Definitions and Concepts:

Radioactivity results from the spontaneous decay of an atom's nucleus.

There are  three main types of nuclear decay: alpha, beta and gamma.  Each results in a characteristic change in the parent nucleus and is characterized by the formation/ejection of a specific particle or electromagnetic radiation.

Describe each type of decay showing the particle or EM radiation ejected.

Alpha:
 

Beta [Beta(-)]:
 

Gamma:
 

There are additional two less common modes of nuclear decay, describe them:

Positron [Beta(+)]:
 

Electron Capture:
 

Because electrons and positrons (anti-electrons) have negligible mass compared to the mass of neutrons and protons, we can view the nuclear transformations as transformations between nucleons (protons & electrons in the nucleus).  Write the equations for the neutron/proton transformations occurring as a result of the following processes:

Beta(-):
 

Beta(+):
 

EC:
 

The symbols for a nuclide (a specific type of nucleus of any given element) is given by an Atomic Symbol bearing a preceding superscript bearing the "nominal mass number" and a preceding subscript bearing the "atomic number".

Elementary Particles (alphas, protons, neutrons, electrons, positrons) also have symbols.  Write them:
 
 

Balanced Nuclear Equations have the same mass and atomic number sums on each side of the equation.  The pre-superscripts (times their coefficients, if any) will add to the same number.  The pre-subscripts (times their coefficients, if any) will add to the same number.

Balance the following adding the appropriate elementary particles:
( "/" => goes to)

Radium(226)/Radon(222)
 

Technecium(95)/Molybdenium(95)
 

Carbon(14)/Nitrogen(14)
 

Potassium(40)/Argon(40)
 

Nuclear Fission:  Uranium(235)/Barium(139)+Krypton(94)
 
 

Nuclear Fusion:  Hydrogen(2)+Hydrogen(3)/Helium(4)
 

Nuclear Bombardment:

Nitrogen(14)+Helium(4)/Oxygen(17)
 

Berylium(9)+Helium(4)/Carbon(12)
 

Nitrogen(14)/Carbon(14)+Hydrogen(1)
 

Carbon(14)/Nitrogen(14)
 

Protons:          2    8    20    28    50    82   (126)
Neutrons:        2    8    20    28    50    82    126    (~183)
(unknown superheavy elements predicted to be unusually stable)

Elements with magic numbers of neutrons and protons, i.e. "double magic" are especially stable:  Helium(4), Oxygen(16), Calcium(40), Lead(208).  Elements with magic numbers of protons will generally have larger numbers of stable isotopes.  Why do you think this is?
 
 

The rate of decay (or activity) of a nuclide is a first-order kinetic process.
Since concentration is not applicable the number of parent nuclei is used, a rate constant applies and a rate of decay (an activity) can be calculated.

Write the rate law as it applies to nuclear decay.
 
 

Activities in Curies [Rates of decay relative to the rate of one gram of Radium(226)] are common units for buying and selling activity.  It is usually the radiation one is after when a radionuclide is purchased.

1 Curie, Ci, = 3.7 x 10exp(10) disintegrations per second, the rate of decay of one gram of Radium(226), a lot of activity, Radium(226) at this level glows white-hot!  Millicuries and microcuries are much more commonly purchased.

Since radioactive decay is a first-order kinetic process the half life and the rate constant are related.  Write that equation:
 
 

The activity of a sample determines the rate of decay in disintegrations per second and divides it by 3.7 x 10exp(10) disintegrations per second to get the activity in Curies.
 
 

The integrated First-order Rate Equation allows one to calculate the fraction or the number of nuclei remaining after an elapsed time.  Write that equation:
 
 

Determine log(Nt/N0) relative to the half-life.  Write that equation:
 
 

Radioactive dating infers the fraction (Nt/N0) from the amounts of daughter and parent nuclides present, i.e. (parent nuclide/parent+daughter) = (Nt/N0) and an elapsed time can be calculated knowing the half-life or the rate constant.
 

Radiocarbon dating is similar except that the amount of carbon(14) in the atmosphere is "constant" = N0, and the amount of carbon(14) remaining in a once-living organism (at equilibrium with atmospheric carbon(14) until its death!) is = Nt.  From this ratio and the known half-life an elapsed time (= age since death) can be calculated.  This is good to about 50,000 years ago (a little less than ten half-lives of 5730 years each).  The constant levels of carbon(14) are formed by solar neutron bombardment of nitrogen(14) in the atmosphere.  The solar flux varies only a little in a cyclic manner and can except for the most accurate dating be considered constant.

Example Problem:
Calculate the age of a jawbone which decays at 4.5 disintegrations per second per gram of contained carbon, and the atmospheric constant rate of 15.3 disintegrations per second per gram of contained carbon.
 
 

Nuclear energy comes from the conversion of matter into energy.  Write the equation for this conversion, the famous Einstein equation:
 

To do these calculations we must use the "exact mass" of nuclides and nucleons and the balanced nuclear equation to calculate a change in mass.  Write the equation for a (delta)m.
 
 

The "binding  energy" of a nuclide is the "mass loss" or "mass defect" between the exact mass of a nuclide and the sum of the free masses of the protons and neutrons which are known to compose it substituted into the Einstein equation.  Write the equation for the binding energy and solve it for the formation of one mole of Helium(4).
 
 
 
 
 
 

The binding energy of a nucleus is almost a billion times greater than a chemical bond, about 100 kJ/mole.
 

The binding energy divided by the number of nucleons (#p+#n) composing it is a common way to describe the stability of nuclides.  This "binding energy per nucleon" can be seen in Figure 19.6 (p 835) in your text.  Iron is the most stable nucleus, magic nuclides and double magic nuclides are more stable than their neighbors, nuclides to the left and right of iron are less stable.  Heavy elements breaking "in half" is the basis for fission energy, light nuclei fusing to heavier nuclides is the basis for fusion energy.

Example Problem:
Show that Helium(4) is more stable than Helium(3).