Solutions Manual for College Physics A Strategic Approach, 4e Randall Knight, Jones, Gield (All
Chapters 1-30, 100% Original Verified, A+ Grade) All Chapters Arranged Reverse: 30-1
Work Books Answers Also Included from page 1125-1435
NUCLEAR PHYSICS
30
Q30.1. Reason: No. While the general trend is that larger atomic masses correspond to larger atomic numbers,
there are many exceptions and it would not be a good idea to count too heavily on the generalization. Most elements
have a number of different isotopes (atoms with the same Z -values but different A-values); and it is often the case
that a heavier isotope of one element may have more mass than a lighter isotope of an atom with a slightly higher
atomic number. For example, 59 Co has a larger atomic mass than 58 Ni, but a smaller atomic number.
Assess: You can find other exceptions in tables or charts of the nuclides.
Q30.2. Reason: The system for assigning the mass of an atom is set up such that the 12
C isotope of carbon is
exactly 12 u. Thus 1/12 of the C mass is exactly 1 u. The mass of the hydrogen atom is greater than this.
12
Assess: Using this system, the proton, neutron, and hydrogen atoms all have masses within 1% of 1 u—which is
why the atomic mass unit is so convenient.
Q30.3. Reason: The element type is determined by the number of protons, which is different for the two atoms.
Thus, these are nuclei of different elements. Specifically, the elements are boron (5 protons) and carbon (6 protons).
Assess: If the nuclei had the same number of protons, but different numbers of neutrons, then they would have been
isotopes of the same element.
Q30.4. Reason: (a) The nucleus 3 He has three nucleons (two protons and one neutron) with a binding energy of
2.5 MeV/nucleon and hence a binding energy of
B = (2.5 MeV/nucleon)(3 nucleons) = 7.5 MeV
(b) The nucleus 6 Li has six nucleons (three protons and three neutrons) with a binding energy of 5.0 MeV/nucleon
and hence a binding energy of
B = (5.0 MeV/nucleon)(6 nucleons) = 30 MeV
(c) Yes. The total binding energy of two 3 He atoms is 15 MeV. If those two nuclei fused to form a 6 Li atom, it
would have a total binding energy of 30 MeV. A higher overall binding energy implies a more tightly bound nucleus
with an overall lower energy. So this process could occur, with the release of about 15 MeV of energy, perhaps as
gamma rays.
(d) No, at least not without putting in a great deal of energy. It could not happen spontaneously, because the 6 Li
atom is more tightly bound than the two 3 He atoms. Energy would have to be put in to elevate the nuclei in energy.
Assess: The binding energy is the product of the binding energy per nucleon and the number of nucleons.
Q30.5. Reason: This is because the isotopes have the same electronic structure and will therefore emit practically
the same spectrum. However, they do have different masses, which is a feature that can be detected in a mass
spectrometer.
Assess: The emitted light from an atom, measurable with an ordinary spectrometer, tells you about its electrons. The
mass, measurable with a mass spectrometer, tells you about the nucleus (since the nucleus has most of the mass of an
atom).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
30-1
,30-2 Chapter 30
Q30.6. Reason: The heating of the material would have no effect on the abundance of the various isotopes.
Heating the sample to 3695 K does not provide anything close to the energy required to produce changes in nuclear
structure.
Assess: One could induce nuclear changes by bombarding the sample with such enormously high energy radiation
(possibly light, or possibly alpha particles) that it would break apart some nuclei.
Q30.7. Reason: It is clear from the graph that 1/4 of the original nuclei are left after 8 days. In general, 1/4 are left
after two half-lives; therefore two half-lives equals 8 days, so one half-life must be 4 days.
Assess: There is no need to do any fancy math on this question. We can see that there would be 500,000 nuclei left
when t = 4 d.
Q30.8. Reason: We can apply Equation 30.16 to relate the current activity to the old activity:
R = R0e-t / t = R0e-t / (1.44) t1/2
R0 = Ret /(1.44) t1/2 = (100 mCi)e (6 days)/ (1.44(3 days)) = 400 mCi
Assess: 6 days ago was two half lives ago. Thus, the amount of radioactive material must have been four times as
great 6 days ago. Since the activity depends on the amount present, we expect the activity should have been four
times as large 6 days ago. Our answer makes sense.
