Question

the half life for radioactive decay of C-14 is 5730 y. An archaeological artefact contained wood had only 80% of the C-14 in a living tree. Estimate the age of the sample​

Answer 1

Answer:

Decay constant k=

t

1/2

0.693

=

5730year

0.693

=1.209×10

−4

/years

The rate of counts is proportional to the number of C-14 atoms in the sample.

N

0

=100,N=80

The age of the sample t=

k

2.303

log(

N

N

0

)

t=

1.209×10

−4

2.303

×log(

80

100

)=1846years

Answer 2

Answer:

• $1825\:years$

Explanation:

Question:

The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

Given:

• $[R_{0}] = 100$

• $[R] = 80$

• Half-life for radioactive decay of$^{14}C= 5730\:years$

We know radioactive decay reaction is the first order

Here, $k = \dfrac{ 0693 }{ t_{ \frac{1}{2}} }$

$= \dfrac{0.693}{5730}years^{-1}$

It is known that,

$t = \dfrac{2.303 }{ k}log\:\dfrac{[R]_{0}}{[R]}$

$= \dfrac{ 2.303 }{ \frac{0.693}{5730} }log\:\dfrac{100}{80}$

$= 1845 \:years$

Hence,

• The age of the sample is 1845 years