Nuclear HalfLife Calculations
Key Questions

Nuclear halflife 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 halflife of the decaying quantity.So, if a problem asks you to calculate an element's halflife, it must provide information about the initial mass, the quantity left after radioactive decay, and the time it took that sample to reach its postdecay 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 halflife:
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 halflife 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 halflife. 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 hallife 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 = halflife.
M. = the beginning amount
M = the ending amountOne example of how to use the equation:
One of the Nuclides in spent nuclear fuel is U234, an alpha emitter with a halflife of 2.44 x10^5 years. If a spent fuel assembly contains 5.60 kg of U234, how long would it take for the amount of U234 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!