Q30.9. Reason: We must realize that when there are 750 B nuclei there are 250 A nuclei. Clearly there are 1/4 of
the original A nuclei left, and that takes two half-lives, so t = 2 ´ 10 s = 20 s.
Assess: We can also do this mathematically by starting with Equation 30.12 and solving for t.
t / t
æ1ö
N = N0 ç ÷
è2ø
t / t
N æ1ö
=
N 0 çè 2 ÷ø
t æ N ö
= log1/ 2 ç ÷
t1/ 2 è N0 ø
æ N ö
t = (t1/ 2 )log1/ 2 ç ÷
è N0 ø
Use the change-of-base formula.
t = (t1/ 2 )
ln ( ) = (10 s) ln ( ) = 20 s
N
N0
250
1000
ln ( 12 ) ln ( 12 )
Rather than use the change-of-base formula, you could also realize that you need to raise 1/2 to the second power to
get 1/4 so that log1/ 2 (250/1000) = 2.
Q30.10. Reason: As stated in the question, the sample has a half-life of 1.0 min and hence each nucleus has a 50%
probability of undergoing decay in the next minute. As the nucleus ages, the half-life does not change. After 15 min
the half life is still 1.0 min and each nucleus has a 50% chance of decaying in the next minute.
Assess: The probability that a nucleus will decay in the next unit of time is not a function of how long the nucleus
has survived—it depends only on the half-life.
Q30.11. Reason: Because there is more 238
U less of it has decayed, so it must have a longer half-life.
Assess: This can be verified by looking up the two half-lives.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Nuclear Physics 30-3
Q30.12. Reason: The quantum mechanical probability that a particle will decay does not depend on its history.
They are equally likely to decay in the next 66 h.
Assess: This problem is related to fundamentals of probability. If you flip a fair coin and obtain heads 10 times, it is
not more likely to turn up tails in the next attempt than a fresh fair coin. This is the because the number of ways of
flipping 10 heads and one tails is the same as the number of ways of flipping 11 heads in a row. So the chances for
either the fresh fair coin or the one that has been flipped 10 times to come up heads is 50%.
Q30.13. Reason: The half-life of 14 C is 5730 years. After ten half-lives one thousandth of the nuclei are left:
10
æ1ö
N = N 0 ç ÷ = 9.8 ´ 10 -4 N 0 » 10 -3 N 0
è2ø
After 10 half-lives or 10t1/ 2 = 10(5730 y) = 57,300 yr one thousandth of the original amount of 14
C is still in
existence. The fossil fuel has been buried for more than 57,300 years.
After 20 half-lives one millionth of the nuclei are left:
20
æ1ö
N = N 0 ç ÷ = 9.5 ´ 10 -7 N 0 » 10 -6 N 0
è2ø
After 20 half-lives or 20t1/ 2 = 20(5730 y) = 114 600 y there is still some 14 C, the fossil fuel has been buried for
more than 114,600 years. In like manner, after 50 half-lives or 286 500 y the number of nuclei left is 10-16 N 0 . We
could continue this process until the number of nuclei of 14 C is so small we agree none exists. The point is that the
fossil fuel has been buried for a long time, in fact so long that no 14 C remains in the fuel.
Assess: The age of the earth is greater than the time it has taken the 14 C to totally decay (none is left).
Q30.14. Reason: Suppose 14
C was less abundant 10,000 years ago than today. If scientists are unaware of this
fact, they will assume it has always had the same abundance. When they compute the age of the artifact, they will
measure a very small 14 C concentration and will assume it is because the artifact is very old. Thus they will
overestimate the age. Their mistake is that the low concentration is really due to a low concentration in living things
10,000 years ago.
Assess: Details like this complicate radiocarbon dating, but you should be aware that scientists are aware of these
issues and account for them. Furthermore, there are other methods of dating things, and they all act as a check and
balance on each other, giving a very good consensus.
Q30.15. Reason: The unknown X may be identified as follows:
(a) 222
86 Rn ® 218
84 Po + 2 X ; X must be an alpha particle, which is a He atom nucleus 2 He.
4 4
(b) 228
88 Ra ® 228
89 Ac + 0
-1 X ; X must be an electron or beta-minus radiation.
(c) 140
54 Xe ® 140
55 Cs + 0
-1 X ; X must be an electron or beta-minus radiation.
