# Homework Help: Natural Log on Radioactive Decay Formula

1. Jan 26, 2013

### jucristina

1. The problem statement, all variables and given/known data

The initial amount of radioactive atoms on a sample of 24Na is 10^10. It's half-life corresponds to 15 hours. Give the amount of 24Na atoms that will disintegrate in 1 day.

2. Relevant equations

I started to solve it using the formula N=Initial Amount of Atoms / e^(λxTime/Half-Life) which ended up as:
10^10 / e^(0.693x24/15)
It is supposed to give me the amount of atoms that did not disintegrate in 1 day so I can continue with the problem but my biggest issue here is understanding how to take the natural log (e) to get to the result which is said to be 0.33x10^10.
Can somedoby please give me easy step by step explanation on the operation above?
Please help before I go crazy. I have an exam soon on which I can NOT use a calculator and it's really important.

3. The attempt at a solution
Final answer is: Initial amount of atoms - Atoms that did not disintegrate= 0.67x10^10. I just do not know how to get there since I got stuck at the formula above. I don't know what to do with the "e" on it.

Last edited: Jan 26, 2013
2. Jan 26, 2013

### Staff: Mentor

"e" is the base of the natural logarithms, e = 2.71828...

Your calculator should have an $e^x$ function button on it.

3. Jan 26, 2013

### rude man

The basic formula is N = N0exp(-t/T). Letting N = N0/2 lets you solve for the half-life t1/2 = T*ln(2). So, given t1/2 you know T, and now use the basic formula with t = 24 hrs to get N(24hrs).

It's better to derive from fundamentals than memorize formulas which may or may not work out.

Of course,your answer needs to be N0 - N(24hrs).

4. Jan 26, 2013

### tiny-tim

welcome to pf!

hi jucristina! welcome to pf!
forget the formula, just look at the obvious meaning of half-life

after 15 hours, it's 1/2, after 30 hours it's 1/4 etc

so after 24 hours it's 1/224/15 = 1/28/5

28 = 256, 35 = 243 (and yes, you should know these!),

so 28/5 must be just over 3

5. Jan 26, 2013

### rude man

Re: welcome to pf!

What sort of arcana is this?

6. Jan 26, 2013

### jucristina

Thank you so much! :!!) Internet is amazing and I really appreciate you smart people behind it who make this a fantastic tool for learning.

So now I know that e = 2.71828 and if I use this value on the exercise I posted I can easily find the correct answer. Alright. Gneill mentioned using a calculator but calculators are NOT allowed on exams. Any good idea of how to solve it without one?
Because right now it seems easy but if I have to find 2.71828^1.1088 during my test I will have no idea of where to start from. :|

I will also try to follow the thinking of the other people who answered my topic.
Thank you so much again!

Last edited: Jan 26, 2013