Radioactive Rate of Decay:
half-life: the time it takes from one-half of the nuclei in a sample to decay. Independent of the amount of sample, denoted t½.
First Order Half Life:
t½ = 0.693 / k
Example:
k = 1.45 x 10-12 s-1 for Ta-183, find t½ in years.
t½ = 0.693 / 1.45 x 10-12 s-1
t½ = 4.78 x 1011 s
Now convert to years:
(4.78 x 1011 s) x (1h / 3600s) x (1 day / 24h) x (1 year / 365 days) = 1.52 x 104 years
Once we know the decay constant (k) for a radioactive isotope, we can calculate the fraction of the radioactive nuclei left after a given time by the following equation:
ln (Nt / No) = –kt
No = the number of nuclei in the original sample
Nt = the number of nuclei left after time t
Never really need to find Nt or No, it is the fraction that is important.
Example:
When t½ = 10.76 years, what is the fraction remaining after 25 years?
Remember k = 0.693 / t½:
k = 0.693 / (10.76 years)
ln (Nt / No) = –(0.693t / t½)
ln (Nt / No) = –(0.693x 25years / 10.76 years)
ln (Nt / No) = –1.61
(Nt / No) = e–1.61
(Nt / No) = 0.200
This means 20% is left.
Example:
How long until there is only 5% left?
(Nt / No) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years
Radioactive Dating:
Carbon–14, t½ = 5730 years
The upper atmosphere is constantly bombarded by cosmic ray radiations from space, which creates high energy neutrons when it reacts with Nitrogen–14 (the most abundant nitrogen nuclide).
147N + 10n → 146C + 11H
Then goes into carbon dioxide:
146CO2 → plants → animals
There is a set amount of Carbon–14 per kg of living matter. When something dies, it no longer takes in Carbon–14. The Carbon–14 then decays:
146C → 147N + 0-1e
In living matter, we measure 15.3 disintegrations per kg of C per minute. Anything lower → must be dead. In this way, this ratio of carbon isotopes becomes a clock measuring the time since death of an organism.
Nt / No = (less than 15.3 dis/kg • min) / (15.3 dis/kg • min)
Example:
You want to carbon date a caveman's foot, where you observe (4.50 dis/kg • min). When did the caveman lose his foot?
ln(Nt / No) = –(0.693t / t½)
ln(4.50 / 15.3) = –(0.693t / 5730 years)
t = 10118 years
Applications of Radioactive Isotopes:
Chemical Analysis:
Radioactive Tracer: a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system.
Evidence of dynamic equilibrium can be seen using radioactive tracers. In a beaker, imagine a solution of PbI2 in contact with the solid. This beaker contains only iodine atoms with nonradioactive isotopes. Now add a radioactive iodide ion to the solution. The solution is still saturated, and the amount of solid remains constant. But, after time, the solid lead iodide, which was initially nonradioactive, is radioactive. Nonradioactive iodide ions in the solid have swapped with radioactive ions in solution.
Tracers are also used to determine chemical mechanisms, like photosynthesis.
Isotope Dilution: a technique to determine the quantity of a substance in a mixture or of the total volume of solution by adding a known amount of an isotope to it. After time, removal of a sample of the mixture and measurement of the fraction by which the isotope has diluted provides a way to determine the quantity of substance or volume of solution.
Neutron Activation Analysis: an analysis of elements in a sample based on the conversion of stable isotopes to radioactive isotopes by bombarding a sample with neutrons. Example: measuring trace amounts of arsenic in human hair. The arsenic is converted into a metastable nucleus via bombardment with a neutron. As the arsenic returns to its ground state, it emits gamma rays. These measurements can be interpreted to find who the hair belongs to, or how much arsenic is in it.
7533As + 10n → 76m33As
76m33As → 7533As + 00γ
Mass-Energy Equivalence:
When nuclei decay, they form products of lower energy. The change of energy is related to changes in mass, according to the mass-energy equivalence derived by Albert Einstein.
E = mc2
Where c = the speed of light = 3.00 x 108 m/s
E = energy
m = mass
For any mass, there is an associated energy. For any energy, there is an associated mass.
When the energy changes by an amount ΔE, the mass changes by an amount Δm.
ΔE = Δmc2
When energy is given off, mass decreases.
Example:
Chemical reaction:
When carbon burns in oxygen, it releases heat energy:
C (gr) + O2 (g) → CO2 (g); ΔH = –393.5 kJ
ΔE = –3.935 x 105 J (converted from kJ to J)
Recall that J = (kg•m2 / s2)
Δm = ΔE / c2
Δm = –(3.935 x 105 kg•m2 / s2) / (3.00 x 108 m/s)2
Δm = –4.37 x 10-12 kg
Mass changes in nuclear reactions are approximately a milion times larger per mole of reactant thatn those in chemical reaction.
Example:
Nuclear Reaction:
Consider the alpha decay of Uranium–238 to Thorium–234
23892U → 23490Th + 42He
Use the nuclear mass in amu to find the change in mass for this reaction. You would think the mass of the reactants is the same as the mass of the products, but not so. There is a mass loss.
23892U = 238.0003 amu
23490Th = 233.9942 amu
42He = 4.00150 amu
Δm = (233.9942 amu + 4.00150 amu) – (238.00003 amu)
Δm = – 0.0046 amu
(Assume 1 mole of isotope.)
Δm = – 0.0046 amu = – 0.0046g = – 4.6 x 10-6 kg
ΔE = Δmc2 = (– 4.6 x 10-6 kg)(3.00 x 108 m/s)2
ΔE = – 4.14 x 1011 J = – 4.14 x 108 kJ
Nuclear Fission: spontaneous break up of isotopes, naturally occurring, does not require energy input. Powers nuclear reactors and bombs.
Nuclear Fusion: Fusing of two nuclei into a larger nucleus. Takes GIGANTIC amounts of energy.
21H 31Th→ 42He + 10n
No comments:
Post a Comment