(d) Cu ® Ni + X ; X must be a positron or beta-plus radiation.
64
29
64
28
0
1
Assess: The equation is balanced by conserving mass and charge.
Q30.16. Reason: (a) No. The alpha particle is added to the wrong nucleus. Th + a would work.
236
92 U® 232
90
(b) No. The original nucleus would need to lose four nucleons, not two.
(c) No. In beta-minus decay, the atomic mass numbers of the parent nucleus and of the daughter nucleus must be
equal. Here A has decreased by one.
Assess: Each reaction must be examined to see that there are the same number of nucleons on each side of the arrow,
and that charge is conserved.
Q30.17. Reason: (a) This nucleus is in an excited state and as it decays the proton in the higher level will drop
down to the lower level. The number of protons and neutrons will remain the same and the nucleus will get rid of the
excess energy by emitting a gamma ray.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 30-4 Chapter 30
(b) This nucleus is in a stable unexcited state.
(c) This nucleus is in an excited state and as it decays the neutron in the higher level will drop down to the empty
lower level proton state. That is, a neutron will become a proton and an electron as follows: n ® p + -10e + energy.
During this process a beta-minus is emitted.
Assess: Nuclear energy–level diagrams provide a lot of information about the nucleus.
Q30.18. Reason: Just as with filled electron shells in chemistry, if the nuclear shells are just filled then the nucleus
is stable and “happy.” For 4 He the n = 1 shell is exactly filled for both protons and neutrons (hence it is doubly
magic), with two each. Likewise, for 16 O the n = 1, 2, 3 shells are all exactly filled for both protons and neutrons,
since the n = 1 shell can accommodate two each, the n = 2 shell can accommodate four each, and the n = 3 shell can
accommodate another two each.
Assess: 12 C and 16 O can be thought of as being made of three and four 4 He nuclei (or alpha particles) respectively.
These nuclei are indeed made that way in stars through the process of nuclear fusion.
Q30.19. Reason: As you radiate the apple, you are putting energy into it. This energy will break molecular bonds
and cause the molecules to vibrate. These vibrations will cause the temperature of the apple to rise. After some time,
the apple will appear almost as if it had been heated in the oven.
Assess: The process is somewhat like a microwave oven. Microwaves are just like gamma rays except for the
amount of energy.
Q30.20. Reason: The Note in the text says, “ionizing radiation causes damage to materials, but objects irradiated
with alpha, beta, or gamma radiation do not become radioactive.” So the patient’s body does not itself become
radioactive.
Assess: However, if radioactive materials are ingested or inserted into the body, they will remain radioactive in the
body.
Q30.21. Reason: In a gamma scan, phosphorus-containing molecules with an attached 99
Tc atom (a gamma
emitter) are administered to the patient. The gamma emitting molecules are absorbed preferentially by the tumor so
that the tumor is highlighted. In an x-ray, the tumor may not show up because it may look like the neighboring
tissues. In this way, the gamma scan actually shows bodily function and not just structure.
Assess: X-ray images only show structure because they merely show whether an x-ray was able to penetrate the
bodily tissue or not. By contrast the gamma scan shows which parts of the body are using certain molecules.
Q30.22. Reason: (a)
Sample Dose in Gy Absorbed Energy in J/kg RBE Dose eq in Sv
A 0.1 0.1 1 0.1
B 0.2 0.2 1 0.2
C 0.1 0.1 2 0.2
D 0.2 0.2 2 0.4
The radiation dose is given in gray, and the gray tells us the energy in J absorbed per kg of absorbing material.
(1 Gy = 1.00 J/kg of absorbed energy). So the dose with the greatest number of Gy has received the most energy.
The samples are rated for energy absorbed as follows: B = D > A = C.
(b) The biological damage is assessed by looking at the dose equivalent in Sv. The dose equivalent in Sv = dose in
Gy ´ RBE. The samples are rated for biological damage as follows: D > B = C > A.
Assess: In order to determine the biological damage, the relative biological effectiveness must be taken into
consideration.
Q30.23. Reason: The dose equivalent (in Sv) is the dose (in Gy) multiplied by the RBE. Two different sources of
radiation could give the same dose equivalent (in Sv) but have different RBE if they give different doses (in Gy). So
the two sources do not necessarily have the same RBE.
Assess: For example, alpha particles can deliver a high dose equivalent even from a low-dose source because its
RBE is so high.
© Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapters 1-30, 100% Original Verified, A+ Grade) All Chapters Arranged Reverse: 30-1
Work Books Answers Also Included from page 1125-1435
NUCLEAR PHYSICS
30
Q30.1. Reason: No. While the general trend is that larger atomic masses correspond to larger atomic numbers,
there are many exceptions and it would not be a good idea to count too heavily on the generalization. Most elements
have a number of different isotopes (atoms with the same Z -values but different A-values); and it is often the case
that a heavier isotope of one element may have more mass than a lighter isotope of an atom with a slightly higher
atomic number. For example, 59 Co has a larger atomic mass than 58 Ni, but a smaller atomic number.
Assess: You can find other exceptions in tables or charts of the nuclides.
Q30.2. Reason: The system for assigning the mass of an atom is set up such that the 12
C isotope of carbon is
exactly 12 u. Thus 1/12 of the C mass is exactly 1 u. The mass of the hydrogen atom is greater than this.
12
Assess: Using this system, the proton, neutron, and hydrogen atoms all have masses within 1% of 1 u—which is
why the atomic mass unit is so convenient.
Q30.3. Reason: The element type is determined by the number of protons, which is different for the two atoms.
Thus, these are nuclei of different elements. Specifically, the elements are boron (5 protons) and carbon (6 protons).
Assess: If the nuclei had the same number of protons, but different numbers of neutrons, then they would have been
isotopes of the same element.
Q30.4. Reason: (a) The nucleus 3 He has three nucleons (two protons and one neutron) with a binding energy of
2.5 MeV/nucleon and hence a binding energy of
B = (2.5 MeV/nucleon)(3 nucleons) = 7.5 MeV
(b) The nucleus 6 Li has six nucleons (three protons and three neutrons) with a binding energy of 5.0 MeV/nucleon
and hence a binding energy of
B = (5.0 MeV/nucleon)(6 nucleons) = 30 MeV
(c) Yes. The total binding energy of two 3 He atoms is 15 MeV. If those two nuclei fused to form a 6 Li atom, it
would have a total binding energy of 30 MeV. A higher overall binding energy implies a more tightly bound nucleus
with an overall lower energy. So this process could occur, with the release of about 15 MeV of energy, perhaps as
gamma rays.
(d) No, at least not without putting in a great deal of energy. It could not happen spontaneously, because the 6 Li
atom is more tightly bound than the two 3 He atoms. Energy would have to be put in to elevate the nuclei in energy.
Assess: The binding energy is the product of the binding energy per nucleon and the number of nucleons.
Q30.5. Reason: This is because the isotopes have the same electronic structure and will therefore emit practically
the same spectrum. However, they do have different masses, which is a feature that can be detected in a mass
spectrometer.
Assess: The emitted light from an atom, measurable with an ordinary spectrometer, tells you about its electrons. The
mass, measurable with a mass spectrometer, tells you about the nucleus (since the nucleus has most of the mass of an
atom).
© Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
30-1
,30-2 Chapter 30
Q30.6. Reason: The heating of the material would have no effect on the abundance of the various isotopes.
Heating the sample to 3695 K does not provide anything close to the energy required to produce changes in nuclear
structure.
Assess: One could induce nuclear changes by bombarding the sample with such enormously high energy radiation
(possibly light, or possibly alpha particles) that it would break apart some nuclei.
Q30.7. Reason: It is clear from the graph that 1/4 of the original nuclei are left after 8 days. In general, 1/4 are left
after two half-lives; therefore two half-lives equals 8 days, so one half-life must be 4 days.
Assess: There is no need to do any fancy math on this question. We can see that there would be 500,000 nuclei left
when t = 4 d.
Q30.8. Reason: We can apply Equation 30.16 to relate the current activity to the old activity:
R = R0e-t / t = R0e-t / (1.44) t1/2
R0 = Ret /(1.44) t1/2 = (100 mCi)e (6 days)/ (1.44(3 days)) = 400 mCi
Assess: 6 days ago was two half lives ago. Thus, the amount of radioactive material must have been four times as
great 6 days ago. Since the activity depends on the amount present, we expect the activity should have been four
times as large 6 days ago. Our answer makes sense.
