Question

An ancient club is found that contains 189g of carbon and has an activity of 6.66 decays per second. Determine its age in years (do not enter units) assuming that in living trees the ratio of ¹⁴C/¹²C atoms is 1.30×10^-12.

Answer 1

here,

the decay constant for 14C,

which has a half-life of 5700 years.

decay constant = In2/ T1/2

decay constant = In2/ (5700 years)

decay constant = 1.2160x10^-4 y^-1

the number of 12C nuclei in 189 g of carbon is ,N(12C)= (6.02 * 10^23 nuclei/mol) * (189g)/ 12.0 g/mol

N(12C) = 9.48 * 10^24 nuclei

Ratio of 14C to 12C is 1.3 * 10^-12

N<original>(14C) = (1.3 * 10^-12) (9.48 * 10^24)

N<original>(14C) = 1.23 * 10^13 nuclei

Therefore, the initial activity of the sample,R<o> = decay constant x N<o>

R<o> = (1.216x10^-4 y^-1) * (1.23 * 10^13 nuclei)

R<o> = 1.5 * 10^9 decays/y

the age of the ancient club using Equation ,R =(R<o>)(e^(-decay constant * time),

R = R<o> * e^(-decay constant * t)

R/ R<o> = e^(-decay constant * t)

R = 6.66 * (3.1526x10^7) decays/y (given)

We found that R<o> = 9.5164x10^8 decays/y

therefore ,

ln(R/R<o>) = In e^(-decay constant * t)

-decay constant * t = ln(R/R<o>)

-1.216* 10^-4 * x = ln(6.66 * (3.15 * 10^7)/1.5 * 10^9)

t = 324666 years

the age in years is 324666