An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of [tex]^{14}\text{C}[/tex] . Estimate the minimum age of the charcoal, noting that [tex]2^{10} = 1024[/tex]
Answer:
57300 years
Step-by-step explanation:
Using the relation of an half-life time in relation to fraction which can be expressed as:
[tex]\dfrac{N}{N_o} = (\dfrac{1}{2})^{\frac{t}{t_{1/2}}[/tex]
here;
N represents the present atom
[tex]N_o[/tex] represents the initial atom
t represents the time
[tex]t_{1/2}[/tex] represents the half - life
Given that:
Its charcoal is found to contain less than 1/1000 the normal amount of [tex]^{14}\text{C}[/tex] .
Then ;
[tex]\dfrac{N}{N_o} = \dfrac{1}{1000}[/tex]
However; we are to estimate the minimum age of the charcoal, noting that [tex]2^{10} = 1024[/tex]
so noting that [tex]2^{10} = 1024[/tex], then:
[tex]\dfrac{1}{1000}> \dfrac{1}{1024}[/tex]
[tex]\dfrac{1}{1000}> \dfrac{1}{2^{10}}[/tex]
[tex]\dfrac{1}{1000}> (\dfrac{1}{2})^{10}[/tex]
If
[tex]\dfrac{N}{N_o} = \dfrac{1}{1000}[/tex]
Then
[tex]\dfrac{N}{N_o} > (\dfrac{1}{2})^{10}[/tex]
Therefore, the estimate of the minimum time needed is 10 half-life time.
For [tex]^{14}\text{C}[/tex] , the normal half-life time = 5730 years
As such , the estimate of the minimum age of the charcoal = 5730 years × 10
= 57300 years