Q30.9. Reason: We must realize that when there are 750 B nuclei there are 250 A nuclei. Clearly there are 1/4 of
the original A nuclei left, and that takes two half-lives, so t = 2 ´ 10 s = 20 s.
Assess: We can also do this mathematically by starting with Equation 30.12 and solving for t.
t / t
æ1ö
N = N0 ç ÷
è2ø
t / t
N æ1ö
=
N 0 çè 2 ÷ø
t æ N ö
= log1/ 2 ç ÷
t1/ 2 è N0 ø
æ N ö
t = (t1/ 2 )log1/ 2 ç ÷
è N0 ø
Use the change-of-base formula.
t = (t1/ 2 )
ln ( ) = (10 s) ln ( ) = 20 s
N
N0
250
1000
ln ( 12 ) ln ( 12 )
Rather than use the change-of-base formula, you could also realize that you need to raise 1/2 to the second power to
get 1/4 so that log1/ 2 (250/1000) = 2.
Q30.10. Reason: As stated in the question, the sample has a half-life of 1.0 min and hence each nucleus has a 50%
probability of undergoing decay in the next minute. As the nucleus ages, the half-life does not change. After 15 min
the half life is still 1.0 min and each nucleus has a 50% chance of decaying in the next minute.
Assess: The probability that a nucleus will decay in the next unit of time is not a function of how long the nucleus
has survived—it depends only on the half-life.
Q30.11. Reason: Because there is more 238
U less of it has decayed, so it must have a longer half-life.
Assess: This can be verified by looking up the two half-lives.
© Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Nuclear Physics 30-3
Q30.12. Reason: The quantum mechanical probability that a particle will decay does not depend on its history.
They are equally likely to decay in the next 66 h.
Assess: This problem is related to fundamentals of probability. If you flip a fair coin and obtain heads 10 times, it is
not more likely to turn up tails in the next attempt than a fresh fair coin. This is the because the number of ways of
flipping 10 heads and one tails is the same as the number of ways of flipping 11 heads in a row. So the chances for
either the fresh fair coin or the one that has been flipped 10 times to come up heads is 50%.
Q30.13. Reason: The half-life of 14 C is 5730 years. After ten half-lives one thousandth of the nuclei are left:
10
æ1ö
N = N 0 ç ÷ = 9.8 ´ 10 -4 N 0 » 10 -3 N 0
è2ø
After 10 half-lives or 10t1/ 2 = 10(5730 y) = 57,300 yr one thousandth of the original amount of 14
C is still in
existence. The fossil fuel has been buried for more than 57,300 years.
After 20 half-lives one millionth of the nuclei are left:
20
æ1ö
N = N 0 ç ÷ = 9.5 ´ 10 -7 N 0 » 10 -6 N 0
è2ø
After 20 half-lives or 20t1/ 2 = 20(5730 y) = 114 600 y there is still some 14 C, the fossil fuel has been buried for
more than 114,600 years. In like manner, after 50 half-lives or 286 500 y the number of nuclei left is 10-16 N 0 . We
could continue this process until the number of nuclei of 14 C is so small we agree none exists. The point is that the
fossil fuel has been buried for a long time, in fact so long that no 14 C remains in the fuel.
Assess: The age of the earth is greater than the time it has taken the 14 C to totally decay (none is left).
Q30.14. Reason: Suppose 14
C was less abundant 10,000 years ago than today. If scientists are unaware of this
fact, they will assume it has always had the same abundance. When they compute the age of the artifact, they will
measure a very small 14 C concentration and will assume it is because the artifact is very old. Thus they will
overestimate the age. Their mistake is that the low concentration is really due to a low concentration in living things
10,000 years ago.
Assess: Details like this complicate radiocarbon dating, but you should be aware that scientists are aware of these
issues and account for them. Furthermore, there are other methods of dating things, and they all act as a check and
balance on each other, giving a very good consensus.
Q30.15. Reason: The unknown X may be identified as follows:
(a) 222
86 Rn ® 218
84 Po + 2 X ; X must be an alpha particle, which is a He atom nucleus 2 He.
4 4
(b) 228
88 Ra ® 228
89 Ac + 0
-1 X ; X must be an electron or beta-minus radiation.
(c) 140
54 Xe ® 140
55 Cs + 0
-1 X ; X must be an electron or beta-minus radiation.
(d) Cu ® Ni + X ; X must be a positron or beta-plus radiation.
