Question

The half-life for the (first order) radioactive decay of 14C is5730

Answer 1

Answer:

The age of this archeological sample is 2737.53 years

Explanation:

Given that:

Half life = 5730  years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{\ln\ 2}{5730}\ year^{-1}$

The rate constant, k = 0.00012 year⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $0.00012$ year⁻¹

It is given that 72 % of the 14C remains. So,

$\frac {[A_t]}{[A_0]}$ = 0.72

Applying values as:-

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.72 =e^{-0.00012\times t}$

$\ln \left(0.72\right)=-0.00012t\ln \left(e\right)$

$\ln \left(0.72\right)=-0.00012t$

$t=2737.53\ years$

The age of this archeological sample is 2737.53 years