Ch19 Solution
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Ch19/P.1
19 Atomic Structure and Radioactive Decay 19.1 The Atomic Model Warm-up 19.1 (p. 226) Number of protons
Check-point 1 (p. 229) 1 2
A B
3
Ca, 20, number of neutrons, 40,
40 20
Ca , 40
Practice 19.1 (p. 230) 1 2 3 4
B A A A
5
(a) (b) (c) (d) (e)
Electrons Protons Electrons Neutrons Electrons
6
(a)
(i) Number of protons = 90 (ii) Number of neutrons = 234 − 90 = 144 (iii) Number of electrons = Number of protons = 90
(b)
238 90Th
7
Atom P has 88 protons and it is a radium atom. Atom Q has 88 electrons. Hence, it has 88 protons and it is a radium atom. Atom R has 38 electrons. Hence, it has 38 protons and it is a strontium atom. Atom S has 90 protons and it is a thorium atom. Atoms P and Q are isotopes.
Ch19/P.2
19.2 Radioactive decay and uses of radioisotopes Warm-up 19.2 (p. 231) The first statement.
Check-point 2 (p. 243) 1 2 3
4
C B Half, undecayed 8, 8 12, 4 14, 2 14 g Halved 150, 600 2 2, 20 hours
Practice 19.2 (p .244) 1
D 24 11 Na
2 3
→
24 12 Mg
+
0 -1 e
B A 90 hours is equal to 6 half-lives. 6
4
1 Expected activity = 500 × = 7.81 Bq 2 D
5
The daughter nuclide is radium-224. 228 90Th
→
224 88 Ra
+ 42 He
6
Tracers have short half-lives so that their effect on the system disappears after a short time.
7
(a) (b)
(d)
C-14 is radioactive. This method could not work if the table were made of iron instead because carbon-14 is not present in iron. No, it is not useful. This is because the life of the Hong Kong Year Book is much shorter than the half-life of C-14. This is because the amount of C-14 left in the material is too little to give an accurate result.
(a)
D1:
238 92 U
D2:
234 91 Pa
(c)
8
(b)
→ →
234 90Th 234 92 U
A = 226, Z = 88
+ 42 He +
0 −1 e
Ch19/P.3
Revision exercise 19 Multiple-choice (p. 247) Section A 1 A 2 B 3 B Number of half-lives passed = 1 Expected activity = 500 × 2 4
B
7.5 = 1.42 5.3 1.42
= 187 Bq
Let n be the number of half-lives passed. n
1 6.25 = 100 × ⇒ n=4 2 It takes 4 half-lives, i.e. 60 hours, for 100 g sodium-24 to decay to 6.25 g. 5 6
A D (HKCEE 2001 Paper II Q40)
Section B 7 A 8 D (HKCEE 1999 Paper II Q37) 9 D (HKCEE 2001 Paper II Q39) 10 B (HKCEE 2002 Paper II Q40)
Conventional (p. 248) Section A 1
(a)
42 19 K
→
42 20 Ca
+
0 −1 e
Correct symbols Correct atomic number and mass number of Ca (b)
(1A) (1A)
Ch19/P.4
(c)
2
(a)
(b)
3
(a) (b)
(c) 4
Correct labelled axes (1A) Showing background radiation on the graph (1A) Correct half-life of the sample (1A) I don't agree. (1A) Since K-42 decays into Ca-42, which has the same mass as K-42, (1A) the balance reading does not change as K-42 decays. No weight-time graph similar to the curve in (b), and hence no half-life of K-42, can be obtained. (1A) β source is suitable. (1A) It is because it is slightly penetrating (1A) but it is not as harmful as γ rays. (1A) The source with half-life of 8 hours is more suitable. (1A) It is because the half-life is long enough to detect the leakage of oil pipes before the activity of the source varies significantly, (1A) and the activity of the source will not stay high over a long time to cause harmful effects to people. (1A) Number of protons = 32 − 17 = 15
(1A)
Correct position of P-32 on the graph It is because P-32 does not fall on the curve of stable nuclei in Figure a.
