In: Chemistry
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 61.7 minutes , what is the half-life of this substance?
Part B; An unknown radioactive substance has a half-life of 3.20 hours . If 27.1 g of the substance is currently present, what mass A0 was present 8.00 hours ago?
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 37.0 % of an Am-241 sample to decay?
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?
Part A
400 counts dropping to 100 counts means two half lives have
passed
400 --> 200 one half life
200 --> 100 one half life
so the half life is 61.7/2 =30.85 mins
Part B
Use this equation below to find out how much of a radioactive
material is left after a given period of time. You can use the
equation in reverse to find out the initial
mass of the radioactive material if you know its current
mass.
A = current mass of radioactive material = 27.1 g
A0 = initial mass of radioactive material = to be determined
T½ = half-life of the radioactive material = 3.20 h
t = elapsed time since the original mass was present = 8.00 h
A = A0e^(-0.693t/T½)
A/A0 = e^(-0.693t/T½)
ln (A/A0) = ln [e^(-0.693t/T½)]
ln A - ln A0 = (-0.693t/T½)
ln A = -0.693t/T½ + ln A0
ln 27.1 = -0.693(8.00 h)/3.20 h + ln A0
3.211 = -1.7325 + ln A0
3.211 + 1.7325 = ln A0
ln A0 = 3.211 + 1.7325
ln A0 =4.9435
A0 = e^4.9435
A0 = 140.26 Initial Mass of Radioactive Material
Part C:
t1/2 = 0.693/K
K = 0.693 / 432
K = 1.6042 * 10^-3
K = 2.303/t log(a / (a-x))
1.6042 * 10^-3 = 2.303 / t log (100/(100-37))
t = 71.74 years.
Part D:
Let the amount (No) of carbon sample be 100g
So the amount left (Nt) = 20% of 100g = 20g
Using the formula ,
Nt = No (1/2)t / t 0.5
where t is the amount of time lapsed
and t 0.5 is the half life period
putting all the values we get ,
20 = 100 (1/2)t / 5730
t = 13304.65 years
So the fossil must be 13304.65 years old.