CHAPTER 23
NUCLEAR CHEMISTRY
Problem Categories
Biological
: 23.88, 23.90, 23.95.
Conceptual
: 23.14, 23.15, 23.16, 23.27, 23.47, 23.49, 23.50, 23.51, 23.52, 23.59, 23.60, 23.61, 23.63, 23.64, 23.65,
23.71, 23.73, 23.76, 23.77, 23.81, 23.82, 23.84, 23.92, 23.96.
Descriptive
: 23.35, 23.36, 23.48, 23.57, 23.58, 23.62, 23.69, 23.75, 23.78.
Difficulty Level
Easy
: 23.5, 23.6, 23.14, 23.15, 23.16, 23.17, 23.18, 23.23, 23.25, 23.28, 23.29, 23.33, 23.34, 23.49, 23.55, 23.57,
23.62, 23.66, 23.68, 23.79, 23.80, 23.82.
Medium
: 23.13, 23.19, 23.20, 23.24, 23.26, 23.27, 23.35, 23.36, 23.47, 23.50, 23.51, 23.56, 23.58, 23.59, 23.60, 23.63,
23.65, 23.67, 23.69, 23.71, 23.72, 23.75, 23.76, 23.77, 23.81, 23.83, 23.84, 23.85, 23.93, 23.96.
Difficult
: 23.30, 23.48, 23.52, 23.53, 23.54, 23.61, 23.64, 23.70, 23.73, 23.74, 23.78, 23.86, 23.87, 23.88, 23.89, 23.90,
23.91, 23.92, 23.94, 23.95.
23.5
(a)
The atomic number sum and the mass number sum must remain the same on both sides of a nuclear
equation.
On the left side of this equation the atomic number sum is 13 (12
+
1) and the mass number
sum is 27 (26
+
1).
These sums must be the same on the right side.
The atomic number of X is
therefore 11 (13
−
2) and the mass number is 23 (27
−
4).
X is sodium
−
23 (
23
11
Na ).
(b)
11
Xis Hor p
(d)
56
26
Xis
Fe
(c)
1
0
Xis n
(e)
0
1
−
β
23.6
Strategy:
In balancing nuclear equations, note that the sum of atomic numbers and that of mass numbers
must match on both sides of the equation.
Solution:
(a)
The sum of the mass numbers must be conserved.
Thus, the unknown product will have a mass number
of 0.
The atomic number must be conserved. Thus, the nuclear charge of the unknown product must be
−
1.
The particle is a
β
particle.
135
135
53
54
IX
e
+
−
⎯⎯→
0
1
β
(b)
Balancing the mass numbers first, we find that the unknown product must have a mass of 40.
Balancing
the nuclear charges, we find that the atomic number of the unknown must be 20.
Element number 20 is
calcium (Ca).
40
0
19
1
K+
−
⎯⎯
→β
40
20
Ca
(c)
Balancing the mass numbers, we find that the unknown product must have a mass of 4.
Balancing the
nuclear charges, we find that the nuclear charge of the unknown must be 2.
The unknown particle is an
alpha (
α
) particle.
59
1
56
27
0
25
Co +
n
Mn +
4
2
α
(d)
Balancing the mass numbers, we find that the unknown products must have a combined mass of 2.
Balancing the nuclear charges, we find that the combined nuclear charge of the two unknown particles
must be 0.
The unknown particles are neutrons.
235
1
99
135
92
0
40
52
U+ n
S
r+
Te+2
1
0
n