the degree of damage.

5

a

2 mSv = 2000 µSv, so at 1000 µSv per day it will take just 2 days to exceed this dose. 50 mSv = 50 000 µSv, so at 1000 µSv per day it will take 50 days to exceed this dose. 803×1000 × 10–6 Sv = 0.803 Sv = 803 mSv

b c 6

Minimise your exposure to medical procedures involving ionising radiation, keep air travel to a minimum, and live in a house with weatherboard walls near the equator.

7

The people working in the immediate vicinity of the reactor would have received massive doses of radiation, even exceeding 8.0 Sv.

8

a b

9

36 Gy h–1 = n(0.40 Gy h–1) n = 90 hours The beta particles have enough penetrating power to treat the tumour without irradiating adjoining healthy tissue. Also the half-life of the emitter must be long enough to ensure that the activity of the sample does not decrease too much during the course of the treatment.

D. Radioactive tracers need to emit radiation with a low quality factor so as not to damage healthy tissue. They should also have a relatively short half-life so as not to remain in the body for too long.

10 Effective dose = ∑(dose equivalent × W) = (35 × 0.20) + (35 × 0.05) + (50 × 0.12) = 15 mSv

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Chapter 1: Nuclear physics and radioactivity

Chapter review 1

a b

17 protons, 18 neutrons, 35 nucleons 88 protons, 138 neutrons, 226 nucleons

2

a b c d e f

X-rays, infrared radiation, microwaves, gamma rays alpha particle alpha particle beta particle α, β, γ, X-rays gamma radiation

3

Balance the nuclear equation in each case. a x = 82, y = 208 b x = 78, y = 176

4

a b c

β– α β+

5

Curve will be negative exponential in shape with vertical axis representing mass and horizontal axis representing time. When time = 0 hours, mass = 60 g, and when time = 8.0 hours, mass = 30 g. 6

a b c d

7

a b c

15 minutes/3 minutes = 5 half-lives then mass remaining = (120 g)/25 = 3.75 g 218 Po → 214Pb + α 214 Pb, 214Bi, 214Po Lead-214 has the longest half-life, which means the atoms of this isotope will remain in the sample for the longest time. 14 6

C !147 N + "01 e + í

80/20 = 4 per minute The activity of the carbon in the bone has dropped from 16 disintegrations per gram per minute to 4 disintegrations per gram per minute, i.e. by a factor of 4. This indicates that the age of the bone = 2 half-lives = 2(5730) ≈ 11 500 years.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Worked solutions

8

a b c d

9

Chapter 1: Nuclear physics and radioactivity

E = 140 × 103 × 1.6 × 10–19 = 2.2 × 10–14 J The half-life is the time at which the activity is half its initial value. t½ = 6 hours. Activity = (4.0 × 106)/2 = 2.0 × 106 Bq The time interval involved = 18 hours = three half-lives. Activity = (4.0 × 106)/23 = 0.5 × 106 Bq

a

24 11

b c

Gamma radiation can travel much greater distances in air than can beta particles. The beta particles would certainly be stopped, even by a steel box of minimal thickness. In order to completely stop the gamma radiation the steel would need to be several centimetres thick. Kinetic energy is carried by the gamma ray, the beta particle and the magnesium-24 nucleus.

d

Na "

24 12

Mg + #01 e + !

10 The woman’s blood cells have been affected. This is a somatic effect. 11 Balance the nuclear equation. The mass number has decreased by 28, so there must be seven alpha decays. This will also decrease the atomic number by 14. The equation shows that atomic number has decreased by 14, so there must also be four beta decays. 12 a b c d

(6.0 × 1010)/2 = 3.0 × 1010 atoms (6.0 × 1010)/4 = 1.5 × 1010 atoms (6.0 × 1010)/8 = 7.5 × 109 atoms (6.0 × 1010)/64 = 9.4 × 108 atoms

13 D. The amount remaining = 60/22 = 60/4 = 15 g. 14 Balance the nuclear equation. The mass number of the stable isotope = 10 – 4 = 6, while the atomic number = 5 – 2 = 3. The isotope is lithium-6. 15 a b

The isotopes are identical, so their half-lives must be the same. The activity of A will always be twice that of B.

16 a b

A proton has transformed into a neutron, a positron and a neutrino. Kinetic energy is carried by the neutrino, the positron and the carbon-12 nucleus.

17 Let the number of decays per second = n. Then 5 × 106 W = n × 1.0 × 106 × 1.6 × 10–19 J so n = 3.1 × 1019 Bq 18 a b

Assuming 250 working days per year; annual dose = 10 × 250 × d = 7500 µSv. So worker receives 3 µSv per photograph. This represents about 4 times background radiation.

19 Effective dose for patient A = ∑(dose equivalent × W) = (5000 × 0.12) + (4000 × 0.12) = 1080 µSv This is greater than the dose received by patient B, so A is at a slightly greater risk of developing cancer from this exposure.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Area of study review: Nuclear physics and radioactivity 1

Beta particles are electrons that are emitted from an unstable nucleus. Beta decay occurs when one of the nucleons decays.

