Page 460 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 36] NUCLEAR PHYSICS 445

In order that the total numbers of protons and neutrons be the same before and after the fission reaction, two
neutrons must be liberated. Hence the reaction is

235 1 140 Xe + Sr + n + n + E
1
1
94
92 U + n → 54 38 0 0
0
In this case E is about 200 MeV.
SOLVED PROBLEM 36.8
When 235 U undergoes fission, about 0.1 percent of the original mass is released as energy. (a) How much
92
energy is released when 1 kg of 235 U undergoes fission? (b) How much 235 U must undergo fission per
92 92
8
day in a nuclear reactor that provides energy to a 100-MW (10 -W) electric power plant? Assume perfect
efficiency. (c) When coal is burned, about 32.6 MJ/kg of heat is liberated. How many kilograms of coal
would be consumed per day by a conventional coal-fired 100-MW electric power plant?
2
13
2
8
(a) E = mc = (0.001 kg)(3 × 10 m/s) = 9 × 10 J
(b) Energy = power × time, and so here
8
12
E = Pt = (10 W)(3600 s/h)(24 h/day) = 8.64 × 10 J/day
Hence the mass of 235 U required is
92
12
8.64 × 10 J/day −2
= 9.6 × 10 kg/day = 96 g/day
9 × 10 J/kg
13
7
(c) The energy liberated per kilogram of coal burned is 3.26 × 10 J. Hence the mass of coal required is
12
8.64 × 10 J/day 5
= 2.65 × 10 kg/day
3.26 × 10 J/kg
7
which is 265 metric tons.
SOLVED PROBLEM 36.9
In the sun and most other stars the principal energy-liberating process is the conversion of hydrogen
to helium in a series of nuclear fusion reactions in the course of which positrons (positively charged
electrons) are emitted. (a) Write the equation for the overall process in which four protons form a helium
1
4
nucleus (alpha particle). (b) How much energy is liberated in each such process? The masses of H, He,
1 2
and the electron are, respectively, 1.007825, 4.002603, and 0.000549 u.
(a) Two positrons must be given off so that charge will be conserved. Hence the overall process is
1
1 H + H + H + H → He + e + e +
4
1
1
+
1 1 1 1 2
(b) Since a helium atom has only two electrons around its nucleus, two electrons as well as two positrons are lost
when each helium atom is formed. The mass change is therefore
m = 4m H − (m He + 4m e ) = (4)(1.007825 u) − [4.002603 u + (4)(0.000549 u)]
= 0.026501 u
and the energy liberated is (0.026501 u)(931 MeV/u)= 24.7 MeV.

RADIOACTIVITY
Certain nuclei are unstable and undergo radioactive decay into more stable ones. The five types of radioac-
tive decay are shown in Fig. 36-2. A gamma ray is a high-energy photon, a positron is a positive electron,
4
and an alpha particle is a nucleus of the helium atom He. Gamma rays have the greatest ability to pene-
2
trate matter before being absorbed and alpha particles are the most readily absorbed. The weak interaction is