Question

The age of a piece of wood from an archeological site is to be determined using the Carbon-14 method. The activity of the sample is measured to be 0.574 times the Carbon-14 activity of living plants. What is the age of the sample in years? (The half-life of the Carbon-14 isotope is 5730 years.)

Answer 1

4589.05 year

Step-by-step explanation:

The relation of activity is given by
`A=A_0e^(\lambda t)` we have given
`(A)/(A_0)=0.574`

Half life
`T=5730\ years`

We know that half life
`T=(0.693)/(\lambda )`

`\lambda =(0.693)/(5730)=1.2094* 10^(-4)\ per\ year`

So
`0.574=e^{-1.2094* 10^(-4)t}`

`e^{1.2094* 10^(-4)t}=(1)/(0.574)=1.742`

`{1.2094* 10^(-4)t}=ln1.742=0.555`

`t=(0.555)/(1.2094* 10_(-4))=4589.05\ year`