What follows is not a calculus-based discussion concerning half-life calculations. Later on, you may learn the calculus-based approach. And that will be a good thing.

Doing half-life problems is focused on using several equations. The order in which you use them depends on the data given and what is being asked. Here is the first equation:

(1/2)^{number of half-lives}= decimal amount remaining

Let us use several different half-lives to illustrate this equation.

(1/2)^{0}= 1

In this example, zero half-lives have elapsed. The 1 represents the decimal fraction remaining. In other words, at the very start, before any decay has taken place, 100% of the material is on hand. Keep in mind that the decimal amount times 100 becomes the percentage.

(1/2)^{1}= 0.5

This second example shows one half-life elapsed. At this point 0.5 of the original amount remains. 0.5 expressed as a percentage is 50%. However, please be aware that it is the decimal amount that will be used in the various calculations, not the percentage.

Just below are the amounts remaining after 2, 3, and 4 half-lives.

(1/2)^{2}= 0.25

(1/2)^{3}= 0.125

(1/2)^{4}= 0.0625

I would like to stress that the decimal amount is the amount that __remains__ after a given number of half-lives. Many times, a problem will specify how much has decayed and you must use that information to determine how much remains. It's the amount that __remains__ that goes into the calculation.

Sometimes, you will see the equation expressed thusly:

1/2^{n}<--- the exponent is applied only to the 2, as in one over 2^{n}

This is often done in order to highlight the amount remaining progression in a fractional way. For example:

1/2^{2}= 1/4

The ChemTeam thinks the following is good advice: learn to recognize which whole-number half-lives are associated with their fractional amounts and the decimal amount. Thusly:

half-lives exponent fraction decimal one 1/2 ^{1}1/2 0.5 two 1/2 ^{2}1/4 0.25 three 1/2 ^{3}1/8 0.125 four 1/2 ^{4}1/16 0.0625 five 1/2 ^{5}1/32 0.03125

Decay problems at the start of your study of these problems will often be whole number half-lives, as in one half-life, two half-lves, three half-lives and so on. However, as you advance, you will see values like 2.45 and 0.5882 for the number of half-lives elapsed. In that case, you need to turn to your calculator and do a calculation using the y^{x} (or x^{y}) key. Thusly:

(1/2)^{2.45}= 0.18301

1/2^{0.5882}= 0.66517

You can raise 0.5 to the proper power, as in the 2.45 problem. If you want, you can raise 2 to the power and then use the 1/x key, as you would in the 0.5882 problem.

The second equation to be aware of concerns the number of half-lives that have occurred. It can be expressed as

total time elasped ÷ length of half-life = number of half-lives elapsed

Often, the problem will tell you the length of the half-life (say, 5730 years, the half-life of carbon-14) and then ask you about how much remains after, for example, 17190 years. In that case, you do this:

17190 yr / 5730 yr = 3 half-lives

Or, for example, the problem gives you 12,300 years elapsed:

12300 / 5730 = 2.1466 half-lives <--- note I dropped the units because I knew they would cancel

Usually, you then go to the first equation discussed above in order to determine the decimal amount remaining. After you've memorized the whole-number half-lives, you'll recognize three half-lives as being associated with 0.125 remaining. For 2.1466 half-lives, you turn to the calculator:

(1/2)^{2.1466}= 0.225844

The third equation is this:

starting amount times decimal amount remaining = remaining amount

To use symbols, we have this:

(N_{o}) (decimal amount remaining) = N

The subscripted 'o' indicates the starting (or original, hence the letter 'o') amount. Normally, N_{o} and N are expressed in grams.

You will sometimes see equation two:

total time elasped ÷ length of half-life = number of half-lives elapsed

inserted as a fraction into the exponent portion of equation one. Here's equation one:

(1/2)^{number of half-lives}= decimal amount remaining

And here is the fraction inserted into equation one:

decimal amount remaining = (1/2)^{total time / half-life time}

You can then substitute into equation three:

(N_{o}) (1/2)^{total time / half-life time}= N

For example, problem #1 just below can be set up like this:

(88.0) (1/2)^{x / 12.3}= 11.0

Just below, I have solved Examples #1 and #3 using both a several-step approach (the ChemTeam's preferred way) and using the single equation just above.

If you want to see more of the one-equation approach, please go here. The link takes you to one example and there are links to nine others in a sidebar.

**Example #1:** How many years will it take for 88.0 grams of tritium to decay to an 11.0 gram sample? (The half-life of tritium is 12.3 years.)

**Solution:**

1) Determine the decimal amount remaining:

11.0 g / 88.0 g = 0.125

2) Determine the number of half-lives:

(1/2)^{n}= 0.125n log 0.5 = log 0.125

n = log 0.125 / log 0.5

n = 3

3) Determine years elapsed:

12.3 yr times 3 = 36.9 yr

Comment: you could recognize 0.125 as being associated with 3 half-lives and avoid the calculation in step 2.

**Solution using the one equation form:**

(N_{o}) (1/2)^{total time / half-life time}= N(88.0) (1/2)

^{x/12.3}= 11.0(1/2)

^{x/12.3}= 0.125log (1/2)

^{x/12.3}= log 0.125(x/12.3) (-0.30103) = -0.90309

x/12.3 = 3

x = 36.9 yr

**Example #2:** How many years will it take for 84.0 grams of tritium to decay to a 23.5 gram sample? (The half-life of tritium is 12.3 years.)

**Solution:**

1) Determine the decimal amount remaining:

23.5 g / 84.0 g = 0.279762Note that I rounded off, but not too much.

2) Determine the number of half-lives:

(1/2)^{n}= 0.279762n log 0.5 = log 0.279762

n = 1.837728

3) Determine years elapsed:

12.3 yr times 1.837728 = 22.6 yr

Comment: in these calculations, I carry several extra digits (called guard digits) from each step to the next. I only round off to the proper number of significant figures at the end of the calculation.

**Example #3:** A 208 g sample of sodium-24 decays to 13.0 g of sodium-24 in 60.0 hr. What is the half-life of this radioactive isotope?

**Solution #1:**

1) Determine the decimal amount remaining:

13.0 g / 208 g = 0.0625

2) Determine the number of half-lives:

recognize 0.0625 as being associated with 4 half-livesor

(1/2)

^{n}= 0.0625n log 0.5 = log 0.0625

n = 4

3) Determine the length of the half-life:

60.0 hr / 4 = 15.0 hr

**Solution #2:**

1) Set it up in one equation:

(208) (1/2)^{60/n}= 13

2) Divide both sides by 208:

0.0625 = (1/2)^{60/n}Note: the 0.0625 comes from 13 divided by 208

3) Take the log of both sides:

log 0.0625 = log (1/2)^{(60/n)}log 0.0625 = (60/n) log 0.5

-1.2041 = (60/n) (- 0.3010)

4) Divide by -0.3010:

4 = 60/n4n = 60

n = 15 hr

**Example #4:** Calculate the half life of a sample of ChemTeamium which decays from 27.5 g to 0.598 g in a period of 574 years.

**Solution:**

0.598 / 27.5 = 0.02174545(1/2)

^{n}= 0.02174545n = 5.52314

574 yr / 5.52314 = 104 yr (to three sig figs)

**Example #5:** The radioactive element carbon-14 has a half-life of 5730. years. The percentage of carbon-14 present in the remains of plants and animals can be used to determine age. How old is an object that has lost 60% of its carbon-14?

**Solution:**

1) Determine how many half-lives have elapsed:

(1/2)^{n}= 0.40n log 0.5 = log 0.40

n = 1.32193

2) Calculate the age of the object:

5730 yr times 1.32193 = 7575 yr

Note that the problem said 60% was lost. Since the solution technique uses the amount remaining, I used 40%. Be aware that this is a common thing: to give you the amount lost, which you have to convert into the amount remaining.

Also, note that I used the decimal equivalent of 40%. Don't use a percentage in the calculation, use the decimal amount, 0.4 in this example.

**Example #6:** An element has a half-life of 25 years. You currently have 20 grams. How many grams did you have 100 years ago?

**Solution:**

100 yr / 25 yr = 4 <--- this is the number of half-lives(1/2)

^{4}= 0.0625 <--- this is the decimal amount remaining after 4 half-lives20 g / 0.0625 = 320 g <--- the answer

Here's the 320 g decaying over four half-lives:

320 ---> 160

160 ---> 80

80 ---> 40

40 ---> 20

**Example #7:** P-32 is radioactive isotope with a half-life of 14.3 days. If you currently have 73.3 g of P-32, how much P-32 was present 4 days ago?

**Solution:**

1) How many half-lives have elapsed in 4 days?

4 days / 14.3 days = 0.27972

2) The decimal amount that remains after 0.27972 half-lives have elapsed:

(1/2)^{0.27972}= 0.82375

3) Determine amount from 4 days prior:

73.3 is to 0.82375 as x is to 1x = 88.9833 g

rounded to three sig figs, 89.0 g

**Example #8:** If a sample contains 144 atoms of Au-179 (half-life = 7.5 s), approximately how many such atoms were present 30 seconds earlier?

**Solution:**

30 s / 7.5 s = 4 <--- four half-lives elapsed during the 30 s(1/2)

^{4}= 0.0625 <--- decimal amount remaining after four half-livesx is to 1 as 144 is to 0.0625

144 / 0.0625 = 2304

**Example #9:** Find the half-life of an element which decays at a rate of 3.402% per day.

**Solution:**

1) After one day, this much remains:

100% minus 3.402% = 96.598%We will use it as 0.96598

2) How many half-lives have elapsed?

(1/2)^{n}= 0.96598n log 0.5 = log 0.96598

n = 0.049935

3) Find the half-life:

1 day is to 0.049935 as x is to 1x = 20 days

Comment: you could set up a spreadsheet and do it by brute force, subtracting 3.402% of the material on hand each day, with the half-life being the number of days needed to arrive at 50%.

Day 19 to 20 would be this:

x - 0.03402x = 50%

where x is equal to 51.76%

**Example #10:** The half-life of In-111 is 0.007685 years; how long (in hours) would it take for the amount of In-11 to decrease to 43.24% of its initial amount?

**Solution #1:**

(1/2)^{n}= 0.4324n log 0.5 = log 0.4324

n = log 0.4324 / log 0.5

n = 1.20956 <--- the number of half-lives

1.20956 times 0.007685 yr = 0.00930 yr

0.00930 yr times (365 day / yr) times (24 hr / day) = 81.468 hr

To three sig figs, this is 81.5 hours

**Bonus Example:** A scrap of paper taken from a Dead Sea scroll was found to have a C-14/C-12 ratio of 0.795 times than found in plants living today. Estimate the age of the scroll.

**Solution:**

(1/2)^{n}= 0.795 <--- n is the half-life, 0.795 is the decimal amount of C-14 remainingn log 0.5 = log 0.795

n = 0.330973 <--- the number of half-lives elapsed

The problem does not provide the half-life of C-14. We look it up and find it to be 5730 year.

0.330973 times 5730 yr = 1896.47529

Rounded off to three significant figures, the answer would be 1.90 x 10

^{3}years. (Using 1900 would be wrong, as that shows only two sig figs. Using 1900. would also be incorrect, as that shows four SF.)