1. Balancing Nuclear Equations
EXAMPLE 4.1
Balancing Nuclear Equations
Write balanced nuclear equations for each of the following processes. In each case, indicate what new element is formed.
a Plutonium-239 emits an alpha particle when it decays.
b Protactinium-234 undergoes beta decay.
c Carbon-11 emits a positron when it decays.
d Carbon-11 undergoes electron capture.
Solution
a. We start by writing the symbol for plutonium-239 and a partial equation showing that one of the products is an alpha particle (helium nucleus).
Mass and charge are conserved. The new element must have a mass of 239 – 4 = 235 and a charge of 94 – 2 = 92. The nuclear charge Z = 92 identifies the element as uranium (U).
b. Write the symbol for protactinium-234 and a partial equation showing that one of the products is a beta particle (electron).
Pu  He + ?
239 94 2 4
Pu  He U
235 92
Pa  e + ?
234 91 –1

2. Balancing Nuclear Equations continued
EXAMPLE 4.1
Balancing Nuclear Equations continued
The new element still has a nucleon number of 234. It must have a nuclear charge Z = 92 in order for the total charge to be the same on each side of the equation. The nuclear charge identifies the new atom as another isotope of uranium.
c. Write the symbol for carbon-11 and a partial equation showing that one of the products is a positron.
To balance the equation, a particle with A = 11 – 0 = 11and Z = 6 – 1 = 5 (boron) is required.
d. We write the symbol for carbon-11 and a partial equation showing it capturing an electron.
To balance the equation, the product must have A = = 11 and Z = 6 + (-1) = 5 (boron).
As we mentioned previously, positron emission and electron capture result in identical changes in atomic number, and therefore the identical elements are formed. Also, as parts (c) and (d) illustrate, carbon-11 (and certain other nuclei) can undergo more than one type of radioactive decay.
Pa  e U
234 91 –1 92
C  e + ? 11 6 +1
C  e + B 5
C + e  ?
C  e  B

3. Balancing Nuclear Equations continued
EXAMPLE 4.1
Balancing Nuclear Equations continued
Write balanced nuclear equations for each of the following processes. In each
case, indicate what new element is formed.
a Radium-226 decays by alpha emission.
b Sodium-24 undergoes beta decay.
c Gold-188 decays by positron emission.
d Argon-37 undergoes electron capture.
Exercise 4.1

4. More Nuclear Equations
EXAMPLE 4.2 5
More Nuclear Equations
In the upper atmosphere, a nitrogen-14 nucleus absorbs a neutron. A carbon-14 nucleus and another particle are formed. What is the other particle?
Solution
We start by writing the symbols for nitrogen-14 and a neutron (from Table 4.4) on the left of an equation, and the symbol for carbon-14 on the right.
The total of nucleon numbers on the left is 15, and on the right is 14, so the missing particle must have a nucleon number of = 1. The total atomic number on the left is 7 and on the right is 6, so the missing particle must have an atomic number of 1.
From Table 4.4, the particle with an atomic number of 1 and a nucleon number of
1 is a proton.
N + n  C + ? 14 7 1 6
N + n  C + ?
N + n  C + p
In one reaction that might be a future energy source, a hydrogen-2 nucleus combines with a hydrogen-3 nucleus, forming a helium-4 nucleus and another particle. What is the other particle formed?
Exercise 4.2

5. EXAMPLE 4.3 5
Half-Lives
You obtain a new sample of cobalt-60, half-life years, with a mass of 4.00 mg. How much cobalt-60 remains after years (three half-lives)?
Solution
The fraction remaining after three half-lives is
The amount of cobalt-60 remaining is ( ) (4.00 mg) = 0.50 mg.
= = = 1 2n 23
2 x 2 x 2 8
You have mg of freshly prepared gold-189, half-life 30 min. How much of the gold-189 sample remains after five half-lives?
Exercise 4.3A
The half-life of phosphorus-32 is 14.3 days. What percentage of an original sample’s radioactivity remains after four half-lives?
Exercise 4.3B

