Chapter 27 - Worked Solutions

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

(a) Mass of components = m_p + m_e + m_n 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 = m_p x 2 + m_e x 2 + m_n 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
 

(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
 

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
 

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
 

D (Both particles have to be deflected; Closer particle is deflected more).
 

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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