Nuclear Chemistry ❮ 267❯ Answers and Explanations
- D—The mass should be 226 – (4 + 4 + 0 + 4) =
214. The atomic number should be 88 – (2 +
2 – 1 + 2) = 83. - B—The calculations are as follows: mass
difference = 234 – 230 = 4; atomic number dif-
ference = 92 – 90 = 2. These correspond to an
a particle. - C—Alpha particles are the least penetrating, and
gamma rays are the most penetrating. - A—The complete symbol for a b particle is –1^0 b or
0
–1e. As indicated by the symbol, the mass number
is zero. Beta particles are electrons, which means
they have a –1 charge. - B—The calculations are as follows: mass
number = 238 – 4 = 234; atomic number =
92 – 2 = 90. - C—After one half-life, 50% would remain.
After another half-life, this amount would be
reduced by one-half to 25%. The total amount
decayed is 75%. Thus, 24.6 years must be two
half-lives of 12.3 years each.
7. D—^1 H has no mass defect because there is noth-
ing to bind to the single proton in the nucleus.
8. B—During electron capture, an electron is
absorbed into the nucleus, which will alter the
mass defect and change the atomic number. Beta
decay also involves electrons.
9. D—The half-life of carbon-14 (5,730 years) is
much too short to determine an age in the billions
of years.
10. C—It is necessary to determine both the mass
number and the atomic number of the nucleus
formed. The mass number difference depends
on the superscripts. The total on each side of
the reaction arrow must be identical. Mass dif-
ference = 236 - 4(1) - 136 = 96. The atomic
number difference depends on the subscripts.
The total on each side of the reaction arrow
must be identical. Atomic number difference =
92 - 4(0) - 53 = 39.
❯ Free-Response Question
You have 10 minutes to answer the following question. You may use a calculator and the
tables in the back of the book.Question
It is possible to determine the age of a geological sample by determining the amounts of
uranium-238 and lead-206 present. Uranium-238 decays to lead-206 with a half-life of
4.47 × 109 years. A mineral sample contains 47.6 mg of uranium-238 and 16.4 mg of
lead-206. Answer the following two questions concerning this sample.
(a) How many milligrams of uranium-238 were originally in the sample?
(b) What is the age of the sample?