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Nuclear Half-Life Calculations

Key Questions

  • Nuclear half-life expresses the time required for half of a sample to undergo radioactive decay. Exponential decay can be expressed mathematically like this:

    #A(t) = A_0 * (1/2)^(t/t_("1/2"))# (1), where

    #A(t)# - the amount left after t years;
    #A_0# - the initial quantity of the substance that will undergo decay;
    #t_("1/2")# - the half-life of the decaying quantity.

    So, if a problem asks you to calculate an element's half-life, it must provide information about the initial mass, the quantity left after radioactive decay, and the time it took that sample to reach its post-decay value.

    Let's say you have a radioactive isotope that undergoes radioactive decay. It started from a mass of 67.0 g and it took 98 years for it to reach 0.01 g. Here's how you would determine its half-life:

    Starting from (1), we know that

    #0.01 = 67.0 * (1/2)^(98.0/t_("1/2")) -> 0.01/67.0 = 0.000149 = (1/2)^(98.0/(t_("1/2"))#

    #98.0/t_("1/2") = log_(0.5)(0.000149) = 12.7#

    Therefore, its half-life is #t_("1/2") = 98.0/(12.7) = 7.72# #"years"#.

    So, the initial mass gets halved every 7.72 years.

    Sometimes, if the numbers allow it, you can work backwards to determine an element's half-life. Let's say you started with 100 g and ended up with 25 g after 1,000 years.

    In this case, since 25 represents 1/4th of 100, two hal-life cycles must have passed in 1,000 years, since

    #100.0/2 = 50.0# #"g"# after the first #t_("1/2")#,

    #50.0/2 = 25.0# #"g"# after another #t_("1/2")#.

    So, # 2 * t_("1/2") = 1000 -> t_("1/2") = 1000/2 = 500# #"years"#.

  • Answer:

    T=th xlog(m./m)/log(2)

    Explanation:

    You could use this formula:
    Where Th = half-life.
    M. = the beginning amount
    M = the ending amount

    One example of how to use the equation:

    One of the Nuclides in spent nuclear fuel is U-234, an alpha emitter with a half-life of 2.44 x10^5 years. If a spent fuel assembly contains 5.60 kg of U-234, how long would it take for the amount of U-234 to decay to 0.35? First, break down the complicated equation:
    T= unknown
    Th= 2.44 x10^5 or 244000
    M.= 5.60
    m= 0.35

    Then plug the number into the equation:

    T= 244000 xlog(5.60/0.35)/log(2)

    T= 244000x1.20412/log(2)

    So the Answer is: T= 9.76x10^5 min.

    Note: The example was a question from my quiz I took three weeks ago!

Questions