(1A) (1A)
Correct labelled axes Correct data Correct curve
(1A) (1A) (1A)
(a)
Ch19/P.5
(b) (c)
The points do not fall on a smooth curve because of the random fluctuation of the decay process. (1A) Correct method to determine the half-life of the sample by using the graph in (a) (1M) Half-life = 80 s (1A)
Ch19/P.6
Section B 5 (HKCEE 2000 Paper I Q11)
Ch19/P.7
6
(HKCEE 2002 Paper I Q10)
7
(a)
(b)
Background radiation has random fluctuations. (1A) Finding the average value of a large number of readings can eliminate the random nature of the readings. (1A) (i) α particle (1A) (Or helium nucleus 1A) (ii)
220 86 Rn
→
Correct
216 84 Po
216 84 Po
Correct 42 He
+ 42 He (1A) (1A)
(iii) Ra-224 decays into Rn-220 and then Rn-220 decays into Po-216. Since the half-life of Rn-220 is much shorter than that of Ra-224, (1A) when a Rn-220 nucleus is produced, it decays quickly into a Po-216 nucleus. (1A) Hence the growth rate of Rn-220 nuclei is slower than the decay rate of Ra-224 nuclei. (iv) When we breathe in radon gas, radon gas decays inside our body and emits α radiation. (1A) Since α radiation has very weak penetrating power, it will be stopped by our body tissue. (1A) And since α radiation has a strong ionizing power, its damage towards our tissue is severe. (1A) Hence, the presence of radon gas in buildings is a health hazard to people.
Ch19/P.8
Physics in articles (p. 251) (a) (b)
(c)
An α source is used because α radiation has strong ionizing power; (1A) it ionizes air and hence a current can be maintained inside the detector. (1A) No, it is not harmful to health. (1A) It is because the range of α radiation is very short (~ 5 cm) and α radiation cannot reach people from the ceiling. (1A) No, a source with short half-life cannot be used. (1A) It is because the activity of the source, and hence the current in the detector, changes sharply within a short time if a source with short half-life is used. This will lead to false alarm. (1A)
19 Atomic Structure and Radioactive Decay 19.1 The Atomic Model Warm-up 19.1 (p. 226) Number of protons
Check-point 1 (p. 229) 1 2
A B
3
Ca, 20, number of neutrons, 40,
40 20
Ca , 40
Practice 19.1 (p. 230) 1 2 3 4
B A A A
5
(a) (b) (c) (d) (e)
Electrons Protons Electrons Neutrons Electrons
6
(a)
(i) Number of protons = 90 (ii) Number of neutrons = 234 − 90 = 144 (iii) Number of electrons = Number of protons = 90
(b)
238 90Th
7
Atom P has 88 protons and it is a radium atom. Atom Q has 88 electrons. Hence, it has 88 protons and it is a radium atom. Atom R has 38 electrons. Hence, it has 38 protons and it is a strontium atom. Atom S has 90 protons and it is a thorium atom. Atoms P and Q are isotopes.
Ch19/P.2
19.2 Radioactive decay and uses of radioisotopes Warm-up 19.2 (p. 231) The first statement.
Check-point 2 (p. 243) 1 2 3
4
C B Half, undecayed 8, 8 12, 4 14, 2 14 g Halved 150, 600 2 2, 20 hours
Practice 19.2 (p .244) 1
D 24 11 Na
2 3
→
24 12 Mg
+
0 -1 e
B A 90 hours is equal to 6 half-lives. 6
4
1 Expected activity = 500 × = 7.81 Bq 2 D
5
The daughter nuclide is radium-224. 228 90Th
→
224 88 Ra
+ 42 He
6
Tracers have short half-lives so that their effect on the system disappears after a short time.
7
(a) (b)
(d)
C-14 is radioactive. This method could not work if the table were made of iron instead because carbon-14 is not present in iron. No, it is not useful. This is because the life of the Hong Kong Year Book is much shorter than the half-life of C-14. This is because the amount of C-14 left in the material is too little to give an accurate result.
