1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    gneill

    User Avatar

    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

    User Avatar
    Homework Helper
    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Natural Log on Radioactive Decay Formula
  1. Radioactive Decay? (Replies: 1)

  2. Radioactive Decay (Replies: 5)

  3. Radioactive decay (Replies: 5)

Loading...