Chapter 27 - Worked Solutions

to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

Question 17
(a) Mass of components = mp + me + mn x 2
= 1.007276 + 0.000549 +1.008665 x 2 = 3.025155 u
D m = 3.016049 - 3.025155 = 9.106 x 10-3
E = 931 x 9.106 x 10-3 = 8.4776 MeV
 
(b) Mass of components = mp x 2 + me x 2 + mn x 1
= 1.007276 x 2 + 0.000549 x 2 + 1.008665 = 3.024315
D m = 3.024315 - 3.016030 = 8.285 x 10-3 u = 7.713 MeV
 
Question 22
(i)
(a) 16 x 1.007276 + 19 x 1.008665 + 16 x 0.000549 = 35.289835 u
D m = 35.289835 - 34.969033 = 0.320802
(b) E = 298.66 MeV = 298.66/35 MeV/nucleon = 8.5 MeV/nucleon
 
(ii)
(a) 92 x 1.007276 = 92.66932
+ 143 x 1.008665 = 144.239095
+ 92 x 0.000549 = 0.050508
Total = 236.958995 u
D m = 1.915055 u
(b) E = 1782.9 MeV = 7.58 MeV/nucleon
 
(iii)
(a) 48 x 1.007276 + 65 x 1.008665 + 48 x 0.000549 = 113.93825 u
D m = 1.034417 u
(b) E = 963 MeV = 8.5 MeV/nucleon
 
(iv)
(a) 3 x 1.007276 + 4 x 1.008665 + 3 x 0.000549 = 7.058135 u
D m = 0.042131 u
(b) E = 39.22 MeV = 5.6 MeV/nucleon
 
Question 25
E = 1200/(2.4 x 10-2) = 50000 Vm-1
m = qE/g = 3 x 1.6 x 10-19 x 50000/9.8 = 2.44 x 10-15 kg
 
Question 26
q = mg/E = 3.52 x 10-14 x 9.8/(9.8 x 10-14) = 3.52 x 10-18 C
number of electrons = 3.52 x 10-18 C/(1.6 x 10-19C) = 22
 
Question 27
D (Both particles have to be deflected; Closer particle is deflected more).
 
Question 28
F (In the first diagram, plate X must have a negative charge, and Y a +ve charge as the nett electrostatic force on the the positively charge oil drop has to be upwards (to counter the downward force of gravity). In the second diagram the electrostatic force is horizontal and to the left (arrow G), and the gravitational force is directly down (arrow E). The resultant force is the vector sum of these which can only be F.
 
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