2

Beam ii is gamma radiation since it is undeflected by magnetic fields. Beam i is alpha particles and beam iii is beta particles. Beta particles are much lighter and so are deflected more strongly.

3

B. A neutron decays into a proton, electron and antineutrino.

4

a b c

nucleus nucleus nucleus

5 Time (Gyears)

No. of K nuclei

0 1.3 2.6 3.9

No. of Ar nuclei

1000 500 250 125

0 500 750 875

Ratio K:Ar 1:1 1:3 1:7

6

The isotopes have identical chemical properties.

7

Each atom of gold-198 has one more neutron in its nucleus, so this isotope will be slightly more dense than the other isotope.

8

7 3 Li

+ 11 H → 2( 42 He )

9

a b c

E = 1.4 × 10–12/1.6 × 10–19 = 8.8 × 106 eV = 8.8 MeV E = 6.7 × 10–14/1.6 × 10–19 = 4.2 × 105 eV = 0.42 MeV E = 8.0 × 10–14/1.6 × 10–19 = 5.0 × 105 eV = 0.50 MeV

10

11 From graph, after 13 minutes, activity is about 320 Bq. 12 Find the time at which activity has been reduced from 800 Bq to 400 Bq: t½ ≈ 10 min. 13 From graph, extrapolate to find activity when t = 30 min. Activity ≈ 100 Bq. Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Area of study review: Nuclear physics and radioactivity

14 B. The probability of a nuclide decaying during the next half-life is always 0.50. 15 Gamma rays are uncharged and travel at the speed of light. This means that they interact with matter infrequently when they collide directly with a nucleus or an electron. The low density of atoms makes this a relatively unlikely occurrence, and so gamma radiation passes through matter very easily and has a low ability to ionise atoms. 16 Alpha particles travel through air at a relatively low speed and have a double positive charge. They interact with just about every atom that they encounter and have enough energy to ionise these. They lose their energy very rapidly and so have a very poor penetrating ability. 17 Energy form each decay = 500 keV = 500 × 103 × 1.6×10–19 = 8.0 × 10–14 J Total energy per second = 8.0 × 106 J Number of decays each second = 8.0 × 106/8.0 × 10–14 = 1.0 × 1020 Bq 18 1 p, 2 n 19 By balancing the nuclear equation, it can be determined that the daughter nuclide has an atomic number of 2 and so is helium. 20 B. A neutron in the hydrogen atom has transformed into a proton and an electron. 22

185 79

Au →

181 77

Ir + 42 !

23 79 p, 106 n 24 No, the alpha particles would not penetrate through the dead cells into the living skin cells in your hand. 25 Gamma rays have the ability to penetrate the skull and pass through the target region. The half-life of 5.3 years means that the source would retain its activity for a long time and would not need replacing very often. 26 Dose equivalent = Absorbed dose × quality factor = 300 × 1 = 300 µSv 27 The time for the activity to reduce from 4000 Bq to 1000 Bq is 120 s. The half-life is 60 s. 28 5 minutes is 5 half-lives. The radioactive sample will reduce by a factor of 25 = 32. Remaining mass = 150/32 = 4.7 g. 29

26 11

0 Na # 26 12 Mg + \$1 " + !

30 Symptoms would include radiation sickness and hair loss. 31 The testes are the most radiosensitive organs. 32 C 33 Effective dose = ∑(dose equivalent × W) = (6000 × 0.12) + (4000 × 0.0.20) = 1520 = 1.5 × 103 µSv 34 These bombarding electrons may not be the product of radioactive decay.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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Area of study review: Nuclear physics and radioactivity

35 The electrons would be repelled by the electron clouds of the atoms. B; the radioisotope with the shorter half-life (Bi-211) is less stable by a factor of four and so will initially have a higher activity by a factor of four. The activity of Bi-211 is 4 times greater at the start, but after 8 minutes (4 half-lives) its activity is reduced by a factor of 16. After 8 minutes the activity of Bi-215 has fallen by half and so it has double the activity of Bi-211 at this time. 38 a b c

Absorbed dose (Gy) = energy(J)/mass (kg) Energy = absorbed dose × mass = 5.0 × 75 = 375 = 380 J Dose equivalent = Absorbed dose × quality factor = 5.0 × 1.0 = 5.0 Sv This massive dose would result in nausea, hair loss, rashes, the development of cancers later in life and possibly death.

39 There are 2500 X-rays taken. The dose from each one = 0.03/2500 = 1.2 × 10–5 Sv Dose = 12 µSv 40 0.03 Sv / 2000 Sv = 15 This is about 15 times greater than the background radiation dose of 2000 µSv.

Heinemann Physics 11 (3e) Teacher’s Resource and Assessment Disk ISBN 9781442501249

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