# Chapter 28 - Worked Solutions

### to selected Complex Reasoning questions from the OUP text Senior Physics - Knowledge, Processes and Reasoning by Walding, Rapkins and Rossiter

Question 43

(a) mr = 4 x 1.007825 = 4.031300
mp = 4.002603 + 2 x 0.000549 = 4.003701
D m = 0.027599 u = 4.58 x 10-29 kg
E = mc2 = 4.123 x 10-12 J

(b) mass of 4 atoms of 1H = 4.0313 x 1.66 x 10-27 kg = 6.69 x 10-27 kg
Therefore, E for 1 kg of 1H = 1/(6.69 x 10-27) x 4.123 x 10-12 = 6.16 x 1014 J

Question 44

Reaction I:
mr = 239.05218 + 1.008665 = 240.06085 u
mp = 136.9100 + 99.9076 + 3 x 1.008665 = 239.8436 u
D m = 0.21725 u = 3.606 x 10-20 kg
E = mc2 = 3.246 x 10-11 J
mass of reactants in kg: 240.06085 x 1.66 x 10-27 = 3.98 x 10-25 kg
Energy/kg = 3.246 x 10-11/3.98 x 10-25 = 8.145 x 1013 j/kg

Reaction II:
mr = 2 x 2.014102 = 4.028204 u
mp = 4.002603 u
D m = 0.025601 u = 4.25 x 10-29 kg
E = mc2 = 3.824 x 10-12 J
mass of reactants in kg: 4.028204 x 1.66 x 10-27 = 6.687 x 10-27 kg
Energy/kg = 3.824 x 10-12/6.687 x 10-27 = 5.72 x 1014 J/kg

NOTE: Answer in back of the 1997 edition of the book says Reaction II produces 9.61 x 1012 J/kg. This is wrong - the correct answer is shown above (5.72 x 1014 J/kg). Thank you to Clint Dempster for his keen eye.

Answer: Reaction II produces more energy per kg than Reaction I.
Reaction I is fission; Reaction II is fusion.

Question 45

Question 46

(a) E = 3.9 x 1023 J
D m = E/c2 = 4.3333 x 106 kg/s

(b) D m = 1 x 1030 kg, therefore it takes 1 x 1030/4.333 x 106 s
= 2.3 x 1023 s = 7.3 x 1015 years

Question 47

URANIUM: m (U-235) = 235 x 1.66 x 10-27 = 3.9 x 10-25 kg
E/kg = 3.5 x 10-11 x 1/3.9 x 10-25 = 8.97 x 1013 J
METHANE: 50 x 106 J/kg
Therefore: EUranium/EMethane = 8.97 x 1013/50 x 106 = 1794400 times (1.8 x 106 times).

Question 48

(a) E = 1.8 x 1014 kJ/s = 1.55 x 1019kJ/day
m = E/c2 = 1.55 x 1019 x 103/(3 x 108)2 = 172800 kg

(b) Coal: 1.55 x 1019 x 103 J/day/(32 x 103 J/g) = 4.84 x 1017 g/day = 4.84 x 1014 kg/day

Question 49

Background = 13 counts/min = 0.2167 Bq

 Day Time elapsed (days) Time for 2000 counts Observed counts/ second Actual counts/ second lnA M 0 76 26.316 26.099 3.26 Tue 1 99 20.202 19.985 2.99 W 2 130 15.385 15.168 2.72 Th 3 171 11.696 11.479 2.44 F 4 220 9.091 8.874 2.18 M 7 508 3.937 3.720 1.31 Tu 8 615 3.252 3.035 1.11 W 9 828 2.415 2.200 0.79 Th 10 1053 1.899 1.683 0.52 F 11 1379 1.450 1.234 0.21

Slope = (lnA - lnA0 ) / t = -3.05/11 = 0.2773 days-1
l = - slope = 0.2773 days-1
t1/2 = 0.693/l = 2.5 days

Question 51:

Number of disintegrations in 10 s = 3.7 x 1013 x 10 s = 3.7 x 1014 disintegrations.

Total energy = 3.7 x 1014 disintegrations x 1.25 x 1.6 x 10-13 = 74J

Energy absorbed =  74J x 0.5 x 0.02 = 0.0074J

AD = 0.0074J/74kg = 0.01057 Gy (0.011 Gy)

For gamma, QF = 1 so Dose Equivalent = 0.01057 Gy x 1 = 0.01057 Sv