College Physics for AP® Courses

# Chapter 31

### 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.

$1.9 fm 1.9 fm size 12{1 "." 9" fm"} {}$

9.

(a) $4.6 fm4.6 fm size 12{4 "." "6 fm"} {}$

(b) $0.61 to 10.61 to 1 size 12{0 "." "61 to 1"} {}$

11.

$85 . 4 to 1 85 . 4 to 1 size 12{"85" "." "4 to 1"} {}$

13.

$12.4 GeV 12.4 GeV size 12{"12" "." "4 GeV"} {}$

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.

$25 50 M 25 → 24 50 Cr 26 + β + + ν e 25 50 M 25 → 24 50 Cr 26 + β + + ν e size 12{"" lSub { size 8{"25"} } lSup { size 8{"50"} } M rSub { size 8{"25"} } rightarrow "" lSub { size 8{"24"} } lSup { size 8{"50"} } "Cr" rSub { size 8{"20"} } +β rSup { size 8{+{}} } +v rSub { size 8{e} } } {}$

21.

$4 7 Be 3 + e − → 3 7 Li 4 + ν e 4 7 Be 3 + e − → 3 7 Li 4 + ν e size 12{"" lSub { size 8{4} } lSup { size 8{7} } "Be" rSub { size 8{3} } +e rSup { size 8{ - {}} } rightarrow "" lSub { size 8{3} } lSup { size 8{7} } "Li" rSub { size 8{4} } +v rSub { size 8{e} } } {}$

23.

$84 210 Po 126 → 82 206 Pb 124 + 2 4 He 2 84 210 Po 126 → 82 206 Pb 124 + 2 4 He 2 size 12{"" lSub { size 8{"84"} } lSup { size 8{"210"} } "Pb" rSub { size 8{"126"} } rightarrow "" lSub { size 8{"82"} } lSup { size 8{"206"} } "Pb" rSub { size 8{"124"} } +"" lSub { size 8{2} } lSup { size 8{4} } "He" rSub { size 8{2} } } {}$

25.

$55 137 Cs 82 → 56 137 Ba 81 + β − + ν ¯ e 55 137 Cs 82 → 56 137 Ba 81 + β − + ν ¯ e size 12{"" lSub { size 8{"55"} } lSup { size 8{"137"} } "Cs" rSub { size 8{"82"} } rightarrow "" lSub { size 8{"56"} } lSup { size 8{"137"} } "Ba" rSub { size 8{"81"} } +β rSup { size 8{ - {}} } + {overline {v rSub { size 8{e} } }} } {}$

27.

$90 232 Th 142 → 88 228 Ra 140 + 2 4 He 2 90 232 Th 142 → 88 228 Ra 140 + 2 4 He 2 size 12{"" lSub { size 8{"90"} } lSup { size 8{"232"} } "Th" rSub { size 8{"142"} } rightarrow "" lSub { size 8{"88"} } lSup { size 8{"228"} } "Ra" rSub { size 8{"140"} } +"" lSub { size 8{2} } lSup { size 8{4} } "He" rSub { size 8{2} } } {}$

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 $γγ size 12{γ} {}$ 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 alignl { stack { size 12{Z+1=Z - 1" before/after; captured "e rSup { size 8{ - 1} } " is last term rhs;"} {} # A=A" ; efn : " left (+1 right )= left (+1 right ) {} } } {}

35.

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

(b) 4.87 MeV

37.

(a) $n→p+β−+ν¯en→p+β−+ν¯e$

(b) ) 0.783 MeV

39.

1.82 MeV

41.

(a) 4.274 MeV

(b) $1.927×10−51.927×10−5 size 12{1 "." "927" times "10" rSup { size 8{ - 5} } u} {}$

(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) $815O7+ e− →715N8+νe815O7+ e− →715N8+νe size 12{"" lSub { size 8{8} } lSup { size 8{"15"} } O rSub { size 8{7} } +e rSup { size 8{ - {}} } rightarrow "" lSub { size 8{7} } lSup { size 8{"15"} } N rSub { size 8{8} } +v rSub { size 8{e} } } {}$

(b) 2.754 MeV

44.

57,300 y

46.

(a) 0.988 Ci

(b) The half-life of $226Ra226Ra size 12{"" lSup { size 8{"226"} } "Ra"} {}$ is now better known.

48.

$1.22 × 10 3 Bq 1.22 × 10 3 Bq$

50.

(a) 16.0 mg

(b) 0.0114%

52.

$1.48 × 10 17 y 1.48 × 10 17 y size 12{ {underline {1 "." "48" times "10" rSup { size 8{"17"} } y}} } {}$

54.

$5.6 × 10 4 y 5.6 × 10 4 y size 12{ {underline {5 "." 6 times "10" rSup { size 8{4} } y}} } {}$

56.

2.71 y

58.

(a) 1.56 mg

(b) 11.3 Ci

60.

(a) $1.23×10−31.23×10−3 size 12{ {underline {1 "." "23" times "10" rSup { size 8{ - 3} } }} } {}$

(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 $αα size 12{α} {}$ 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) $1.46×10−8u1.46×10−8u$ vs. 1.007825 u for $1H1H$

(b) 0.000549 u

(c) $2.66×10−52.66×10−5 size 12{2 "." "66" times "10" rSup { size 8{ - 5} } } {}$

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.

### Test Prep for AP® Courses

1.

(c)

3.

(a)

5.

When $95 241 Am 95 241 Am MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaI1aaabaGaaGOmaiaaisdacaaIXaaaaOGaaeyqaiaab2gaaaa@3B96@$ undergoes α decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore $93 237 Np 93 237 Np MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaaiaaiMdacaaIZaaabaGaaGOmaiaaiodacaaI3aaaaOGaaeOtaiaabchaaaa@3BA9@$.

7.

During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1.

9.
1. No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 ≠ 88+2).
2. Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is conserved (88 = 82 + 6).
3. Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron’s charge is properly counted (6 = 7 + (-1)).
4. No. Nucleon number is not conserved (24 ≠ 23). The positron released counts as a charge to conserve charge, but it doesn’t count as a nucleon.
11.

This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from:

$N( t )= N 0 e( −0.693t t 1 2 )=3.4× 10 17 e( −( 0.693 )( 0.035 ) 0.00173 ) N( t )= N 0 e( −0.693t t 1 2 )=3.4× 10 17 e( −( 0.693 )( 0.035 ) 0.00173 ) MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6233@$

$N( t )=4.1× 10 11 N( t )=4.1× 10 11 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeOtamaabmaabaGaamiDaaGaayjkaiaawMcaaiabg2da9iaaisdacaGGUaGaaGymaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaIXaGaaGymaaaaaaa@41A6@$ nuclei

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