The Advanced Placement
Examination in Chemistry
Part II - Free Response Questions & Answers
1970 to 2007
Nuclear
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Advanced Placement Examination in Chemistry. Questions
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Nuclear Chemistry
page 2
1989 D
The carbon isotope of mass 12 is stable. The carbon isotopes of mass 11 and mass 14 are unstable. However, the
type of radioactivity decay is different for these two isotopes. Carbon-12 is not produced in either case.
(a) Identify a type of decay expected for carbon-11 and write the balanced nuclear reaction for that decay process.
(b) Identify the type of decay expected for carbon-14 and write the balanced nuclear reaction for that decay
process.
(c) Gamma rays are observed during the radioactive decay of carbon-11. Why is it unnecessary to include the
gamma rays in the radioactive decay equation of (a)?
(d) Explain how the amount of carbon-14 in a piece of wood can be used to determine when the tree died.
Answer:
11
11
0
(a) Positron decay: 6C 5B1
OR
11
0  11
Electron capture: 6C1e  5B
14
14
0
(b) Beta decay: 6C 7N1
(c) Gamma rays have no mass or charge (or they are energy) so they need not be shown in nuclear equations.
(d) Measure the amount of C-14 in the dead wood. Compare with the amount of C-14 in a similar living object.
1991 D
Explain each of the following in terms of nuclear models.
(a) The mass of an atom of 4He is less than the sum of the masses of 2 protons, 2 neutrons, and 2 electrons.
(b) Alpha radiation penetrates a much shorter distance into a piece of material than does beta radiation of the
same energy.
(c) Products from a nuclear fission of a uranium atom such as 90Sr and 137Ce are highly radioactive and decay by
emission of beta particles.
(d) Nuclear fusion requires large amounts of energy and to get started, whereas nuclear fission can occur spontaneously, although both processes release energy.
Answer:
(a) When nucleons are combined in nuclei, some of their mass (mass defect) is converted into energy (binding energy) which is released and stabilizes the nucleus.
(b) Alpha particles have a greater mass than beta particles. Thus, their speed (penetrating potential) is less.
(c) The neutron/proton ratio in Sr-90 and Cs-137 is too large and they emit beta particles (converting neutrons
into protons) to lower this ratio.
(d) Large amounts of energy are needed to initiate fusion reactions in order to overcome the repulsive forces
between the positively charged nuclei. Large amounts of energy are not required to cause large unstable nuclei to split apart.
1997 D
Answer each of the following questions regarding radioactivity.
(a) Write the nuclear equation for decay of 234
94 Pu by alpha emission.
(b) Account for the fact that the total mass of the products of the reaction in part (a) is slightly less than that of
the original 234
94 Pu .
Nuclear Chemistry
page 3
(c) Describe how , , and  rays each behave when they pass through an electric field. Use the diagram below
to illustrate your answer.
(d) Why is it not possible to eliminate the hazard of nuclear waste by the process of incineration?
Answer:
4
230
(a) 234
94 Pu  2  + 92 U
Due to a printing error, the student’s answer booklet had the Pu-239 isotope. Therefore, the following is a valid response.
239
94
Pu  24 +
235
92
U
(b) This mass defect has been converted into energy. E = mc2
(c) An alpha particle,  or He nuclei, has a 2+ charge and would be attracted to the (-) side of the electric field.
A beta particle, , or electron, has a single negative charge and is attracted to the positive side of the electric field, but since it is much lighter and faster than an alpha it would not be as strongly deflected. Gamma,
, rays are not charged and, therefore, not deflected by the electric field.
(d) The half-life of a radionuclide is independent of its environment. Incineration will neither accelerate its decay nor render it non-radioactive. Half-life is a function of its nucleus, incineration is a function of its electrons.
2007 part A, form B, question #2 (repeated in the atomic theory section)
Answer the following problems about gases.
(a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two common isotopes of
naturally occurring neon as indicated in the table below.
Isotope
Mass (amu)
Ne-20
19.99
Ne-22
21.99
Nuclear Chemistry
page 4
(i) Using the information above, calculate the percent abundance of each isotope.
(ii) Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occurring neon.
(b) A major line in the emission spectrum of neon corresponds to a frequency of 4.341014 s-1. Calculate
the wavelength, in nanometers, of light that corresponds to this line.
(c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as
shown by the equation below. Ozone serves to block harmful ultraviolet radiation that comes from the
Sun.
O3(g) UV
 O2(g) + O(g)
A molecule of O3(g) absorbs a photon with a frequency of 1.001015 s-1.
(i) How much energy, in joules, does the O3(g) molecule absorb per photon?
(ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kJ mol-1. Does a photon
with a frequency of 1.001015 s-1 have enough energy to break this bond? Support your answer with a
calculation.
Answer:
(a) (i) let X = decimal percentage of Ne-20, then (1-X) = decimal percentage of Ne-22
19.99X + 21.99(1-X) = 20.18; X = 0.905 or 90.50% Ne-20 and 9.500% Ne-22
(ii) (12.55 g)(9.5%) 
1 mol Ne 6.02  10 23 atoms
= 3.5571022 atoms

20.18 g
1 mol

3.0  10 8 m s-1 10 9 nm
(b) c = ;  = 
= 690 nm

c 4.34  1014 s-1
1m
(c) (i) E = h = (6.6310-34 J s)(1.001015 s-1) = 6.6310-19 J
6.02  10 23 1 kJ
(ii) 6.6310-19 J 
 3 = 399 kJ; this is more than 387 kJ, so there is enough energy to
1 mol
10 J
break the bond.
Copyright ⌐ 1970 to 1998 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom teachers
are permitted to reproduce the questions. Portions copyright ⌐ 1993-8 Unlimited Potential, Framingham, MA 01701-2619.