Nuclear Half-Life Calculations
Key Questions
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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), whereA(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 firstt_("1/2") ,50.0/2 = 25.0 "g" after anothert_("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 amountOne 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= 0.35Then 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!