Chapter 28 - Worked Solutions

to selected extension questions from the OUP text Senior Physics - Concepts in Context 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

 


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