Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Nuclear Physics

  1. May 31, 2008 #1
    1. The problem statement, all variables and given/known data
    The Age of the Earth is 4.5 × 109 y. Use the information given in the following table to
    calculate the ratio of the number of 238U nuclei at the present time to the number of 235U nuclei
    at the time of the formation of the Earth.

    Isotope Half-life Current natural abundance(%)
    238U 4.468 ×109 y 99.274
    235U 7.04 ×108 y 0.72

    2. Relevant equations
    I used T1/2= ln(2)/lamda to work out lambda for both 235 and 238Uranium.

    I converted time in years to seconds by no. of years * 3.156x10^7 ( I saw it done in my book dont really understand it but anyway :) )

    I used N(t)=N0*e^-lambda*t to work out the number of radioactive nuclei at a given time.

    3. The attempt at a solution

    So basically just using those two equations I came to
    238U N(t) and present to be = 0.498N0
    and 235UN(t) at beginning, (hence t=0) to be N0,
    so i said the ratio is 0.498... However upon doing this I realised I didnt take into account the last column of the table given in the quetsion, and seeing as the question says taking into account the following information, I figure I have gone about this the wrong way...
    Any help much appreciated :)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 31, 2008 #2


    User Avatar
    Homework Helper

    You have the current U-238/U-235 ratio of 99.274/0.72 . You can take this to represent a ratio of individual nuclei, rather than percentages. From the half-life information, you can figure out a multiplier for how many times more of each isotope there would have been 4.5 Gyr ago. If you multiply the numerator and denominator of your current ratio by the appropriate multipliers, you will have the ratio for that time in the past.
  4. Jun 4, 2008 #3
    Ok well to work out the multipliers, Im not sure I have done it right, for 238U I divided 4.5e9 by 4.468e9 to see how many times it goes into it, then i multiplied that by two? because were going backwards from halfing them... and i did the same for 235U, is this the right method? I ended up with 238U past : 199.970009 and 235U past : 9.204.
  5. Jun 4, 2008 #4


    User Avatar
    Homework Helper

    You are correct in dividing 4.5·10^9 by 4.468·10^9 to find the number of U-238 half-lives, which is 1.0072. But you would then raise 2 to that power to get the number of times more U-238 that there was 4.5 Gyr ago. (The amount is halved with each passing half-life, so it would be doubled in going back into the past with each half-life.)

    Similarly, you would find that there are 6.392 U-235 half-lives, so this is the power by which you would raise 2 to find the multiplier for the U-235 abundance for 4.5 Gyr in the past.

    Your abundance ratio for 4.5 billion years ago will have a number close to 200 on top for U-238, but the denominator for U-235 will be substantially larger than 9...
  6. Jun 5, 2008 #5
    So I ened up with 199.54/60.46 for the ratio of them in the past.
    So do I just put the number of u238 present (99.274) over 60.46?
    It seems a pretty pointless calculation..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook