Carbon-14 has a half-life of 5730 years. In a plant fossil, you find that the 14C has decayed to 1/4.00 of the original amount. How long ago was this plant alive?

We use the radioactive decay formula to solve this problem. It is expressed as:

An = Aoe^-kt

where An is the amount left in time t, Ao is the initial amount, k is the constant.

We first find the value of k from the half-life data. We do as follows:

An = Aoe^-kt
0.5 = e^-k(5730)
k = 1.21x10^-4

An = Aoe^-kt
1/4 = e^-1.21x10^-4t
t = 11456.98 years

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Annainn Annainn · Answer 2 · 2019-08-13T07:08:19+02:00

Answer:

The plant was alive for 1155 years

Explanation:

Any radioactive decay process follows the exponential law which is mathematically expressed as:

[tex]N(t)= N(0)e^{-kt}------(1)[/tex]

where N(0) = initial amount of the radioactive material

N(t) = amount at time t

k = decay constant

The decay constant is in turn related to the half-life t1/2

[tex]k = \frac{0.693}{t1/2}----(2)[/tex]

It is given that t1/2 of 14C = 5730yrs. Therefore:

[tex]k = \frac{0.693}{5730}=1.21*10^{-4}yr-1[/tex]

In addition it is mentioned that:

N(t) = 1/4*N(0)

substituting this in equation (1) gives:

[tex]1/4*N(0)= N(0)e^{-1.21*10^{-4}t}[/tex]

t = 1155 years