Page 300 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 19] NUCLEAR REACTIONS 289

19.15. Complete the following nuclear equations:

94
1
1
1
4
(a) 14 N + α −→ ? + H (c) 235 U + n −→ 140 Ba + Kr + ? n
7 2 1 92 0 56 36 0
1
4
1
1
(b) 27 Al + α −→ ? + n (d) 235 U + n −→ 90 Sr + ? + 3 n
13 2 0 92 0 38 0
1
1
4
94
1
Ans. (a) 14 7 N + α −→ 17 8 O + H (c) 235 U + n −→ 140 Ba + Kr + 2 n
2
92
1
56
0
36
0
1
1
1
4
(b) 27 Al + α −→ 30 P + n (d) 235 U + n −→ 90 Sr + 143 Xe + 3 n
15
38
54
0
0
92
0
2
13
NUCLEAR ENERGY
19.16. Calculate the energy of the reaction of a positron with an electron.
−
+ β + e −→ energy
Ans. Data from Table 19-3 are used. The sum of the masses of the reactants is converted to energy with Einstein’s
equation.
m = 2(0.00054858 amu) = 0.0010972 amu
This mass is changed to kilograms to calculate the energy in joules:

1g 1kg
0.0010972 amu = 1.822 × 10 −30 kg
23
6.022 × 10 amu 1000 g
The energy is given by
2
8
2
E = mc = (1.822 × 10 −30 kg)(3.000 × 10 m/s) = 1.640 × 10 −13 J
19.17. Calculate the energy of the reaction that the free neutron undergoes if it does not encounter a nucleus to
react with.
n −→ p + e + energy
−
Ans. The data from Table 19-3 are used. The sum of the masses of the products minus the mass of the neutron,
converted to energy with Einstein’s equation, yields the energy produced.
m = [1.008665 − (1.00728 + 0.00054858)] amu = 0.00084 amu

1g 1kg
0.00084 amu = 1.4 × 10 −30 kg
23
6.022 × 10 amu 1000 g
The energy is given by
8
2
2
E = mc = (1.4 × 10 −30 kg)(3.00 × 10 m/s) = 1.3 × 10 −13 J
14
14
19.18. Calculate the energy of the reaction of one atom C to yield N and a beta particle. The atomic masses
14
are C = 14.003241 amu and 14 N = 14.003074 amu.
Ans. The difference in masses of products and reactants for the nuclear equation
14 14
C −→ N + β
is m = m N nucleus + m beta − m C nucleus
That is equal to
(m N atom − 7m e ) + m beta − (m C atom − 6m e )
The mass of the seven electrons cancels out, and the difference in mass between the reactants and products
in the nuclear equation is equal to the difference in masses of the atoms.
m = 14.003241 amu − 14.003074 amu = 0.000167 amu
1g 1kg

0.000167 amu = 2.77 × 10 −31 kg
6.022 × 10 amu 1000 g
23
The energy is given by
2
8
2
E = mc = (2.77 × 10 −31 kg)(3.00 × 10 m/s) = 2.49 × 10 −14 J