Results 1 to 8 of 8

Math Help - Radioactive decay

  1. #1
    Junior Member
    Joined
    Oct 2008
    From
    Green Bay
    Posts
    30

    Radioactive decay

    A sample of a radioactive substance decayed to 91.5% of its original amount after a year. What is the half life of the substance? Also have to find the amount of time for it to decay to 30% of the original amount.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by GreenandGold View Post
    A sample of a radioactive substance decayed to 91.5% of its original amount after a year. What is the half life of the substance? Also have to find the amount of time for it to decay to 30% of the original amount.
    Radioactivity can be defined using this model:

    N=N_0e^{-kt}, where

    N is the current amount of the substance
    N_0 is the initial amount of the substance
    k is the decay constant
    t is the length of time

    Now, we are told that we have .915N_0 after 1 year.

    Thus, .915N_0=N_0e^{-k}

    Solving for k, we see that k=-\ln(.915)\approx 0.0888

    Now, we can find the half life:

    Half life is defined as \lambda=\frac{\ln 2}{k}, where

    \lambda is the half life
    k is the decay constant

    Thus, \lambda=\frac{\ln 2}{.0888}\approx \color{red}\boxed{7.803\text{ years}}

    Now, our decay model is N=N_0e^{-0.0888t}

    Can you find t when N=.3N_0?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    From
    Green Bay
    Posts
    30
    take the natural log of both sides?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by GreenandGold View Post
    take the natural log of both sides?
    Yup. Now what do you get for your answer?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    From
    Green Bay
    Posts
    30
    I got -13 but i did something wrong. is it 7.803(ln(.3)/ln(2))
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2008
    From
    Green Bay
    Posts
    30
    ok i added a negative to the decay since its negative. so i got 1.35535
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by GreenandGold View Post
    ok i added a negative to the decay since its negative. so i got 1.35535
    Hmmm...

    N=N_0e^{-0.0888t}\implies.3N_0=N_0e^{-0.0888t} \implies \ln(.3)=-0.0888t\implies t=-\frac{\ln(.3)}{0.0888}\approx\color{red}\boxed{13.  558\text{ years}}

    --Chris
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2008
    From
    Green Bay
    Posts
    30
    ok i see now!!! thanks...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Radioactive decay
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 22nd 2011, 12:45 AM
  2. Radioactive decay
    Posted in the Differential Equations Forum
    Replies: 10
    Last Post: November 11th 2010, 08:57 PM
  3. Radioactive Decay
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: January 28th 2010, 10:18 PM
  4. Radioactive Decay
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 12th 2009, 01:51 PM
  5. Radioactive Decay
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 14th 2009, 05:22 PM

Search Tags


/mathhelpforum @mathhelpforum