Part A You are using a Geiger counter to measure the activity of a radioactive s
ID: 834659 • Letter: P
Question
Part A
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 42.2minutes , what is the half-life of this substance?
Express your answer with the appropriate units.
Part B
An unknown radioactive substance has a half-life of 3.20hours . If 44.8g of the substance is currently present, what mass A0 was present 8.00 hours ago?
Express your answer with the appropriate units.
Part C
Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 41.0% of an Am-241 sample to decay?
Express your answer with the appropriate units.
Part D
A fossil was analyzed and determined to have a carbon-14 level that is 20% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?
Express your answer with the appropriate units.
t1/2 =Explanation / Answer
1) Ao = 400 counts
A[t] = 100 counts
For first order rxn,
ln[Ao/A[t]] = k*t
ln4 = k*42.2
k = ln4/42.2 minutes^(-1)
For half life
ln2 = ln4/42.2*t
t1/2 = 21.1 minutes
2) Half life = 3.20 hours
ln2 = k*3.2
k = ln2/3,2 hours^(-1)
Now A[o] = 44.8 gms , A[t] amount present after 8 hours
ln[Ao/A[t]] = ln2/3.2*8
ln[A[o]/A[t]] = 2.5ln2
Ao/A[t] = 5.65
A[t] = 44.8/5.65 = 7.929 gms of substance
3) Half life = 432 years
k = ln2/432 years^(-1)
A[o], A[t] = 0.59 A[o]
ln(1/0.59) = ln2/432*t
treq = 0.527/0.693*432 = 328.51 years
4) Half life = 5730 years
A[o] initial conc
A[t] = 0.2A[o]
ln[Ao/At] = ln2/5730*t
ln5 = ln2/5730*t
t = 5730*ln5/ln2 = 133.04.12 years