(a)
D1:
238 92 U
D2:
234 91 Pa
(c)
8
(b)
→ →
234 90Th 234 92 U
A = 226, Z = 88
+ 42 He +
0 −1 e
Ch19/P.3
Revision exercise 19 Multiple-choice (p. 247) Section A 1 A 2 B 3 B Number of half-lives passed = 1 Expected activity = 500 × 2 4
B
7.5 = 1.42 5.3 1.42
= 187 Bq
Let n be the number of half-lives passed. n
1 6.25 = 100 × ⇒ n=4 2 It takes 4 half-lives, i.e. 60 hours, for 100 g sodium-24 to decay to 6.25 g. 5 6
A D (HKCEE 2001 Paper II Q40)
Section B 7 A 8 D (HKCEE 1999 Paper II Q37) 9 D (HKCEE 2001 Paper II Q39) 10 B (HKCEE 2002 Paper II Q40)
Conventional (p. 248) Section A 1
(a)
42 19 K
→
42 20 Ca
+
0 −1 e
Correct symbols Correct atomic number and mass number of Ca (b)
(1A) (1A)
Ch19/P.4
(c)
2
(a)
(b)
3
(a) (b)
(c) 4
Correct labelled axes (1A) Showing background radiation on the graph (1A) Correct half-life of the sample (1A) I don't agree. (1A) Since K-42 decays into Ca-42, which has the same mass as K-42, (1A) the balance reading does not change as K-42 decays. No weight-time graph similar to the curve in (b), and hence no half-life of K-42, can be obtained. (1A) β source is suitable. (1A) It is because it is slightly penetrating (1A) but it is not as harmful as γ rays. (1A) The source with half-life of 8 hours is more suitable. (1A) It is because the half-life is long enough to detect the leakage of oil pipes before the activity of the source varies significantly, (1A) and the activity of the source will not stay high over a long time to cause harmful effects to people. (1A) Number of protons = 32 − 17 = 15
(1A)
Correct position of P-32 on the graph It is because P-32 does not fall on the curve of stable nuclei in Figure a.
(1A) (1A)
Correct labelled axes Correct data Correct curve
(1A) (1A) (1A)
(a)
Ch19/P.5
(b) (c)
The points do not fall on a smooth curve because of the random fluctuation of the decay process. (1A) Correct method to determine the half-life of the sample by using the graph in (a) (1M) Half-life = 80 s (1A)
Ch19/P.6
Section B 5 (HKCEE 2000 Paper I Q11)
Ch19/P.7
6
(HKCEE 2002 Paper I Q10)
7
(a)
(b)
Background radiation has random fluctuations. (1A) Finding the average value of a large number of readings can eliminate the random nature of the readings. (1A) (i) α particle (1A) (Or helium nucleus 1A) (ii)
220 86 Rn
→
Correct
216 84 Po
216 84 Po
Correct 42 He
+ 42 He (1A) (1A)
(iii) Ra-224 decays into Rn-220 and then Rn-220 decays into Po-216. Since the half-life of Rn-220 is much shorter than that of Ra-224, (1A) when a Rn-220 nucleus is produced, it decays quickly into a Po-216 nucleus. (1A) Hence the growth rate of Rn-220 nuclei is slower than the decay rate of Ra-224 nuclei. (iv) When we breathe in radon gas, radon gas decays inside our body and emits α radiation. (1A) Since α radiation has very weak penetrating power, it will be stopped by our body tissue. (1A) And since α radiation has a strong ionizing power, its damage towards our tissue is severe. (1A) Hence, the presence of radon gas in buildings is a health hazard to people.
Ch19/P.8
Physics in articles (p. 251) (a) (b)
(c)
An α source is used because α radiation has strong ionizing power; (1A) it ionizes air and hence a current can be maintained inside the detector. (1A) No, it is not harmful to health. (1A) It is because the range of α radiation is very short (~ 5 cm) and α radiation cannot reach people from the ceiling. (1A) No, a source with short half-life cannot be used. (1A) It is because the activity of the source, and hence the current in the detector, changes sharply within a short time if a source with short half-life is used. This will lead to false alarm. (1A)
