Radioactive decay?

Question

Hey, I'm struggling on a couple of physics questions regarding radioactive decay, if you could help me out that'd be great

1. The radioactive isotope 198Au has a half-life of 64.8 hr. A sample containing this isotope has an initial activity (t = 0) of 40.0 µCi. Calculate the number of nuclei that decay in the time interval between t1 = 10.0 hr and t2 = 12.0 hr.

2. In a piece of rock from the Moon, the 87Rb content is assayed to be 1.81×1010 atoms per gram of material, and the 87Sr content is found to be 1.06×109 atoms per gram. Calculate the age (in years) of the rock. Do not enter unit. 87Rb decays to 87Sr by beta decay with a half life of 4.75×1010 years.

Any help would be much appreciated.

Thanks

Answer 1

It is convenient in problems of this sort to convert the half-life to decay period by multiplying by 1/ln(2), or 1.4427. For 198Au, this would be 93.48 hours. The amount of material remaining after ten hours would be exp(-10/93.48) = 0.8986, while after twelve hours would be 0.8795. The difference is 0.0190. Multiplying by the initial activity, and converting curies into actual disintegrations gives the answer.

The decay constant of the rubidium is 6.8528E10 years, so if we assume that all of the strontium came from rubidium decay (NOT a given), the initial concentration was 18.1E9 + 1.06E9 = 19.16E9 atoms per gram. Of the original 19.16, there is now left 18.1, or 94.47%. Taking the natural log, this means that 0.0569 decay periods have elapsed, or 3.9 billion years.

Answer 2

First question:

Radioisotope:
Half-life
t(1/2) = 64.8 hr = 233280 s
Mean lifetime
τ = t(1/2) / ln 2 = 336552 s
Decay constant
Λ = 1/ τ = 2.97 * 10^(-6) Bq
--------------------
At the beginning:
Activity
A(0) = 40 μCi = 1.48 * 10^6 Bq
Total number of particles
N(0) = A(0) * τ = 4.98 * 10^11
--------------------
After 10 hours:
Elapsed time
t(1) = 10 hour = 36000 s
Activity
A(1) = A(0) * e^(- Λ*(t(1)) = 1.33 * 10^6 Bq
Total number of particles
N(1) = N(0) * e^(- Λ*(t(1)) = 4.48 * 10^11
--------------------
After 12 hours:
Elapsed time
t(2) = 12 hour = 43200 s
Activity
A(2) = A(0) * e^(- Λ*(t(2)) = 1.30 * 10^6 Bq
Total number of particles
N(2) = N(0) * e^(- Λ*(t(2)) = 4.38 * 10^11
--------------------
Answer:
The number of nuclei that decay in the time interval between 10 hr and 12 hr is
N(1) - N(2) = 9.473 * 10^9
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