Question

What is the age of the charcoal BP if currently living organisms decay at the rate of 15.3 disintegrations/min per gram of carbon?

The half-life of 14C is 5715 years.

Answer 1

1. Calculate the decay constant

      λ= 0.693/t_1/2 =

       = 0.693/(5715 years x 365 day/y x 24 h/d x 3600s/d)=

         = 3.84x10^-12 s^-1

1. In 1 g carbon (the sample) there are

        (1 g/ 12 g/mol ) x 6.02x10²³ nuclei/mol = 0.50 x x10²³ nuclei (¹²C and ¹⁴C)

3. The present activity is

          R= 15.3 dis/min / 60 s/min = 0.255 dis/s

          R = N λ  

          N = R/ λ = 0.255 dis/s / 3.84x10^-12 s^-1 = 6.64x10¹⁰ active nuclei (¹⁴C)

4. In living organisms, the ¹⁴C /¹²C ratio is known to be 1.3x10^-12 .

    The number of active nuclei in the plant when it died was

N_o = 0.50 x x10²³ nuclei (¹²C and ¹⁴C) x 1.3x10^-12 = 6.5x10¹²

5. Use the general eq. to find t:

      N= N_o e^{- λt}

     ln (N/N_o) = - λt

     t = - ( ln (N/N_o) )/ λ

        = - (ln (6.64x10¹⁰ / 6.5x10¹² ) )/ 3.84x10^-12 s^-1 =

        = 4.58/ 3.84x10^-12 s^-1

        = 1.19 x 10¹² s

      = 1.19 x 10¹² s /(365 day/y x 24 h/d x 3600s/d) =

       = 37 700 years.