20.2.1 Rates of Disintegration Reactions Flashcards
Rates of Disintegration Reactions
- Just as for any first-order reaction, half-lives can be used for calculations involving nuclear disintegration reactions.
- Activity is the rate of nuclear decay.
- Measurements of activities and half-lives can be used to make calculations for nuclear disintegration reactions.
note
- All nuclear disintegration reactions are first-order. Therefore, the quantity of parent nuclei determines the rate that daughter nuclei are produced.
- For example, strontium-90 undergoes beta decay, producing yttrium-90 and a beta particle. This reaction has a half-life of 29 years. If 10 g of 90 Sr are present to begin with, 5 g 90 Sr will be present after 29 years, 2.5 g 90 Sr after another 29 years, etc.
- Just as for any first-order reaction, half-lives can be used for calculations involving nuclear disintegration reactions.
- Many radioactive nuclei have extremely long half-lives. This is the heart of the nuclear waste problem—once nuclear waste is formed, it will remain radioactive for a large number of years.
- For a first-order reaction, the concentration of reactant at time t is related to the initial concentration of reactant by the equation [reactant]t = [reactant] 0 e –kt .
- For a nuclear disintegration reaction, the number of nuclei is used instead of the concentration of reactant. Therefore, N = N i e –kt .
- Activity (A) is the rate of nuclear decay (A = kN). Since A is related to N by a constant, A = A i e –kt .
- The SI unit for activity is the becquerel (Bq).
1 Bq = 1 disintegration / s. - Another common unit for activity is the curie (Ci). 1 Ci = 3.7 x 10 10 disintegrations / s. Since 1 Ci is so large, mCi (1/1000 of a curie) are often used.
- Activity is measured using a Geiger counter. Geiger counters contain an anode surrounded by argon gas.
When ionizing radiation such as a beta particle strikes an argon atom, an argon ion (Ar + ) is created. This allows for the completion of the circuit, and a sound is produced. - Problem: A sample of 245 Am with a half-life of 7.37 x 10 3 y has an activity of 1.00 Bq. How many 245 Am nuclei are present?
- Since A = kN and k = 0.693 / t 1/2 , N = A · t 1/2 / 0.693.
- The half-life must first be converted to seconds. Plugging this value into the equation N = A · t 1/2 / 0.693 yields the number of 245 Am nuclei (3.35 x 10 11 nuclei).
note 2
- Problem: How many 245 Am nuclei remain after 100 years?
- The number of nuclei at a given time (N) is related to the initial number of nuclei (N i ) by the equation N = N i e –kt . The half-life is used to solve for k (9.4 x 10 –5 y –1 ). Plugging the known values into the equation yields the number of 245 Am nuclei after 100 years (3.32 x 10 11 nuclei). About 99% of the nuclei remain. This makes sense, since the half-life of 245 Am is over 7000 years—after only 100 years, many more than half of the initial 245 Am nuclei remain.
Which of the following statements about nuclear reactions is not correct?
At = Ai e^k t
Which of the following radioactive isotopes would be in greatest abundance after about 28 years?
100 g of 210Pb, whose t½ = 22.3 y.
A sample of 245Am has a t½ of 7.37 × 103 y and an activity value, A, of 1.00 Bq. What is the value of k, and how many nuclei are there initially and after 500 years?
k = 9.4 × 10−5 / y; Ni = 3.35 × 1011 nuclei; N500 = 3.20 × 1011 nuclei.
Which statement about the kinetics of nuclear reactions is not correct?
t½ = ln (2) k = 0.693k.
Which statement about reactions is not correct?
The term half-life refers exclusively to first-order nuclear reactions.
Which statement about the activity term, A = kN = Aie−kt is not correct?
A is expressed in units of disintegrations.
Which statement about nuclear reactions is not correct?
Almost all nuclear reactions are second-order reactions.
Nitrogen-13 decays by beta emission to carbon-13
13 7 N -> 13 6 C + 0 +1 e
Suppose that you start with 16 g of 13 N 7 and that its half-life is 10 minutes. Which statement about this decay reaction is not correct?
After six half-lives, there are about 0.125 g of 13 7 N left
Suppose that you start with 40 g of 90Sr at the beginning of the reaction below
90Sr -> 90 Y + 0 -1 e
Which statement about this reaction is not correct?
90Y is definitely a stable isotope.
These are the half-lives of some heavy isotopes.
238U = 4.5 × 10^9 y 235U = 7.1 × 10^8 y 239Pu = 2.4 × 10^4 y
Which statement about half-lives is not correct?
After about five or six half-lives, virtually all of a starting isotope will have decayed.