(a) Find the number of carbon atoms in the sample. (b) Find the number of carbon
ID: 2140149 • Letter: #
Question
(a) Find the number of carbon atoms in the sample.(b) Find the number of carbon-14 atoms in the sample.
(c) Find the decay constant for carbon-14 in inverse seconds.
s?1
(d) Find the initial number of decays per week just after the specimen died.
decays/week
(e) Find the corrected number of decays per week from the current sample.
decays/week
(f) From the answers to parts (d) and (e), find the time interval in years since the specimen died.
yr Please show all of your working A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life = 5 730 yr) for every 7.70 times1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22O11) contains 22.4 mg of carbon. When the sample is placed inside a shielded beta counter with 88.0% counting efficiency, 839 counts are accumulated in one week. We wish to find the age of the sample. Find the number of carbon atoms in the sample. Find the number of carbon-14 atoms in the sample. Find the decay constant for carbon-14 in inverse seconds. s?1 Find the initial number of decays per week just after the specimen died. decays/week Find the corrected number of decays per week from the current sample. decays/week From the answers to parts (d) and (e), find the time interval in years since the specimen died. yr
Explanation / Answer
a)
Nc =(22.4*10^-3g/12g/mol)*(6.02*10^23 molecules/mol)
Nc =1.124*10^21 carbon atoms
b)
Nc-14 =1.124*10^21/(7.7*10^11
Nc-14 =1.46*10^9
c)
lambda =ln(2)/5730
lambda =1.21*10^-4 /year =1.21*10^-4*(1/365*24*60*60)
lambda =3.83*10^-12 /sec
d)
Ro =lambda*Nc-14 =3.83*10^-12/sec*1.46*10^9*[7*(86400/week]
Ro =3382 decays/week or 3.382*10^3 decays/week
e)
R =839/0.88
R=953.41 decays/week
f)
R=Roe^(-lambda*t)
=>t =(-1/lambda)*ln(R/Ro)
t =(-1/1.21*10^-4 years)*ln(953.41/3382)
t=10464.27 years or 1.046 *10^4 years
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