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Homework Help: Natural Log on Radioactive Decay Formula

  1. Jan 26, 2013 #1
    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. jcsd
  3. Jan 26, 2013 #2


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    Staff: Mentor

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

    Your calculator should have an ##e^x## function button on it.
  4. Jan 26, 2013 #3

    rude man

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    Gold Member

    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).
  5. Jan 26, 2013 #4


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    welcome to pf!

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

    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 :smile:
  6. Jan 26, 2013 #5

    rude man

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    Re: welcome to pf!

    What sort of arcana is this? :bugeye:
  7. Jan 26, 2013 #6
    Thank you so much! :!!) Internet is amazing and I really appreciate you smart people behind it who make this a fantastic tool for learning. :smile:

    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
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