Nuclear Chemistry Worksheet

1. Nuclear ReactionsNuclear 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 ParticlesThe 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: A

ZX.a) α particles are helium nuclei: 4

2He 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 ( 1

0n), or capture of a core electron (0

-1e ‒)

3. Nuclear Decay ReactionsThe most common form of nuclear decay reactions are the following:

a) α decay – emission of an α particle from the nucleus: 23592U xxv 4

2He 2+ + 23190Th

Exercise: predict the products of the following reactions:

b) β decay - emission of a β particle from the nucleus: 7

2He xxv 0-1e ‒ +

c) β+ decay - emission of a positron (β+) particle from the nucleus: 8

5B xxv 01e + +

d) “isomeric transition” or IT : γ decay - loss of a γ particle: 34

17Cl xxv00γ +

e) “electron capture” or EC: capture of a core electron of the atom by the nucleus 7

4Be + 0-1e ‒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 241Am 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).

Activity = −Δ NΔ t

= k N = k wM

N A where M is the atomic mass

OR

Activity =wM

kN A

3.70× 1010 s−1Ci

Exercise: Show that the specific activity (the activity per gram) of americium-241 is 3.43 Ci/g

5. Nuclear Reactionsa) nuclear synthesisb) spontaneous fissionc) induced fissiond) fusion

Exercise: balance the following reactions:i) 249

98Cf + xxv 263106Sg + 4 1

0nii) 3

1H + 21H xxv10n +

iii) 23892U xxv 141

57La + 10n +

iv) 24195Am xxv + 8 4

2He + 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 = (Δm)c2

E measured in Joulesm measured in kilograms

c = speed of light measured in metersseconds

= 2.998 × 108 metersseconds

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 reactionsi) Fission

One reaction in nuclear reactors is:1n + 235U → 236U → 92Kr + 141Ba + 3 1n

Given the following masses, determine the energy available from this process.

Isotope Atomic Mass (amu)

1n 1.00866235U 235.043991Kr 90.9234

142Ba 141.9164

ii) FusionThe easiest fusion reaction to initiate is

2H + 3H → 4He + 1n

Calculate the energy released per nucleus of 4He produced and per mole produced.

Isotope Atomic Mass (amu)

1n 1.008662H 2.014103H 3.01605

4He 4.00260