Nuclear Chemistry Worksheet
1.
Nuclear Reactions
Nuclear reactions are rearrangements of the constituents of one nucleus into another nuclear species.
During the reaction, three quantities are conserved: energy, charge, and baryon number (mass number).
2.
Nuclear Decay Particles
The forms of radiation of interest to us are α particles, β particles, β+ particles, and γ particles.
In addition, the constituent particles of the nucleus, protons and neutrons, are necessary components of
nuclear reactions. Each particle has a specific mass number (A) and charge (Z) associated with its
symbol in the usual way: AZX.
a) α particles are helium nuclei: 42He 2+
b) β particles are high-energy electrons: 0-1e ‒
c) β+ particles are positrons, the anti-particles of electrons: 01e +
d) γ particles are high-energy photons: 00γ
In addition, nuclear reactions may involve protons ( 11p.), neutrons ( 10n), or capture of a core electron
(0-1e ‒)
3.
Nuclear Decay Reactions
The most common form of nuclear decay reactions are the following:
a) α decay – emission of an α particle from the nucleus: 23592U xxv 42He 2+ + 23190Th
Exercise: predict the products of the following reactions:
b) β decay - emission of a β particle from the nucleus:
7
0
‒
2He xxv -1e +
c) β+ decay - emission of a positron (β+) particle from the nucleus:
8
0
+
5B xxv 1e +
d) “isomeric transition” or IT : γ decay - loss of a γ particle:
34
0
17Cl xxv 0γ +
e) “electron capture” or EC: capture of a core electron of the atom by the nucleus
7
0
‒
4Be + -1e xxv
4.
Kinetics of Nuclear Decay
All spontaneous nuclear decay processes are 1st order. This makes it convenient to discuss the kinetics
in terms of half-lives and activity (the number of decays per second). For example, the half-life of
241
Am is 432 years, or 1.364×1010 s.
From the relation between the rate constant and half-life of a 1st order reaction ( k τ½ = ln 2), the rate
constant is k = 5.082×10–11 s–1.
From the rate law, we can determine the activity, that is, the number of decays, from a mass of w
grams of 241Am (241.05682 amu). This is usually expressed in terms of Curies (1 Ci = 3.70×1010 s–1).
ΔN
w
Activity = −
= k N = k
NA
where M is the atomic mass
Δt
M
OR
NA
w
k
Ci
M
3.70 × 1010 s−1
Exercise: Show that the specific activity (the activity per gram) of americium-241 is 3.43 Ci/g
Activity =
5.
Nuclear Reactions
a) nuclear synthesis
b) spontaneous fission
c) induced fission
d) fusion
Exercise: balance the following reactions:
i) 24998Cf +
xxv 263106Sg + 4 10n
ii) 31H + 21H xxv10n +
iii) 23892U xxv 14157La + 10n +
iv) 24195Am xxv
+ 8 42He + 4 0-1e
6.
Thermodynamics of Nuclear Reactions
Nuclear energy changes are related to changes in mass through a familiar relation:
E = m c2
→→→
Δ E = (Δ m) c 2
E measured in Joules
m measured in kilograms
c = speed of light measured in
= 2.998 × 10 8
meters
seconds
meters
seconds
Although the mass change may be small, the corresponding energy change are large.
a) Nuclear Binding Energies
As a nucleus is formed, mass is converted to energy; this energy loss is the stabilization of the of
the nucleons (protons and neutrons) within the nucleus.
Consider a single 19F atom:
i) Given the mass of a proton is 1.0072765 amu and the mass of a neutron is 1.0086649 amu,
determine the expected mass of a 19F atom.
ii) The experimentally determined mass of the fluorine-19 nucleus is 18.9984032amu; what is
the mass lost as the nucleus forms?
iii) Determine the energy which corresponds to this loss of mass for one nucleus. for a mole of
nuclei.
iv) Determine the binding energy for F-19 per nucleon.
One useful unit of energy used in nuclear processes is the electron volt (eV). It is the amount of
kinetic energy an electron gains as it accelerates between two charged plates separated by 1 cm
with a 1 volt difference on the plates. 1 eV is 1.6021766 × 10 ‒19J; in more familiar terms, 1eV
corresponds to 96.485 kJ/mol.
Exercise: Show that the binding energy of Iron-56 ( 55.9349 amu) is 8.5537 MeV per nucleon.
b) Energies of nuclear reactions
i) Fission
One reaction in nuclear reactors is:
1
n + 235U → 236U → 92Kr + 141Ba + 3 1n
Given the following masses, determine the energy available from this process.
Isotope Atomic Mass
(amu)
1
n
1.00866
235
U
235.0439
91
Kr
90.9234
142
Ba
141.9164
ii) Fusion
The easiest fusion reaction to initiate is
2
H + 3H → 4He + 1n
Calculate the energy released per nucleus of 4He produced and per mole produced.
Isotope Atomic Mass
(amu)
1
n
1.00866
2
H
2.01410
3
H
3.01605
He
4.00260
4