6. EXAMPLE 4.4 5
Half-Lives
You obtain a 2.60-mg sample of mercury-190, half-life 20 min. How much of the mercury-190 sample remains after 2.0 h?
Solution
There are 120 min and thus ( ) = 6 half-lives in 2.0 h. The fraction remaining after six half-lives is
The amount of mercury-190 remaining is ( )(2.60) mg = mg.
= = = 1 2n 26
2 x 2 x 2 x 2 x 2 x 2 64
120 20
A sample of 16.0 mg of nickel-57, half-life 36.0 h, is produced in a nuclear reactor. How much of the nickel-57 sample remains after 7.5 days?
Exercise 4.4A
Technetium-99 decays to ruthenium-99 with a half-life of 210,000 y. Starting with 1.00 mg of technetium-99, how long will it take for 0.75 mg of ruthenium-99 to form?
Exercise 4.4B

7. EXAMPLE 4.5 5 Radioisotopic Dating
An old wooden implement has carbon-14 activity one-eighth that of new wood. How old is the artifact? The half-life of carbon-14 is 5730 years.
Solution
Using the relationship
we see that one-eigth is , where n = 3; that is, the fraction is The carbon-14 has gone through three half-lives. The wood is therefore about 3 x 5730 = 17,190 years old.
Fraction remaining = 1 2n 8 23
How old is a piece of a fur garment that has carbon-14 activity that of living tissue? The half-life of carbon-14 is 5730 years.
Exercise 4.5A 1 16
Strontium-90 has a half-life of 28.5 years. How long will it take for the strontium-90 now on Earth to be reduced to of its present amount?
Exercise 4.5B 32 1

8. Artificial Transmutation Equations
EXAMPLE 4.6 5
Artificial Transmutation Equations
When potassium-39 is bombarded with neutrons, chlorine-36 is produced. What other particle is emitted?
K + n  Cl + ? 19 39 1 17 36
Solution
Write a balanced nuclear equation. To balance the equation, we need four mass units and two charge units (that is, a particle with and A = 4 and Z = 2—an alpha particle).
K + n  Cl + He 19 39 1 17 36 2 4
Mo + H  Tc + ?
Technetium-97 is produced by bombarding molybdenum-96 with a deuteron (hydrogen-2 nucleus). What other particle is emitted?
Exercise 4.6A 42 96 1 2 43 97
Some scientists want to form uranium-235 by bombarding another nucleus with an alpha particle. They expect a neutron to be produced along with the uranium-235 nucleus. What nucleus must be bombarded with the alpha particle?
Exercise 4.6B

9. Positron Emission Equations
EXAMPLE 4.7 5
Positron Emission Equations
One of the isotopes used for PET scans is oxygen-15, a positron emitter. What new element is formed when oxygen-15 decays?
Solution
First write the nuclear equation
The nucleon number A does not change, but the atomic number Z becomes 8 – 1 = 7; and so the new product is nitrogen-15.
O  e + ? 15 8 +1
O  e + N 7
Phosphorus-30 is a positron-emitting radioisotope suitable for use in PET scans. What new element is formed when phosphorus-30 decays?
Exercise 4.7

10. CONCEPTUAL EXAMPLE 4.8 Radiation Hazard
Most modern fire detectors contain a tiny amount of americium-241, a solid element that is an alpha emitter. The americium is in a chamber that includes thin aluminum shielding, and the device is usually mounted on a ceiling. Rate the radiation hazard of this device in normal use as either (a) high, (b) moderate, or (c) very low, and explain your choice.
Solution
Alpha particles have very little penetrating ability—they are stopped by a sheet of paper or a layer of skin. The emitted alpha particles cannot exit the chamber, because even thin metal is more than sufficient to absorb the alpha particles. Even if alpha particles could escape, they would likely be stopped by the dead cells of your hair, or by the air after a short distance. Therefore, the radiation hazard is probably best rated as (c)—very low.
Radon-222 is a gas that can diffuse from the ground into homes. Like americium-241, it is an alpha emitter. Is the radiation hazard from radon-222 likely to be higher or lower than that from a smoke detector? Explain.
Exercise 4.8