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

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

Karen C Timberlake

Solution

a. An alpha (α) particle, , has two protons and two neutrons.

b. A beta (β) particle, , 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

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

Sample Problem 5.1 Radiation Particles

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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)

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

Sample Problem 5.2 Writing an Equation for Alpha Decay

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Karen C Timberlake

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

Study Check 5.2

Write the balanced nuclear equation for the alpha decay of Po-214

Answer

Continued

Sample Problem 5.2 Writing an Equation for Alpha Decay

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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 = ?

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)

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

Sample Problem 5.3 Writing a Nuclear Equation for Beta Decay

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Karen C Timberlake

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

Study Check 5.3

Write the balanced nuclear equation for the beta decay of chromium-51

Answer

Continued

Sample Problem 5.3 Writing a Nuclear Equation for Beta Decay

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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 = ?

25 = ? + 1

25 – 1 = ?

25 – 1 = 24 (atomic number of new nucleus) Write the balanced nuclear equation for manganese-49, which decays by emitting a positron

Sample Problem 5.4 Writing a Nuclear Equation for Positron Emission

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Karen C Timberlake

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

Study Check 5.4

Write the balanced nuclear equation for xenon-118, which undergoes positron emission

Answer

Continued

Sample Problem 5.4 Writing a Nuclear Equation for Positron Emission

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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 = ?

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) Write the balanced nuclear equation for the bombardment of nickel-58 by a proton, , which produces a radioactive isotope and an alpha particle

Sample Problem 5.5 Writing an Equation for an Isotope Produced by Bombardment

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Karen C Timberlake

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

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

Continued

Sample Problem 5.5 Writing an Equation for an Isotope Produced by Bombardment

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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

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?

Sample Problem 5.6 Radiation Measurement

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Karen C Timberlake

Solution

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?

Sample Problem 5.7 Using Half-Lives of a Radioisotope

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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

Continued

Sample Problem 5.7 Using Half-Lives of a Radioisotope

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Karen C Timberlake

Solution

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?

Sample Problem 5.8 Dating Using Half-Lives

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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

Continued

Sample Problem 5.8 Dating Using Half-Lives

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Karen C Timberlake

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

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

Sample Problem 5.9 Medical Applications of Radioactivity

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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

Continued

Sample Problem 5.9 Medical Applications of Radioactivity

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Karen C Timberlake

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

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

Sample Problem 5.10 Identifying Fission and Fusion

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