half-life physics?

Question

living plants and trees absorb carbon dioxide which contains a small amount of radioactive carbon-14 which has a half-life of 5600 years. when the plant dies, the proportion of radioactive C to non-radioactive C in the plant decreases as the C-14 decays. the activity of a living plant is 15 Bq for each gram of carbon. if the activity of a piece of wood containing 4g of carbon is 30Bq, how old is the wood?

please help.. i'm very confused

Answer 1

The formula for radio-active decay is :

dN/dt = -lambda * N

where lambda is the decay constant and N is the number of radioactive atoms remaining at any time t

Solving we get
t = -(1/lambda) * ln(N/N_0)
where N_0 = number of radio-carbon atoms at =0 i.e. the origin of the disintegration time.

The age of the sample is
t_age = -(1/lambda) * ln(N/N_0)

Also, the average and the half-lifes are
t_avg = 1/lambda
t_(1/2) = t_avg * ln(2)

For C14 we get,
t_avg = 8033 years (Libby value)
t_(1/2) = 5568 years (Libby value)

Combining the formulas we get:
t = - t_avg * ln(N/N_0)
= - [t_(1/2)/ln(2)] * ln(N/N_0)



Activity:
The unit of activity is "Becquerel" (symbol: Bq); it has replaced the formerly used unit "Curie". 1 Bq activity means that in a particular substance one nucleus disintegration occurs per second. The activity can also be expressed in relation to a volume of one litre or a mass of one kilogram. In that case we obtain the activity concentration (Bq/l) and specific activity (Bq/kg), respectively. For example, U-238 has a specific activity of 12,400 Bq/g. Prefixes are used to express several orders of magnitude:

e.g. 1 mBq/l means 0.001 Bq per litre

1 kBq/l means 1000 Bq per litre

1 MBq/g means 1,000,000 Bq per g

1 GBq means 1,000,000,000 Bq (1 billion Bq)



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In your problem,
t_(1/2) = 5600 years

Specific activity of living plant = 15 Bq/g

For 4 grams of carbon in a living plant , activity is
4 * specific activity of living plant
=4*15 Bq = 60 Bq
For a piece of wood weighing 4 g the activity is
30 Bq (given)

Now the activity is (-dN/dt) which is proportional to the number of radioactive atoms present. Hence, in the piece of wood the ratio of the C14 atoms that remained compared to that were present when the plant was alive is:

N/N_0 = 30Bq/60Bq = 1/2

Hence the age is

t_age = - [t_(1/2)/ln(2)] * ln(N/N_0)
= - 5600 yeras * (1/ln(2)) * ln(1/2)
= -5600 years * [1/ln(2)] * [-ln(2)]
= 5600 years

Cheers.

Answer 2

You got a very detailed and precise answer but in this case, the calculations are pretty simple. Let us see how:

First understand what is half life. The quantity of a radioactive substance decreases by half in a time eriod T. In this case, T is 5600 years. So we have for example:

Time, Activity

0, 15

5600, 7.5

11200, 3.75

16800, 1.875

and so on. You got the idea.

In our case, we have 15 Bq. to start with for each gram and we have 4 grams to start with. So we have 4 X 15 = 60 Bqs at time = 0

After a time t years, we found that the 4g piece is having only 30 Bq of radioactivity left. That correspons to 7.5 Bq for each gram. We see the table above and find that t = T, i.e 5600 years have passed after the plant has become dry wood and stopped absorption.

The logarithms are needed and are useful for other periods and that way the other answer is very useful. There are graphs available if you can't use logartihms yet.

Answer 3

ok, 0.5 existence applies to components that are radioactive (volatile) The 0.5 existence is the time it takes for 0.5 the radioactive atoms that are present day to decay. operating example, Substance X has a 0.5 existence of one hundred years. contained in the year 0, it had 2 hundred radioactive atoms. Then one hundred years later, 0.5 of those would have decayed, so it would then have one hundred radioactive atoms. Then one hundred years extra later, 0.5 of the single hundred atoms would have decayed, so there will be 50 radioactive atoms left. Then yet another one hundred years later, 0.5 of the remaining radioactive atoms would have decayed, so there would merely be 25 radoiactive atoms left. you'll hit upon about it on BBC bitesize it truly is a really solid information superhighway web site for revision :)