64
29
64
28
0
1
Assess: The equation is balanced by conserving mass and charge.
Q30.16. Reason: (a) No. The alpha particle is added to the wrong nucleus. Th + a would work.
236
92 U® 232
90
(b) No. The original nucleus would need to lose four nucleons, not two.
(c) No. In beta-minus decay, the atomic mass numbers of the parent nucleus and of the daughter nucleus must be
equal. Here A has decreased by one.
Assess: Each reaction must be examined to see that there are the same number of nucleons on each side of the arrow,
and that charge is conserved.
Q30.17. Reason: (a) This nucleus is in an excited state and as it decays the proton in the higher level will drop
down to the lower level. The number of protons and neutrons will remain the same and the nucleus will get rid of the
excess energy by emitting a gamma ray.
© Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 30-4 Chapter 30
(b) This nucleus is in a stable unexcited state.
(c) This nucleus is in an excited state and as it decays the neutron in the higher level will drop down to the empty
lower level proton state. That is, a neutron will become a proton and an electron as follows: n ® p + -10e + energy.
During this process a beta-minus is emitted.
Assess: Nuclear energy–level diagrams provide a lot of information about the nucleus.
Q30.18. Reason: Just as with filled electron shells in chemistry, if the nuclear shells are just filled then the nucleus
is stable and “happy.” For 4 He the n = 1 shell is exactly filled for both protons and neutrons (hence it is doubly
magic), with two each. Likewise, for 16 O the n = 1, 2, 3 shells are all exactly filled for both protons and neutrons,
since the n = 1 shell can accommodate two each, the n = 2 shell can accommodate four each, and the n = 3 shell can
accommodate another two each.
Assess: 12 C and 16 O can be thought of as being made of three and four 4 He nuclei (or alpha particles) respectively.
These nuclei are indeed made that way in stars through the process of nuclear fusion.
Q30.19. Reason: As you radiate the apple, you are putting energy into it. This energy will break molecular bonds
and cause the molecules to vibrate. These vibrations will cause the temperature of the apple to rise. After some time,
the apple will appear almost as if it had been heated in the oven.
Assess: The process is somewhat like a microwave oven. Microwaves are just like gamma rays except for the
amount of energy.
Q30.20. Reason: The Note in the text says, “ionizing radiation causes damage to materials, but objects irradiated
with alpha, beta, or gamma radiation do not become radioactive.” So the patient’s body does not itself become
radioactive.
Assess: However, if radioactive materials are ingested or inserted into the body, they will remain radioactive in the
body.
Q30.21. Reason: In a gamma scan, phosphorus-containing molecules with an attached 99
Tc atom (a gamma
emitter) are administered to the patient. The gamma emitting molecules are absorbed preferentially by the tumor so
that the tumor is highlighted. In an x-ray, the tumor may not show up because it may look like the neighboring
tissues. In this way, the gamma scan actually shows bodily function and not just structure.
Assess: X-ray images only show structure because they merely show whether an x-ray was able to penetrate the
bodily tissue or not. By contrast the gamma scan shows which parts of the body are using certain molecules.
Q30.22. Reason: (a)
Sample Dose in Gy Absorbed Energy in J/kg RBE Dose eq in Sv
A 0.1 0.1 1 0.1
B 0.2 0.2 1 0.2
C 0.1 0.1 2 0.2
D 0.2 0.2 2 0.4
The radiation dose is given in gray, and the gray tells us the energy in J absorbed per kg of absorbing material.
(1 Gy = 1.00 J/kg of absorbed energy). So the dose with the greatest number of Gy has received the most energy.
The samples are rated for energy absorbed as follows: B = D > A = C.
(b) The biological damage is assessed by looking at the dose equivalent in Sv. The dose equivalent in Sv = dose in
Gy ´ RBE. The samples are rated for biological damage as follows: D > B = C > A.
Assess: In order to determine the biological damage, the relative biological effectiveness must be taken into
consideration.
Q30.23. Reason: The dose equivalent (in Sv) is the dose (in Gy) multiplied by the RBE. Two different sources of
radiation could give the same dose equivalent (in Sv) but have different RBE if they give different doses (in Gy). So
the two sources do not necessarily have the same RBE.
Assess: For example, alpha particles can deliver a high dose equivalent even from a low-dose source because its
RBE is so high.
© Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
