Question

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

years. An archeological sample
contained wood with only 72% of the 14C found in living trees.
Whatwas the age of this archeological sample?

Answer 1

The age of this archeological sample is 2737.53 years

Step-by-step 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=(\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* t)`

`0.72 =e^(-0.00012* 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