Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Fall 2003
Lecture Notes::Lec 40_12 December	© R. Paselk 2003
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Nuclear Chemistry

Radioactive Decay, cont.

Radioactivity Demo - Geiger counter and radioactive objects. (Geiger counter overhead). Note differentiation of gamma and beta decay by shielding.

We looked at the three most common forms of decay last time: a-decay (A decreases by 2, Z decreases by 4), b-decay (A constant, Z increases by 1), and g-decay. Can also have positron (positive electron) emission (A constant, Z decreases by 1) and K(inner electron)-capture (A constant, Z decreases by 1). See Table 21.2 in text.

Nuclear Decay Series: The first two reactions we looked at last time demonstrated the beginning of a common phenomena, a nuclear decay series, wherein a decay process involves a series of sequential decays (overhead - note alpha [diagonal] vs. beta [horizontal] decays):

Rates of Nuclear Decay

In parallel with chemical systems we've seen in the past, the stability of a nucleus is characterized by two factors. Last time we looked at thermodynamic stability, that is the amount of energy given off when the nucleus decays, and thus how favorable the reaction is. We now will focus on kinetic stability, that is how rapidly a collection of nuclei decays. (Note that for an individual nucleus this is expressed as the probability it will decay during a given time interval. For a long lived species, the probability of decay is very low and vice-versa.)

The decay of nuclides is independent of factors outside the individual nucleus - it is unaffected by environment. This means it must be a first order process in which the rate depends only on the concentration of unstable nuclei. If we think back to chemical kinetics in Chem 109:

Recall for first order decay reactions:

• Rate = -dN/dt = -kN, where k is a rate constant characteristic to the system, and the negative sign tells us the concentration is decreasing.
• This gives rise to a logarithmic decay curve (overhead) with a constant half-life, t_1/2.
• First order reactions will give a linear plot with a negative slope when ln [A] is plotted against time.
  • ln [A] = -kt + ln[A]_o
  • Note that for first order only, the half life, t_1/2, is independent of [A] and therefore of time.
  • t_1/2 = ln 2 / k = 0.693/k.

Example: What is the age of a meteorite if it has a ratio of ²³⁸₉₂U to ²⁰⁶₈₂Pb of 1:1. t_{1/2, ²³⁸92U} = 4.51 x 10⁹ years? Assume the only source of ²⁰⁶Pb is uranium decay.

The first thing we want to do is determine if it has decayed a whole number of half-lives, because if it has the solution becomes trivial. If the current ratio of U:Pb is 1:1, then the original amount of uranium, U_o = U + Pb = 1 + 1 = 2. Excellent, it has decayed a whole number of half=lives!

The uranium has decayed by one half-life and the meteorite is 4.51 x 10⁹ years old.

Example: What is the age of a rock sample if it has a ²³⁸₉₂U to ²⁰⁶₈₂Pb ratio of 0.906?

In this case U/Pb = 0.906, and U = 0.906Pb. For Pb = 1, U_o = U + Pb = 1 + 0.906 = 1.906, and we do NOT have a whole number of half-lives. What now?

Since r = -kN, can say r_o = -kN_o. If we divided both sides of the equation by r_o (recall in algebra can do anything we like as long as we do it to both sides) we get:

r/r_o = N/N_o = kN/kN_o = 0.906/1.906 = 0.475

Now recall that the integrated rate law is lnN = -kt + lnN_o, subtracting initial from final conditions:

lnN - lnN_o = -kt -(-kt_o) + lnN_o - lnN_o

assuming t_o = 0, dating from initial time 0

ln(N/N_o) = -kt

and since t_{1/2, ²³⁸92U} = 4.51 x 10⁹ years, can substitute into:

t_1/2 = 0.693/k and solve for k

k = 0.693/(4.51 x 10⁹y) = 1.540 x 10^-10y^-1

Going back and substituting into ln(N/N_o) = -kt

ln(N/N_o) = ln(0.475) = -0.7444 = -kt

Finally, rearranging and substituting:

t = -(-0.7444)/k = 0.7444/(1.540 x 10^-10y^-1) = 4.83 x 10⁹ years.

Carbon-14 dating: The half-life for carbon-14, t_{1/2, ¹⁴6C} = 5.74 x 10³ years. It is quite useful for dating human artifacts etc. over the past 50,000 years or so. However, a number of caveats apply to this use.

• Assume the ¹⁴₆C levels of living organisms will be in equilibrium with atmospheric ¹⁴₆C levels since they are constantly exchanging carbon in the form of carbon dioxide.
  • The ¹⁴₆C will then start decaying when the organism dies, since it will no longer exchange carbon dioxide!
• Assume a known initial concentration of 14₆C. This seems reasonable since it is produced in the atmosphere at a continuous rate so that it should reach a nice steady-state concentration due to production and decay:
  • ¹⁴₇N + ¹₀n Æ ¹⁴₆C + ¹₁H
  • ¹⁴₆C Æ ¹⁴₇N + ⁰_-1e
• However, it turns out the rate of production turns out to depend of the Solar cycle, since most of the neutrons are from the sun. But we haven't been measuring solar activity for very long, so what to do?
  • Look at ¹⁴₆C levels in objects of known age!
    • Bristle cone pine of High Sierra's most well known and most important as Earth's oldest living things.
    • If one takes a core through a tree one can count the growth rings to go back in time to a known date and take ¹⁴₆C samples for each year to create a corrected ¹⁴₆C level for each year.
• Can also have problems in specific circumstances when a second source of C is incorporated, e.g. snail shells in snails living on ancient limestone.

The material below was not covered F 2003

Nuclear Binding Energy and Nuclear Stability: Note figures in text. Note peaks and abundances in Universe, e.g. Fe, O, C, relatively common because of stability. Note Fission vs. Fusion as energy yielding processes on diagram.

Fusion: Fusion involves the combining of nuclei to create larger nuclides. It can yield an immense amount of energy, and is the source of energy in stars and the H-bomb.

We will look at the most ancient fusion process, used by the oldest stars as well as the Sun for energy production: the Proton-Proton Cycle:

Late stars such as the Sun can also use the Carbon cycle in which carbon-12 acts as a catalyst to speed the same result.

Fission: Large nuclei can be induced to break into smaller nuclides by neutron bombardment. One of the most important of these process is the fission of ²³⁵₉₂U:

Note that more neutrons are given off than are required. This makes it possible to have a nuclear chain-reaction if the uranium concentration is high enough.

Huge amounts of energy are released in fission reactions. Thus in the fission of ²³⁵₉₂U 0.1933 g/mol of mass is converted into energy in its fission to strontium and xenon (an alternate fission path with the release of three neutrons). This corresponds to 2.1 x 10¹³J/mol, as opposed to 8 x 10⁵J/ml for the combustion of methane. This is approximately 10⁶ times more energy release by fission than combustion on a mass:mass basis. (About 7/8 of this energy is released as KE, the remainder as EM radiation.)

Note terms for nuclear reactions and nuclear reactors:

• Chain reaction
• Subcritical, critical and supercritical.
• Core (reactor core)
• Moderator
• Control rods.

© R A Paselk

Last modified 12 December 2003