20.4-20.7: Radioactive Rate of Decay-Nuclear Fission and Nuclear Fusion

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 t_½.

First Order Half Life:
t_½ = 0.693 / k

Example:

k = 1.45 x 10^-12 s^-1 for Ta-183, find t_½ in years.

t_½ = 0.693 / 1.45 x 10^-12 s^-1
t_½ = 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 (N_t / N_o) = –kt

N_o = the number of nuclei in the original sample
N_t = the number of nuclei left after time t

Never really need to find N_t or N_o, it is the fraction that is important.

Example:

When t_½ = 10.76 years, what is the fraction remaining after 25 years?

Remember k = 0.693 / t_½:

k = 0.693 / (10.76 years)
ln (N_t / N_o) = –(0.693t / t_½)
ln (N_t / N_o) = –(0.693x 25years / 10.76 years)
ln (N_t / N_o) = –1.61
(N_t / N_o) = e^–1.61
(N_t / N_o) = 0.200

This means 20% is left.

Example:

How long until there is only 5% left?

(N_t / N_o) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years


Radioactive Dating:

Carbon–14, t_½ = 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.

N_t / N_o = (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(N_t / N_o) = –(0.693t / t_½)
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

13.4-13.5 (Half Life-Transition State Theory)

First Order Half Life:

Half Life: time it takes for the reactant concentration to decrease to one half its initial value (t_½) .
Value of “t” when [A]_t = (½) [A]₀
So ln (½) = -kt → 0.693 = kt → t_½ = 0.693 / k

Simply: t_½ = 0.693 / k

The half life of a first order reaction is independent of initial concentration:

Example:

*SO₂Cl₂ decomposes in a first order reaction to produce SO₂ and Cl₂. What is the half life if the rate constant at a given temperature is 2.20 x 10^-5 s^-1 ?

t_½ = 0.693 / k
t_½ = 0.693 / 2.20 x 10^-5 s^-1
t_½ = 3.15 x 10⁴ s

*How long would it take for 75% of the SO₂Cl₂ to decompose?

75% is half of the initial concentration plus half of the concentration left over after the first half life, so it is two times the half life.

Second Order Half Life:

t_½ = 1 / k[A]₀

This time, the half life depends on initial concentration and each subsequent half life becomes larger as time progresses.



Zeroth Order Half Life:

t_½ = [A]₀ / 2k

Again, the half life is dependent upon initial concentration of the reactant, but instead of becoming larger as the reaction progresses, as a zeroth order reaction proceeds, each half-life gets shorter.




Graphing of Kinetic Data:

It is possible to determine the order of a reaction by graphical plotting of the data for a particular experiment.

If you plot the data for ln[A]_t on the vertical axis against time “t” on the horizontal axis, a first order reaction will give you a straight line.

If you plot the data for 1 / [A]_t on the vertical axis against time “t” on the horizontal axis, a second order reaction will give you a straight line.

If a curved line results for either graph, the reaction order is not the one that matches that graph.

An organized explanation of the key points in the chapter thus far is found in Table 13.2, pg 544.


Collision Theory: (3 Stooges Universe)



Three things must happen between objects before there is a reaction:
1. Objects must “hit” (Collision).
2. Must collide with enough energy to break bonds (Activation Energy = E_a).
3. Objects must smack into each other the right way (Orientation).

Example:

O=N + Cl ̶ Cl → [O=N ̶ Cl···Cl]ǂ
This ǂ symbol denotes a highly unstable, activated state that quickly falls apart to
O=N ̶ Cl + Cl
Notice the orientation of N to Cl. If O smacked into Cl:

N=O + Cl ̶ Cl → No Reaction

Nitrogen must collide with Chlorine, not Oxygen in order for a reaction to occur here.

Enzymes function the same way: