Using the half – lives of radioactive elements

In this presentation we will learn:

1. That there is an isotope of carbon that is useful for dating materials that have once been alive many years ago.

2. That there is a useful relationship between the radioactive decay constant and half-life that we can call upon to work out the age of ancient artifacts.

3. That we can call upon Avogadro’s number to quickly calculate the number of atoms in any given mass of a sample of known atomic mass.

You’ve probably heard of radiocarbon dating.

This is how it works.

The CO2 that plants take in is slightly radioactive. This is due to some radioactive Carbon formed in the atmosphere by cosmic rays. 12C is turned, magically, into a radioisotope 14C.

When alive, plants have a constant ratio of these isotopes in their tissue. So do you – you’re breathing!

When they die, 14C is no longer replenished. It starts to decay. The nuclei obey our model rule – a constant probability of a decay in a fixed time.

And what’s the time scale?

In this case, few thousand years.

Atoms of 14C

Time in thousands of years

14C when alive

14C after death: decay by half every 5570 years.

5570 y

In the next 5570 years, HALF of what there is of the 14C will decay.

In the following 5570 years, half of what’s left decays (leaving ¼) and so on.

This is the HALF – LIFE of 14C.

Half – life is NOT half the total life. It’s the time taken for half of what there is to decay.

A half-life of thousands, or billions, of years can be found in a few minutes!

1. Find the activity of the sample of a KNOWN number of radioactive nuclei.

2. From this find the decay constant . activity = N.

3. Use this simple formula:

0.693

=t 1/2

Half - life

ln 2

=

Half – life tells you how long the substance lasts.

The decay constant tells you how fast it decays.

Another RECIPROCAL relationship!

Radioactive clock:

In L half – lives, number of nuclei N is reduced by a factor 2L.

e.g. in 3 half-lives, N is reduced by 23 = 8.

•Find activity.

•Find factor F by which activity has been reduced.

•Calculate L so that 2L = F

•L =log2 F

•Age = t½ L

ExampleThe carbon in an axe handle was found to contain 1 part in 40 000 000 Carbon14.

How old is the axe?Originally, the C14 was one part in 10 000 000.

After one half-life, it’s down to one part in

20 000 000.

After 2 half-lives, the C14 is down to one part in

40 000 000.So, the age of the axe must be………………………….

……………….2 x 5600 years = 11200 years old.

Finding ages of rocksUranium isotopes have long half-lives and decay into LEAD, which is stable.

e.g. U238 half-life = 4.5 billion years.

Initially the rock contains 100% U, 0% Pb.

After one half-life, it’s 50% U, 50% Pb.

After 2 half-lives, it’s 25% U, 75% Pb.

After 3 half-lives, it’s 12.5% U, 87.5% Pb……and so on..

(you don’t usually go more than 3 half-lives).

RATIO of U to Pb:Initially:

1:0 U to Pb

After one half-life:

1:1

After 2 half-lives:

1:3

After 3 half-lives:

1:7

LEARN THESE RATIOS!!!

The same thing applies to Potassium 40.

It decays into Argon 40, which is stable.

It’s used to date IGNEOUS rocksIt’s used to date IGNEOUS rocks

The same ratios apply:

Initially 1:0 40K:40Ar

After one half-life 1:1

After 2 half-lives 1:3

After 3 half-lives 1:7

Some typical AS and A2 examples of radioactivity:

Examples:

Q1. 90Sr has a half-life of 27 years. 24Na has a half-life of 15 hours.

a) Which is more active? Why?

b) After how long will their activities be equal?

Q2. 239Pu has a half-life of 24000 year and an activity of 1.5 x 1013 Bq. Each alpha particle has an energy of 5 Mev. (1 year = 3 x 107 s).

a) What’s the power output of 239Pu?

b) What’s its approximate power output after 50000 years?

c) How much energy does it emit in one year?

What assumption have you made?

Answers:Answers:

1 a) 24Na, because in one second more Na atoms will disintegrate than Sr atoms.

1 b) 27 years is 1.6 x 104 longer than 15 hours.

Initially, Na will be 1.6 x 104 times more active than Sr.

After n periods of 15 hours it will have fallen by a factor of 1.6 x 104

Where 2n = 1.6 x 104

n is approximately 14

14 x 15 hours = 9 days.

Answers, continued:

2 a) Power = 5 x 106 x 1.6 x 10-19 x 1.5 x 1013 Js-1

= 12 W

2 b) Power is approximately ½ , since this is after 1 half-life.

2 c) Power over 1 year = 3 x 107 x 12

= 3.6 x 108 W, assuming power is constant over one year.

Q3. A sample of radioactive material contains 1018 atoms. Its half-life is 2 days.

a) What is the fraction of atoms remaining after 5 days?

b) What is the activity of the sample after 5 days?

Q4. 87Kr has a half-life of 78 minutes. What is the activity of 10 g of 87Kr?

(L = 6.0 x 1023 mol-1)

Q5. A radioactive material has an activity of 9.0 x 1012 Bq. Its half-life is 80.0 s. How long will it take for its activity to reach 2.0 x 1012 Bq?

Answers:

3 a) N/No = e -t

= ln2/half-life = 0.693/2.0 days-1

t = 0.693 x 5.0/2.0 = 1.73

So N/No = e –1.73 = 0.177

3 b) dN/dt = - N

N = 0.177 x 1018

= 0.693/2 x 24 x 3600 s-1

So dN/dt = - (0.693 x 0.177 x 1018)/2 x 24 x 3600

= - 7.09 x 1011 s-1.

Answers:

4. 10 g of 87K contains 10 x 10-6 x 6 x 1023/87 atoms

= 6.9 x 1016 atoms

dN/dt = - N

= 0.693/ 78 x 60 s-1

So dN/dt = 0.693 x 6.9 x 1016/ 78 x 60

= 1.02 x 1013 Bq

5. N/No = e-t, so A/Ao = e-t since N is proportional to A.

So ln (A/Ao) = - t

And therefore ln (2 x 1012)/(9 x 1012) = -0.693 x t/80

t = 174 s

Radioactive dating questions.Q 1. 14C has a half-life of 5570 years. It’s present as CO2 in the

atmosphere.

The rate of production of 14C and therefore its concentration, has stayed fairly constant over the past 10000 years. All living organisms contain a small amount of 14C when they are alive. When they die, they stop exchanging CO2 with the atmosphere.

a) When an organism is alive, it contains 1 atom of 14C to every 1010 atoms of 12C. In one second, a fraction 4 x 10-12 of the atoms of 14 C may decay. Estimate the activity of one gram of 14C from a living organism. ( L = 6.0 x 1023 mol-1)

b) What activity would you expect from the bones of a bison eaten by man 11 000 years ago, during the Ice Age?

c) It is found that the ratio of 14C to 12C in a piece of wood is 0.2 x

10-12, whereas that in living wood is 0.8 x 10-12. When did the wood die?

Q2. A rock contains radioactive Potassium, 40K, which decays to a stable isotope of Argon, 40Ar.

The decay rate in the sample is 0.16 Bq.

Mass of Potassium in the sample is 0.6 x 10-6 g

Mass of Argon in the sample is 4.2 x 10-6 g

a) Find for the Potassium and its half-life.

b) Find the age of the rock, assuming there was no Argon in the rock originally.

c) Write about difficulties involved in measuring an activity of 0.16 Bq.

Answers to radioactive dating questions:

Q1 a) 1 g of 12C contains 0.5 x 1023 atoms.

So 1 g C contains 0.5 x 1023 atoms of 14C.

Fraction of atoms decaying per s = 4 x 10-12

So number of 14C atoms decaying per s = 0.5 x 1013 x 4 x 10-12

= 20 Bq

b) 11000 y is approx 2 half-lives. Activity is approx 5 Bq.

c) 14C/12C(old) = 0.2 x 10-12

14C/12C(new) = 0.8 x 10-12

So the factor relating to the time taken to decay is 0.8/0.2 = 4. This amounts to a factor of 22 or 2 half-lives. The age of the wood is 11000 years.

Q2 a) dN/dt = - N

So 0.16 = - x 0.6 x 10-6 x 6 x 1023 /40

= - x 0.09 x 1017.

Thus = 1.8 x 10-17 s-1.

And the half-life is 0.693/ = 1.3 x 109 years.

b) Total 40K originally is 0.6 x 10-6 g, or 1/8 of the original.

Since 1/8 is 1/23, the rock is 3 half-lives old, or about 4 x 109 years old.

c) The background count could be bigger than 0.16 Bq, leading to measurement problems.