# Chapter 27 - Worked Solutions

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

My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.

Solutions-Ch27-Finch.pdf

The rest are by me, Richard Walding:

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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