General orrganic and biological chemistry structures off liffe 5th by karen timberlake05 worked examples

Sample Problem 5.1 Radiation Particles
Identify and write the symbol for each of the following types of radiation:
a.

contains two protons and two neutrons

b.

has a mass number of 0 and a 1– charge

Solution
a.

An alpha (α) particle,

b.

A beta (β) particle,

, has two protons and two neutrons.
, is like an electron with a mass number of 0 and a 1– charge.

Study Check 5.1
Identify and write the symbol for the type of radiation that has a mass number of zero and a 1+ charge.

Answer
A positron,

, has a mass number of 0 and a 1+ charge.

General, Organic, and Biological Chemistry: Structures of Life, 5/e

Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.2 Writing an Equation for Alpha Decay
Smoke detectors that are used in homes and apartments contain americium-241, which undergoes alpha decay. When alpha particles collide with air molecules, charged particles are produced that
generate an electrical current. If smoke particles enter the detector, they interfere with the formation of charged particles in the air,
and the electrical current is interrupted. This causes the alarm to sound and warns the occupants of the danger of fire. Write the balanced nuclear equation for the decay of americium-241.

Solution

Step 1

Write the incomplete nuclear equation.

Step 2

Determine the missing mass number. In the equation, the mass number of the americium, 241, is equal to the sum of the mass numbers of the new nucleus and the alpha particle.
241

=?+4

241 – 4 = ?
241 – 4 = 237 (mass number of new nucleus).

Step 3

Determine the missing atomic number. The atomic number of americium, 95, must equal the sum of the atomic numbers of the new nucleus and the alpha particle.
95

=?+2

95 – 2 = ?
95 – 2 = 93 (atomic number of new nucleus)

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.2 Writing an Equation for Alpha Decay
Continued
Step 4

Determine the symbol of the new nucleus. On the periodic table, the Element that has atomic number 93 is neptunium, Np. The nucleus of this isotope of Np is written as

Step 5

Complete the nuclear equation.

Study Check 5.2
Write the balanced nuclear equation for the alpha decay of Po-214.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.3 Writing a Nuclear Equation for Beta Decay
The radioactive isotope yttrium-90, a beta emitter, is used in cancer treatment and as a colloidal injection into large joints to relieve the pain caused by arthritis. Write the balanced nuclear equation
for the beta decay of yttrium-90.

Solution

Step 1

Write the incomplete nuclear equation.

Step 2

Determine the missing mass number. In the equation, the mass number of the yttrium, 90, is equal to the sum of the mass numbers of the new nucleus and the beta particle.
90

=?+0

90 – 0 = ?
90 – 0 = 90 (mass number of new nucleus)
Step 3

Determine the missing atomic number. The atomic number of yttrium, 39, must equal the sum of the atomic numbers of the new nucleus and the beta particle.
39

=?–1

39 + 1 = ?

39 + 1 = 40 (atomic number of new nucleus)

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.3 Writing a Nuclear Equation for Beta Decay
Continued
Step 4

Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 40 is zirconium, Zr. The nucleus of this isotope of Zr is written as

Step 5

Complete the nuclear equation.

Study Check 5.3
Write the balanced nuclear equation for the beta decay of chromium-51.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.4 Writing a Nuclear Equation for Positron Emission

Write the balanced nuclear equation for manganese-49, which decays by emitting a positron.

Solution

Step 1

Write the incomplete nuclear equation.

Step 2

Determine the missing mass number. In the equation, the mass number of the manganese, 49, is equal to the sum of the mass numbers of the new nucleus and the positron.
49

=?+0

49 – 0 = ?
49 – 0 = 49 (mass number of new nucleus)

Step 3

Determine the missing atomic number. The atomic number of manganese, 25, must equal the sum of the atomic numbers of the new nucleus and the positron.
25

=?+1

25 – 1 = ?
25 – 1 = 24 (atomic number of new nucleus)

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.4 Writing a Nuclear Equation for Positron Emission
Continued
Step 4

Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 24 is chromium (Cr). The nucleus of this isotope of Cr is written as

Step 5

Complete the nuclear equation.

Study Check 5.4
Write the balanced nuclear equation for xenon-118, which undergoes positron emission.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.5 Writing an Equation for an Isotope Produced by Bombardment
Write the balanced nuclear equation for the bombardment of nickel-58 by a proton,

, which produces a radioactive isotope and an alpha particle.

Solution

Step 1

Write the incomplete nuclear equation.

