Scientific Tutor
Chem – Calculating Radioactive Half-Life
Author: Andrew
How do you calculate the radioactive half-life?
Radioactive half-life can be used to describe how old something is by measuring how much of a particular isotope is remaining after a certain amount of time. The mathematics of how radioactive decay works can be a little confusing. If you went over the integrated rate law section, then you will know the calculations coming up are called first order equations. If you did not go over this section in you class don’t worry I can explain it here.
Radioactive decay actually functions as a probability. That is that no one nucleus of an atom is certain to decay at a certain time point. However, if you take a lot of atoms of one element and one specific isotope and measure how fast all of them decay, you will get a consistent answer. That answer follows what we call a half-life reaction. That means that half of the amount or sample that you start with will always disappear or be broken down over the half-life time. This means you are always cutting your sample in half at every half-life and technically the entire sample can never actually disappear. It is what math students would call an inverse square. There are 2 important formulas for half-life that students have to know. They are written below.
Formula 1:
| Time = | Number of Half-Lives |
| Half-Life |
| Time = | n |
| Half-Life |
In the above formula, we have two time numbers we are looking at and some students can get them confused. The time in the top left box is how long we are measuring with a timing device like a stopwatch or clock. The half-life time in the bottom left box is how long it takes for half the sample you have to break down and become something else. The number of half-lives is telling the person how many times it has undergone a half-life. You can also represent the number of half-lives with the letter n. This is the second way to write the formula I have shown above.
Formula 2:
| Amount at Start * ( 0.5n) = | Amount at End |
| 1 |
In this formula, the n is representing the same thing as in the previous formula. It is the number of half-lives that the sample has undergone. This means that when you find the number of half-lives, you then may be able to solve either formula. The amount at the start is always going to be the larger of the amounts you are given. For example, if a problem has the numbers and units of 50g and 12.5g, the 50g is your amount at start. You may have also guessed that your amount at end is always going to be the smaller number. So the 12.5g would be your amount at end in the same example I just gave.
Examples: Solve the half-life reactions below.
If a certain isotope has a half-life of 4500 years and has been around for 22500 years, then how many half-lives has it gone through?
Answer: 5 half lives
If we start out with 40g of an isotope, how much will we have after 4 half-lives?
Answer: 2.5g
If silver-108 has a half-life of 418 years, how long would it take for a sample to go from 8g to 0.0625?
Answer: 2926 years.
VIDEO Half-Life Demonstrated Example 1: If argon-39 has a half-life of 269 years, how many half-lives has it gone through after 1076 years?
What information does the problem give you?
Answer:
Half-life = 269 years
Time = 1076
YOU DO NOT NEED the isotope information (argon-39) EVER with half-life problems.
What does the question ask you?
Answer: number of half lives = ?
How do we set up the problem?
Answer: Start with the equation that includes everything you have and need.
| Time = | Number of Half-Lives |
| Half–Life |
What can we fill in?
Answer: the information that the problem gives you.
| 1076 years = | Number of Half-Lives |
| 269 years |
How do I do the calculations?
Answer: 1076 / 269 = 4
| 1076 years = | 4 half-lives |
| 269 years |
What is the complete answer?
COMPLETE ANSWER: 4 half lives
VIDEO Half-Life Demonstrated Example 2: If we start out with 200g of an isotope, how much will we have after 6 half-lives?
What information does the problem give you?
Answer:
amount at start = 200g
number of half-lives (n) = 6
What does the question ask you?
Answer: amount at end = ?
How do we set up the problem?
Answer: Start with the equation that includes everything you have and need.
| Amount at Start * ( 0.5n) = | Amount at End |
What can we fill in?
Answer: the information that the problem gives you.
| 200g * ( 0.56) = | Amount at End |
Apply the exponent to get
| 200g * ( 0.015625) = | Amount at End |
Multiply the left side to get
| 3.125g = | Amount at End |
What is the complete answer?
COMPLETE ANSWER: 3.125g
VIDEO Half-Life Demonstrated Example 3: If a sample of carbon-14 goes from 50g to 1.5625g and its half-life is 5700 years, how long has it been?
What information does the problem give you?
Answer:
amount at start = 50g
amount at end = 1.5625g
half-life = 5700 years
YOU DO NOT NEED the isotope information (carbon-14) EVER with half-life problems.
What does the question ask you?
Answer: time = ?
How do we set up the problem?
Answer: Start with the equation that you have more information for.
| Amount at Start * ( 0.5n) = | Amount at End |
| 1 |
What can we fill in?
Answer: the information that the problem gives you.
| 50g * ( 0.5n) = | 1.5625g |
| 1 |
What next?
Answer: We want to start getting the n alone, so divide both sides by 50g.
| 50g * ( 0.5n) = | 1.5625g |
| 50g | 50g |
Cross out 50g on the left side
| 50g * ( 0.5n) = | 1.5625g |
| 50g | 50g |
Simplify
| ( 0.5n) = | 1.5625g |
| 50g |
Divide 1.5625 by 50 to get
| ( 0.5n) = | 0.03125 |
| 1 |
THIS IS WHERE LOTS OF STUDENTS GET CONFUSED! To solve for an exponent you must use the NATURAL LOG function on your calculator that usually looks like (ln). What you do is take the natural log of both sides. So that means take the natural log of 0.5 ( = -0.6931 ) and then in a separate calculation take the natural log of 0.03125 ( = -3.4657 ). This turns the problem into a multiplication like below.
Take the natural log of both sides to get
| -0.6931 * n = | -3.4657 |
| 1 |
Now divide both sides by -0.6931
| -0.6931 * n = | -3.4657 |
| -0.6931 | -0.6931 |
Cross out -0.6931 on the left side
| -0.6931 * n = | -3.4657 |
| -0.6931 | -0.6931 |
Simplify
| n = | -3.4657 |
| -0.6931 |
Divide right side. You have to estimate a little.
| n = | 5 |
| 1 |
This means you have 5 half-lives but you are not done yet.
Now we move to the second equations both with the information of 5 half-lives and with some of our original information of 5700 years for one half-life.
| Time = | Number of Half-Lives |
| Half-Life |
Fill in the information
| Time = | 5 |
| 5700 years |
Multiply both sides by 5700 years
| 5700 years * Time = | 5 * 5700 years |
| 5700 years |
Cross out 5700 years on the left side
| 5700 years * Time = | 5 * 5700 years |
| 5700 years |
Simplify
| Time = | 5 * 5700 years |
| 1 |
Multiply the right side to get
| Time = | 28500 years |
| 1 |
What is the complete answer?
COMPLETE ANSWER: 28500 years
PRACTICE PROBLEMS: Solve the half-life reactions below.
If we start out with 80g of an isotope, how much will we have after 3 half-lives?
Answer: 10g
If a certain isotope has a half-life of 1300 years and has been around for 11700 years then how many half-lives has it gone through?
Answer: 9 half-lives
How much would be left of 400g of titanium-44 after 441 years, if it had a half-life of 63 years?
Answer: 3.125g
If silicon-32 has a half-life of 170 years, how long would it take for a sample to go from 12g to 0.1875g?
Answer: 1020 years
Posted in Chemistry
