Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo

Problems & Exercises

1.

1.67 × 10 4 1.67 × 10 4

5.
m = ρV = ρd 3 a = m ρ 1/3 = 2.3 × 10 17 kg 1000 kg/m 3 1 3 = 61 × 10 3 m = 61 km m = ρV = ρd 3 a = m ρ 1/3 = 2.3 × 10 17 kg 1000 kg/m 3 1 3 = 61 × 10 3 m = 61 km
7.

1.9 fm 1.9 fm

9.

(a) 4.6 fm4.6 fm

(b) 0.61 to 10.61 to 1

11.

85 . 4 to 1 85 . 4 to 1

13.

12.4 GeV 12.4 GeV

15.

19.3 to 1

17.
1 3 H 2 2 3 He 1 + β + ν ¯ e 1 3 H 2 2 3 He 1 + β + ν ¯ e
19.
25 50 M 25 24 50 Cr 26 + β + + ν e 25 50 M 25 24 50 Cr 26 + β + + ν e
21.
4 7 Be 3 + e 3 7 Li 4 + ν e 4 7 Be 3 + e 3 7 Li 4 + ν e
23.
84 210 Po 126 82 206 Pb 124 + 2 4 He 2 84 210 Po 126 82 206 Pb 124 + 2 4 He 2
25.
55 137 Cs 82 56 137 Ba 81 + β + ν ¯ e 55 137 Cs 82 56 137 Ba 81 + β + ν ¯ e
27.
90 232 Th 142 88 228 Ra 140 + 2 4 He 2 90 232 Th 142 88 228 Ra 140 + 2 4 He 2
29.

(a) charge: +1+1=0; electron family number:+1+1=0; A: 0+0=0charge: +1+1=0; electron family number:+1+1=0; A: 0+0=0

(b) 0.511 MeV

(c) The two γγ rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest.

31.
Z = Z + 1 1; A = A ; efn : 0 = + 1 + 1 Z = Z + 1 1; A = A ; efn : 0 = + 1 + 1
33.
Z - 1 = Z 1; A = A; efn : + 1 = + 1 Z - 1 = Z 1; A = A; efn : + 1 = + 1
35.

(a) 88226Ra138 86222 Rn136+ 24 He2 88226Ra138 86222 Rn136+ 24 He2

(b) 4.87 MeV

37.

(a) np+β+ν¯enp+β+ν¯e

(b) ) 0.783 MeV

39.

1.82 MeV

41.

(a) 4.274 MeV

(b) 1.927×1051.927×105

(c) Since U-238 is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample.

43.

(a) 815O7+ e 715N8+νe815O7+ e 715N8+νe

(b) 2.754 MeV

44.

57,300 y

46.

(a) 0.988 Ci

(b) The half-life of 226Ra226Ra is now better known.

48.

1.22 × 10 3 Bq 1.22 × 10 3 Bq

50.

(a) 16.0 mg

(b) 0.0114%

52.

1.48 × 10 17 y 1.48 × 10 17 y

54.

5.6 × 10 4 y 5.6 × 10 4 y

56.

2.71 y

58.

(a) 1.56 mg

(b) 11.3 Ci

60.

(a) 1.23×1031.23×103

(b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly αα particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector.

62.

(a) 1.68 × 10 5 Ci 1.68× 10 5 Ci

(b) 8.65 × 10 10 J 8.65× 10 10 J

(c) $ 2.9 × 10 3 $2.9× 10 3

64.

(a) 6.97 × 10 15 Bq 6.97× 10 15 Bq

(b) 6.24 kW

(c) 5.67 kW

68.

(a) 84.5 Ci

(b) An extremely large activity, many orders of magnitude greater than permitted for home use.

(c) The assumption of 1.00 μA 1.00μA is unreasonably large. Other methods can detect much smaller decay rates.

69.

1.112 MeV, consistent with graph

71.

7.848 MeV, consistent with graph

73.

(a) 7.680 MeV, consistent with graph

(b) 7.520 MeV, consistent with graph. Not significantly different from value for 12C12C, but sufficiently lower to allow decay into another nuclide that is more tightly bound.

75.

(a) 1.46×108u1.46×108u vs. 1.007825 u for 1H1H

(b) 0.000549 u

(c) 2.66×1052.66×105

76.

(a) –9.315 MeV–9.315 MeV

(b) The negative binding energy implies an unbound system.

(c) This assumption that it is two bound neutrons is incorrect.

78.

22.8 cm

79.

(a) 92235 U 143 90 231 Th 141 + 2 4 He 2 92235 U 143 90 231 Th 141 + 2 4 He 2

(b) 4.679 MeV

(c) 4.599 MeV

81.

a) 2.4 × 10 8 2.4 × 10 8 u

(b) The greatest known atomic masses are about 260. This result found in (a) is extremely large.

(c) The assumed radius is much too large to be reasonable.

82.

(a) –1.805 MeV–1.805 MeV

(b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous.

(c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect.

84.

(a) r= r 0 A 1/3 =1.2 (235) 1/3 =7.4 fm   V= 4π r 3 3 =1700  fm 3   ρ= m v =0.14  u/fm 3 r= r 0 A 1/3 =1.2 (235) 1/3 =7.4 fm   V= 4π r 3 3 =1700  fm 3   ρ= m v =0.14  u/fm 3

(b) For barium: r= r 0 A 1/3 =1.2 (142) 1/3 =6.3 fm r= r 0 A 1/3 =1.2 (142) 1/3 =6.3 fm
V= 4π r 3 3 =1047  fm 3 ρ= m v =0.14  u/fm 3 V= 4π r 3 3 =1047  fm 3 ρ= m v =0.14  u/fm 3
For krypton: r= r 0 A 1/3 =1.2 (92) 1/3 =5.4 fm r= r 0 A 1/3 =1.2 (92) 1/3 =5.4 fm
V= 4π r 3 3 =660  fm 3 ρ= m v =0.14  u/fm 3 V= 4π r 3 3 =660  fm 3 ρ= m v =0.14  u/fm 3
To two significant figures, they are all alike.

Order a print copy

As an Amazon Associate we earn from qualifying purchases.

Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information Citation information

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.