Physics, 29.03.2020 22:34 snowprincess99447

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13. Radio-carbon dating is used to estimate the age of objects that were once living as all living
things contain a small amount of Carbon-14, which has a half-life of 5700 yrs. By measuring the
activity (counts min") of the dead tissue its approximate age can be determined. Show all the
steps in your working out.
a. A sample of living wood has an activity of 16 counts min per gram. Calculate the activity
of a 20g sample of living wood.
b. Calculate the activity of 20g sample of wood from a tree that died 17100 years ago.
State the correct units in your answer.
C. A5g of old wood from an archaeological site has an activity of 20 counts min?, what was
the activity of the wood when it was living? How long ago did the tree died?

Answers: 3

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Answers

Answer from: cddominguez63

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is m_1 = 1 g , the activity is A_1=16 count/min

- When the mass is m_2=20 g , the activity is A_2

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

A_0 is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

A_0 = 320 count/min (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

We know that a sample of living wood has an activity of

A=16 count/min per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of A=20 count/min : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

Answer from: Quest

f=ma newton's second law 7x8=56n

Answer from: Quest

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