What follows does not explicitly use the general chemical concept of kinetics. Later on, you may learn that approach to half-life calculations, one that uses calculus to develop the concept. 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:

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

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

1/22 = 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 1/21 1/2 0.5 two 1/22 1/4 0.25 three 1/23 1/8 0.125 four 1/24 1/16 0.0625 five 1/25 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 yx (or xy) key. Thusly:

(1/2)2.45 = 0.18301
1/20.5882 = 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:

(No) (decimal amount remaining) = N

The subscripted 'o' indicates the starting (or original, hence the letter 'o') amount. Normally, No 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:

(No) (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 = 0.125

n 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:

(No) (1/2)total time / half-life time = N

(88.0) (1/2)x/12.3 = 11.0

(1/2)x/12.3 = 0.125

log (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 = 0.279762

Note that I rounded off, but not too much.

2) Determine the number of half-lives:

(1/2)n = 0.279762

n 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-lives

or

(1/2)n = 0.0625

n 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/n

4n = 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 = 0.02174545

n = 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 = 0.40

n 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-lives

20 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 1

x = 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-lives

x 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 = 0.96598

n log 0.5 = log 0.96598

n = 0.049935

3) Find the half-life:

1 day is to 0.049935 as x is to 1

x = 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 = 0.4324

n 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 remaining

n 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 103 years. (Using 1900 would be wrong, as that shows only two sig figs. Using 1900. would also be incorrect, as that shows four SF.)