Question

An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of ^{14}\text{C} ​14 ​​ C. Estimate the minimum age of the charcoal, noting that

Answer 1

An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of $^{14}\text{C}$ ​. Estimate the minimum age of the charcoal, noting that  $2^{10} = 1024$

Answer:

57300 years

Step-by-step explanation:

Using the relation of an half-life time in relation to fraction which can be  expressed as:

$\dfrac{N}{N_o} = (\dfrac{1}{2})^{\frac{t}{t_{1/2}}$

here;

N represents the present atom

$N_o$ represents the  initial atom

t represents the time

$t_{1/2}$ represents the half - life

Given that:

Its charcoal is found to contain less than 1/1000 the normal amount of $^{14}\text{C}$ ​.

Then ;

$\dfrac{N}{N_o} = \dfrac{1}{1000}$

However; we are to  estimate the minimum age of the charcoal, noting that  $2^{10} = 1024$

so noting that $2^{10} = 1024$, then:

$\dfrac{1}{1000}> \dfrac{1}{1024}$

$\dfrac{1}{1000}> \dfrac{1}{2^{10}}$

$\dfrac{1}{1000}> (\dfrac{1}{2})^{10}$

If

$\dfrac{N}{N_o} = \dfrac{1}{1000}$

Then

$\dfrac{N}{N_o} > (\dfrac{1}{2})^{10}$

Therefore, the estimate of the minimum time needed is 10 half-life time.

For $^{14}\text{C}$ , the normal half-life time = 5730 years

As such , the estimate of the minimum age of the charcoal =  5730 years × 10

= 57300 years