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Homework Help: Half life :s

  1. Oct 29, 2008 #1
    can help me with this question :S?
    If 70% of a radioactive substance remains after one year, find its half-life.

    and

    Strontium-90 is a radioactive isotope with a half-life of 29 years. If you begin with a sample of 800 units, how long will it take for the amount of radioactivity of the strontium sample to be reduced to
    (a) 400 units
    (b) 200 units
    (c) 1 unit


    *it has to do wit exponential growth and decay and maybe... with differential equations
     
  2. jcsd
  3. Oct 29, 2008 #2

    CompuChip

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    Welcome to PF kbutto.
    The half-life t1/2 is the time until only 50% remains. After two half-lifes, 25% remains, after three, 12,5%, etc.
    The general formula we can infer is then: after n half life periods the sample has been reduced by a factor 2n, so the percentage that is left is then 1/2n (after 0 half lifes, this gives 1 = 100%, after 1 half-life it gives 1/2, after 2 it gives 1/4, etc.)

    Using this, you can solve both questions.

    a) After how many half-lives is there 70% left?
    b) How many half-lives have passed once you have gone from 800 to 400 units?
     
  4. Oct 29, 2008 #3

    HallsofIvy

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    70%= 0.7= (1/2)x. What is x?

    400= (1/2)x*800. 200= (1/2)y*800. 1= (1/2)[z*800. What are x, y, and z?


    Yes, so you may need to use logarithms.
     
  5. Oct 29, 2008 #4
    I worked it out this way, is it correct?

    a) If only 70% of a substance remains, then the final amount y = 70%yo, and it's given that time is one year.

    y = 70%yo = yoe^(-k(1))

    0.70yo = yoe^(-k)

    0.70 = e^(-k)

    k = -ln(0.70) = 0.3567 ; Half-life constant of the substance.

    Now use this to find the half-life, the final amount should be half original:

    y = (1/2)yo = yoe^(-0.3567t)

    (1/2)yo = yoe^(-0.3567t)

    0.5 = e^(-0.3567t)

    t = ln(0.5) / -0.3567 ≈ 1.9 ≈ 2 years

    b) For strontium, use the values given to first find the half-life constant of the strontium.

    y = (1/2)yo = yoe^(-29t)

    0.5 = e^(-29t)

    t = ln(0.5) / -29 ≈ 0.024 ; Half-life constant of strontium

    now use this to find the time for each case:

    1) y = yoe^(-0.024t)

    400 = 800e^(-0.024t)

    t = ln(0.5) / -0.024 ≈ 28.88 ≈ 29 years

    2) y = yoe^(-0.024t)

    200 = 800e^(-0.024t)

    t = ln(0.25) / -0.024 ≈ 57.76 ≈ 58 years

    3) y = yoe^(-0.024t)

    1 = 800e^(-0.024t)

    t = ln(1/800) / -0.024 ≈ 278.52 ≈ 279 years

    yo= y subscript0
     
  6. Oct 30, 2008 #5

    HallsofIvy

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    Yes, that's correct but isn't this simpler? Saying the half life is T means the quantity is multiplied by 1/2 every half-life. There are t/T "half-lives" in time t. In one year, t= 1 so 0.7= (0.5)1/T so ln(0.7)= ln(0.51/T)= ln(0.5)/T. T= ln(0.5)/ln(0.7)= 1.94

    But WHY use exponentials for this simple problem?
    400= 1/2 (800) so this requires exactly one half-life. Which you were told is 29 years.

    200 is 1/2(400) so this is another half-life: 29+ 29= 58 years.
    Or 200= (1/4)(800)= (1/2)2(800) so 2 half-lives: 58 years.

    The last is a little harder: 1= 800(1/2)x so (1/2)x= 1/800, which is not an integer power of 2. x ln(1/2)= ln(1/800) x= ln(1/800)/ln(1/2)= ln(800)/ln(2)= 9.64 half-lives or 9.64(29)= 280 years.
     
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