Question

A fossil gives 435 disintegrations per g of carbon during 1 h of counting. The age of the fossil given that the living matter contains 1.5x10^-12 g of ¹⁴C per g of carbon and that the half life of beta decay of ¹⁴C is 5680yr, will be

Answer 1

N_t = N_oe^-$\lambda$ ^t

$\dfrac{Nt}{No} =$ e^-$\lambda$ ^t (but Nt/No = 1/2)

$\dfrac{1}{2} =$ e^-$\lambda$ ^t

to cancel e

ln ($\dfrac{1}{2} ) =$ ln(e^-$\lambda$ ^t )

ln ($\dfrac{1}{2}$ ) = -$\lambda$t

-0.693 = -$\lambda$t

Also

Given ;

Mass of ¹⁴C = 1.5x10^-12g

No of moles

$Moles = \dfrac{mass}{RAM} = \dfrac{1.5x10^-12g}{14g/mol}$

n = 1.07x10^-13moles

Number of atoms of 14C in living matter = n x N_A

= 1.07x10^-13moles x 6.022x10^-23 atoms

N= 6.445x10¹⁰ atoms

Activity in living matter is given by;

$(\dfrac{-dN}{dt}) = \lambda$N = $\dfrac{0.693N}{t1/2}$

Since t_1/2 = 5680yrs

Activity of ¹⁴C in living matter is given by;

$\lambda$ $=\dfrac{0.693}{t1/2}N$

$\lambda = \dfrac{0.693}{5680 x 365 x 24hrs} x 6.445x10^10$

= 897.6 disintegrations per gram per hour

Now we have

N₀ = initial activity = 897.6 disintegrations per hr

N_t = final activity = 435 disintegrations per hr

From the equation;

t_1/2 = $\dfrac{0.693}{\lambda}$

$\lambda$ = $\dfrac{0.693}{5680 yrs} = 1.22x10^-4 yrs$

Now, from the equation

N_t = N_oe^-$\lambda$ ^t

$\dfrac{Nt}{No} =$ e^-$\lambda$ ^t

$\dfrac{435}{897.6} = e$ ^{-(1.22x10-4) x t}

0.4846 = e^{-1.22x10-4 x t}

t = ln ($\dfrac{0.4846}{-122x10^-4})$

t = 5937.629 years