Question

An Archaeologist in Turkey discovers a spear head that contains 51% of its original amount of C-14. Find the age of the spear head to the nearest year.

Answer 1

Answer: 5566 years

Explanation:

We know that the half-life of carbon-14 is 5730 years.

Let x be the original amount of C-14 , then the current amount of C-14 =
`0.51x`

The radioactive half-life formula for C-14 is given by :-

`P=P_0(0.5)^{\frac{t}{t_{(1)/(2)}}}`, where P is the amount of C-14 at time t and
`P_0` is the original amount.

`\Rightarrow\ 0.51x=x(0.5)^{(t)/(5730)}\\\\\Rightarrow\ 0.51=(0.5)^{(t)/(5730)}`

Taking log on both the sides , we get

`\log(0.51)=(t)/(5730)\log(0.5)\\\\\Rightarrow\ -0.292429823902=(t)/(5730)*-0.301029995664\\\\\Rightarrowt=(5730*0.292429823902)/(0.301029995664)=5566.29875791\approx5566\text{years}`

Hence, the age of the spear head is 5566 years.

Answer 2

Use given half-life for C-14 of 5,730 years.

The exponential decay function:

`y = e^(-kt)`
plug in half life, solve for 'e^(-k)'

`0.5 = e^(-5730k) \\ \\ 0.5 = (e^(-k))^(5730) \\ \\ e^(-k) = (0.5)^(1/5730)`

Sub into original decay function, plug in 0.51 for 'y' and solve for 't'


`y = e^(-kt) = (e^(-k))^t \\ \\ 0.51 = (0.5)^(t/5730) \\ \\ ln(0.51) = (t)/(5730) ln(0.5) \\ \\ t = 5730 (ln 0.51)/(1n 0.5)`