Step 2

Determine the missing mass number. In the equation, the sum of the mass numbers of the proton, 1, and the nickel, 58, must equal the sum of the mass numbers of the new nucleus and the
alpha particle.
1 + 58 = ? + 4
59 – 4 = ?
59 – 4 = 55 (mass number of new nucleus)

Step 3

Determine the missing atomic number. The sum of the atomic numbers of the proton, 1, and nickel, 28, must equal the sum of the atomic numbers of the new nucleus and the alpha particle.
1 + 28 = ? + 2
29 – 2 = ?
29 – 2 = 27 (atomic number of new nucleus)

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.5 Writing an Equation for an Isotope Produced by Bombardment
Continued
Step 4

Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 27 is cobalt, Co. The nucleus of this isotope of Co is written as

Step 5

Complete the nuclear equation.

Study Check 5.5
The first radioactive isotope was produced in 1934 by the bombardment of aluminum-27 by an alpha particle to produce a radioactive isotope and one neutron. Write the balanced nuclear equation for this
bombardment.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.6 Radiation Measurement
One treatment for bone pain involves intravenous administration of the radioisotope phosphorus-32, which is incorporated into bone. A typical dose of 7 mCi can produce up to 450 rad in the bone. What is
the difference between the units of mCi and rad?

Solution
The millicuries (mCi) indicate the activity of the P-32 in terms of nuclei that break down in 1 s. The radiation absorbed dose (rad) is a measure of amount of radiation absorbed by the bone.

Study Check 5.6
For Sample Problem 5.6, what is the absorbed dose of radiation in grays (1Gy)?

Answer

4.5 Gy

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.7 Using Half-Lives of a Radioisotope
Phosphorus-32, a radioisotope used in the treatment of leukemia, has a half-life of 14.3 days. If a sample contains 8.0 mg of phosphorus-32, how many milligrams of phosphorus-32 remain after
42.9 days?

Solution

Step 1

State the given and needed quantities.

Step 2

Write a plan to calculate the unknown quantity.

Step 3

Write the half-life equality and conversion factors.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 5.7 Using Half-Lives of a Radioisotope
Continued
Step 4

Set up the problem to calculate the needed quantity. First, we determine the number of half-lives in the amount of time that has elapsed.

Now we can determine how much of the sample decays in three half-lives and how many grams of the phosphorus remain.

Study Check 5.7
Iron-59 has a half-life of 44 days. If a nuclear laboratory receives a sample of 8.0 μg of iron-59, how many micrograms of iron-59 are still active after 176 days?

Answer
0.50 μg of iron-59

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.8 Dating Using Half-Lives
Carbon material in the bones of humans and animals assimilates carbon until death. Using radiocarbon dating, the number of half-lives of carbon-14 from a bone sample determines the age of the
bone. Suppose a sample is obtained from a prehistoric animal and used for radiocarbon dating. We can calculate the age of the bone or the years elapsed since the animal died by using the half-life
of carbon-14, which is 5730 yr. If the sample shows that four half-lives have passed, how much time has elapsed since the animal died?

Solution

Step 1

State the given and needed quantities.

Step 2

Write a plan to calculate the unknown quantity.

Step 3

Write the half-life equality and conversion factors.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.8 Dating Using Half-Lives
Continued
Step 4

Set up the problem to calculate the needed quantity.

We would estimate that the animal lived 23 000 yr ago.

Study Check 5.8
Suppose that a piece of wood found in a tomb had one-eighth of its original carbon-14 activity. About how many years ago was the wood part of a living tree?

Answer
17 200 yr

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.9 Medical Applications of Radioactivity
In the treatment of abdominal carcinoma, a person is treated with “seeds” of gold-198, which is a beta emitter. Write the balanced nuclear equation for the beta decay of gold-198.

Solution

We can write the incomplete nuclear equation starting with gold-198.

In beta decay, the mass number, 198, does not change, but the atomic number of the new
nucleus increases by one. The new atomic number is 80, which is mercury, Hg.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.9 Medical Applications of Radioactivity
Continued

Study Check 5.9
In an experimental treatment, a person is given boron-10, which is taken up by malignant tumors. When bombarded with neutrons, boron-10 decays by emitting alpha particles that destroy the surrounding
tumor cells. Write the balanced nuclear equation for the reaction for this experimental procedure.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 5.10 Identifying Fission and Fusion
Classify the following as pertaining to fission, fusion, or both:
a.

A large nucleus breaks apart to produce smaller nuclei.

b.

Large amounts of energy are released.

c.

Extremely high temperatures are needed for reaction.

Solution
a.

When a large nucleus breaks apart to produce smaller nuclei, the process is fission.

b.

Large amounts of energy are generated in both the fusion and fission processes.

c.

An extremely high temperature is required for fusion.

Study Check 5.10
Classify the following nuclear equation as pertaining to fission, fusion, or both:

Answer
When small nuclei combine to release energy, the process is fusion.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.