Exam Style Answers 29 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 29 c

Exam-style questions

 

1

D 

[1]

2

A 

[1]

3

a

Using ∆E = ∆mc2, ∆m = 2 × 1.67 × 10−27  [1]

 

energy released = 2 × 1.67 × 10−27 ×  (3.0 × 108)2  −10

= 3.0 × 10  J 

4

5

[1]  23

 

−

 

BE per nucleon = 1.23 × 10−12 J 

a

decrease in mass = 3.015 500 + 2.013 553 − 4.001 506 − 1.007 276  [1]

 

= 0.020 271 u

 

= 0.020 271 × 1.660 × 10−27 = 3.365 ×  10−29 kg 

[1] [1]

[1]

[1]

1 mole contains N A particles (6.02 × 10 )   [1]

b

energy released = ∆mc2 

 

energy released = 3.0 × 10−10 × N A 

 

= 3.365 × 10−29 × (3.0 × 108)2 = 3.028 ×  10−12 J  [1]

 

= 3.0 × 10−10 × 6.023 × 1023 = 1.8 × 1014 J [1]

c

energy released per mole = energy per reaction × N A  [1]

 

= 3.028 × 10−12 × 6.023 × 1023 =  1.823 × 1012 J 

[1]

a

1 mole contains N A atoms 

[1]

 

Using A = λ  λN 

 

λ = 

1.0

[1]

Using ∆E = ∆mc2, ∆m =  (3. 0 × 108 )2  

[1]

 change in mass = 1.1 × 10−17 kg  change

[1]

Using E = mc2, energy released per second = 70 × 10−6 × (3.0 × 108)2  [1] [1]

9

[1]

change in mass = 221.970 − 217.963 − 4.002, change in mass = 0.005 u  [1]  

A

 

[1]

N 

8.0 02 2 × 1021

 = 1.33 × 10−2 s−1 

 

= 

b

 λt1/2 = 0.693, so t1/2 = 

 

half-life = 

6.023 × 10

23

= 0.005 × 1.660 × 10−27 kg = 8.30 × 10  kg 

 energy released = 8.30 × 10−30 × (3 × 108)2  energy

[1]

 energy released = 7.47 × 10−13 J  energy

[1]

The  T he energy is released as kinetic energy of the α-particle  [1]

0.693   × 10 1.3 33 3  

 = 52.0 s 

[1] [1]

mass defect in u = 6 × (1.007 276 + 1.008 6 65 + 0.000 548) − 12.000 = 0.098 934 u  [1]

 

 

mass defect in kg =  0.098 934 × 1.660 × 10−27 = 1.64 × 10−28 kg   [1]

b

 λt1/2 = 0.693, so λ = 

 

 

 

[1]

−

a

 and electromagnetic radiation (the γ-ray).   [1] and

10 a

2

 

λ 

From the graph, 2 × t1/2 = 28 s  28 so t1/2 =   = 14 s  2 (or use the point (14, 80))

b  binding energy = mass defect × c2 

[1]

0.693 [1]

−30

0.693

 

[1]

[1]

t1/ 2

69 693 3 decay constant =  0.14  = 4.95 × 10−2 s−1  [1]

[1]

= 1.64 × 10−28 × (3.0 × 108)2  = 1.48 × 10−11 J 

1

×

b

but  b ut energy per second = power, so power = 6.3 GW 

7

8

[1]

= 6.3 × 109 J 

6

number of nucleons = 12; BE per nucleon 11 =  1.4488 1210   [1]

[1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

11 a

 λ1/2 = 

0.693 t1/ 2

 = 

0.693 4. 9  × 109

 

= 1.4 × 10−10 y−1 

b

Using ln

  N        N 0  

b  

[1] [1]

 = − λt 

i 

Electrostatic forces larger as the charge on each nucleus is twice that on hydrogen (H)  [1]

 ii  ii

∆m = 12.000000 − (3 × 4.001 506) =  0.004 518 u  [1]

 

∆m = 0.004 518 × 1.660 × 10−27 =  7.500 × 10−27 kg 

[1]

 

ln 0.992 = −1.4 × 10−10t 

[1]

 

t = 5.7 × 107 y 

[1]

 

∆E  = 7.500 × 10−27 × (3.0 × 108)2

Graph dra drawn wn using these figures, single smooth line, points plotted as crosses crosses,, suggested scales: activity ( y  y-axis) 50 Bq per cm time (x-axis) 2 minutes per cm   [1]

 

∆E  = 6.75 × 10−13 J 

12 a

i

 ii There is a random element of ii radioactive decay, which becomes more apparent at lower levels of activity. 

15 a

[1]

[1]

Alpha-particles have a very low penetration and those from outside the body are stopped by the layer of dead skin cells.  [1]

 

Dust can be inhaled, bringing the α-particles inside the t he body, body, where they are very dangerous.  [1]

b

N   =

[1]

2.4

×

6.022 ×  10 23 = 6.63 × 1021 

[1]

 

2.18 λ = (ln 2)/183 = 3.78 × 10−3 s−1 

[1]

 

A = λ  λN  = 3.78 × 10−3 × 6.63 × 1021 

but the the timesame. for the remain   rate to halve would [1]

 

[1]

92 protons, 143 neutrons 

c

A = 2.50 × 10  Bq   A 0    1019    5 0 ×  λt = ln   = ln  2.50      A     10   = ln (2.50 × 1018) 

[1]

b

From the graph, t1/2 ≈ 3.8 minutes 

[1]

c

All count rates would be greater 

 

[1] [1]

19

13 a b

V  = 

4

3

π r  = 

(

4 1.4 1×1 10 0

89 9 ×10 density =  3.8 2.7 76 6 ×10

3

235

)

  [1]

25 42

 m3 

[1]

c

When the nucleons are combined in the nucleus they have less energy ( = binding energy) than when separated.   [1]

 

Less energy means less mass (or energy has mass).  [1] Sum the total mass of the separate protons and neutrons. 

[1]

Subtract from that the mass of the uranium nuclide. 

[1]

 

Apply ∆E  = ∆mc2 

[1]

 

∆E  is  is the binding energy 

[1]

14 a

t = ln (2.50 × 1018)/3.78 × 10−3 = 11 200 s [1]

 

t = 3.1 hours 

[1]

Kinetic energy 

[1]

 

of the fission fragments 

[1]

b

i 

16 a

−

density = 1.41 × 1017 ≈ 1.4 × 1017 kg m−3 [1]

 

 

−

 

d

Nuclear fusion is the joining together of two (or more) light nuclei to form a heavier nucleus.  [1]

 

The repulsive electrostatic forces between nuclei must be overcome.  [1]

 

High temperatures mean particles moving very fast / high energy.  

2

×

3

3 = 2.76 × 10−42 m3   

15

−

[1] 3

change in mass = 3.90 ×  10−25 − (1.44 × 10−25 + 2.42 ×  10−25 + 1.67 × 10−27 × 2) = 6.60 × 10−28 kg 

 

[1]

energy released =  6.60 × 10−28 × (3.0 × 108)2   energy released = 5.94 × 10−11 J ≈  5.9 × 10−11 J  [1] 200 × 106  ii  fissions per second =  =  ii 11 5 94 9 4 . × 10 18 3.37 × 10   [1]  iii  fissions per year =  iii 3.37 × 1018 × 3.15 × 107 = 1.06 × 1026 [1]  

   

moles per year

=

1.0 06 6 ×10 26

6.0 02 22 ×1023 mass per year = 176 × 235 =  4.14 × 104 g ≈ 41 kg 

−

 = 176 [1] [1]

[1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020