Nuclear Half-Life Calculations

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Key Questions

  • Answer:

    Half-life is the time it takes for half the population of a sample to change into something else.
    See Below


    It is mostly used for radioactive decay. So when an isotope is unstable in it's nucleus (because of too many neutrons, for example), the atom will undergo a process that alters the nucleus (number of protons and electrons).

    Radioactive Carbon is an example, C-14. It has 6 protons and 8 neutrons. Its nucleus is unstable because of the ratio of neutrons to protons. In order to become more stable, one of the neutrons (we are going to say a neutron is composed of a proton+electron....this is what our simplistic neutron is) is going to break apart. The electron that is bound in the neutron comes 'flying out', and this is called a beta particle.

    The C-14 atom has just undergone radioactive decay. Now the atom has only 7 neutrons...and 1 extra proton (that it got from the breaking apart neutron). So now the Carbon atom is actually a Nitrogen atom with 7 protons and 7 neutrons.....and it is stable. This is radioactive decay.

    It is very hard to determine when 5 atoms change, 10 atoms change...but it is easier to see when half the population changes.
    If you have a sample of Carbon-14, it will undergo this radioactive decay, and the time it takes for HALF of the Carbon-14 atoms to change into Nitrogen is called the Half Life of Carbon14.

    So, radioactive elements have half lives. It is the time it takes for half of them to decay into a new element.

    As an aside, half-life is also sometimes used when talking about things like drugs. If you take Drug X, it will be in your blood, for example..and will slowly (or quickly) start to fall apart in the body. THe time it takes for half the drug to change into something else (usually modified by the body or labile bonds get hydrolized), is the drugs half-life. In this case, it is not radioactive, but rather just falling apart inside the body.

  • 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)


    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!


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