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Half-life of radioactive substance

  1. Feb 8, 2011 #1
    1. The problem statement, all variables and given/known data

    If 20% of a radioactive substance disappears in 70 days, what is its half-life?

    2. Relevant equations

    y = C*e^(k*t)
    where t is time in days
    k is the constant of proportionality?
    y is the current amount of substance

    3. The attempt at a solution

    20% disappears, so 100% - 20% = 80% remaining = 0.80 after 70 days
    .8*C = C*e^(70*k)
    Solve for k
    k = -3.188 * 10^-3

    Plug k value back into equation to find t at the substance's half-life (.5)
    .5*C = C*e^((-3.188 * 10^-3)*t)
    Solve for t
    t = 217.44

    217.44 days is the substance's half-life.

    Is that correct?
    Thank you.
  2. jcsd
  3. Feb 8, 2011 #2
    It matches what I get.
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