# Homework Help: Tutorial: Half-Lives and Radioactive Decay

1. Jan 11, 2009

### jacksonpeeble

After asking my own question on half-lives and radioactive decay despite having read the library article on Radioactive Decay, I felt that I should post a less formal, more in-depth explanation of how to actually solve the equations.

There are two primary equations that I use when dealing with standard radioactive decay.
1. N=N0ek*t
2. r=ln2/k

N represents the final amount remaining.
N0 represents the initial amount.
e is the standard variable for ~2.718
k is the constant exponent for that sample.
t represents the amount of time that passed.
r is the half-life of the sample.
ln stands for natural logarithm.

Depending on which variable you're solving for, you'll need to set up your work in one of several different ways. I've posted each different type of problem, as well as examples, below.

You must usually first determine k before solving a problem (often that will be your problem). Set up your equation r=ln2/k, or switch it around to k=(-ln2)/r if that is more convenient and works for you. From there, you can plug in r (which must be given) and solve for k. Sometimes you may have simply been required to find r, in which case you must be given k (unless you're insanely intelligent and know of some way that I don't).

From there, you simply plug in your values to the other equation, N=N0ek*t. You should now be able to solve for any sort of problem without hassle, as you have the k value in addition to other already provided values by your instructor (remember, you can only solve for one variable at a time unless you have a system of equations).

Example:
The half-life of radium-226 is 1600 years. Suppose you have a 22mg sample. After how long will only 18mg of the sample remain?

k=(-ln2)/r
1600=(-ln(2))/k)
k=(-ln2)/1600
k=~-0.000433216988

N=N0ek*t
18=22*e-0.000433216988*t
t=~463.210587512

463.211 years

As always, when you have your answer, remember to round to correct number of significant digits in chemistry, or typically three decimal places in mathematics. Make sure to include units.

2. Jan 12, 2009

### HallsofIvy

All exponentials are equivalent. Since this is "half-life", it can be done simply by using base 1/2. If the half life is 1600 years, then the amount would be multiplied by 1/2 every 1600 years. In t years, there will be t/1600 "1600 year long" units: If A is the amount at time t= 0, after t years the amount will be A(1/2)t/1600.

The example was: "The half-life of radium-226 is 1600 years. Suppose you have a 22mg sample. After how long will only 18mg of the sample remain?"
So you need to solve the 22(1/2)t/1600= 18. (1/2)t/1600= 18/22= 9/11. Taking logarithms of both sides, (t/1600)log(1/2)= log(9/11) so t= (1600)(log(9/11)/log(1/2)= (1600)(-0.08715/-0.3010)= 463.25 years. My answer differs from Jackson Peebles' in the hundreths place because of round off errors.

3. Jan 12, 2009

### Gokul43201

Staff Emeritus
There's a small issue with signs here. From #1, k would have to be a negative number. But from #2, that makes r, the half-life, also negative, which is kind of goofy. So you will have to replace one of the two $k$s with a $-k$. The standard notation involves positive values of $k$, so it would be better to stick in "- sign" into the exponent of #1.

4. Jan 12, 2009

### jacksonpeeble

Thank you for the correction; apparently I implemented it in my example, but left it out of the step-by-step. Unfortunately, it doesn't look like I can revise it... would a moderator mind doing this for me?

Last edited: Jan 12, 2009