In consultation with the other members of your group, solve the problem outlined below. If you finish your assigned problem quickly, try one of the other problems as well.
| Parent | Daughter | Half Life |
| 40K | 40Ar | 1.3 x 109 yr |
| 238U | 206Pb | 4.5 x 109 yr |
| 87Rb | 87Sr | 47. x 109 yr |
1. A rock sample has a 40Ar-to-40K ratio of 6.0 (i) What is the age of the rock? (ii) If you could measure the 87Sr-to-87Rb ratio, what would you expect it to be?
(i) Age?
N/No = exp(- ln(2) t / T)
No = N + Ndecay = N + 6N = 7N
N/No = (1/7) = exp(- ln(2) t / T)
t = -[ln(1/7)/ln(2)] x 1.3 x 109 yr
t = 3.6 x 109 yr
(ii) Ratio 87Sr/87Rb?
N/No = exp(- ln(2) t / T) = 0.95
No=N + Ndecay
Therefore, Ndecay = No - N
Ndecay/N = (No - N)/N = No/N - 1. = 0.055
2. Imagine that you have discovered a new radioactive parent-daughter pair with a half life of 3.0 x 109 yr. You are able to measure the daughter/parent ratio in a unique rock sample (thought to be a remnant of a meterorite). You find that 26% of the parent nuclei have decayed. (i) What is the age of the meteorite? (ii) If you could measure the 40Ar-to-40K ratio, what would you expect it to be?
(i) Age?
N/No = exp(- ln(2) t / T)
26% decay, therefore 74% remain
N/No = 0.74 = exp(- ln(2) t / T)
t = -[ln(0.74)/ln(2)] x 3.0 x 109 yr
t = 1.3 x 109 yr
(ii) Ratio 40Ar/40K?
N/No = exp(- ln(2) t / T) = 0.50
No=N + Ndecay
Therefore, Ndecay = No - N
Ndecay/N = (No - N)/N = No/N - 1. = 1.0
3. A rock sample has a 87Sr-to-87Rb ratio of 0.051. (i) What is the age of the rock? (ii) If you could measure the 40Ar-to-40K ratio, what would you expect it to be?
(i) Age?
N/No = exp(- ln(2) t / T)
No = N + Ndecay = N + 0.051N = 1.051N
N/No = (1/1.051) = exp(- ln(2) t / T)
t = -[ln(1/1.051)/ln(2)] x 47. x 109 yr
t = 3.4 x 109 yr
(ii) Ratio 40Ar/40K?
N/No = exp(- ln(2) t / T) = 0.16
No=N + Ndecay
Therefore, Ndecay = No - N
Ndecay/N = (No - N)/N = No/N - 1. = 5.0
4. Suppose you determine that 33% of the parent nuclei have decayed for a radioactive pair with a half life of 6.4 x 109 yr. (i) What is the age of the meteorite? (ii) If you could measure the 206Pb-to-238U ratio, what would you expect it to be?
(i) Age?
N/No = exp(- ln(2) t / T)
33% decay, therefore 67% remain
N/No = 0.67 = exp(- ln(2) t / T)
t = -[ln(0.67)/ln(2)] x 6.4 x 109 yr
t = 3.7 x 109 yr
(ii) Ratio 206Pb/238U?
N/No = exp(- ln(2) t / T) = 0.56
No=N + Ndecay
Therefore, Ndecay = No - N
Ndecay/N = (No - N)/N = No/N - 1. = 0.77