Radioactive decay question.....

Question

77)

One of the nuclides in spent nuclear fuel is U-234, an alpha emitter with a half-life of 2.44x10^5 years. If a spent fuel assembly contains 2.80 kg of U-234, how long will it take for the amount U-234 to decay to less than .10 kg?

WHOAA... Can someone please explain so that i can understand in the future? thanks...

Answer 1

will you marry me ?    

Answer 2

0.10kg is ((1/2)^4.8 )times2.8 kg.  I got this result by dividing log 28 by log 2 to get 4.8.  This means that if you halve 2.8 kg 4.8 times, you get 0.1 kg.  So it will take 4.8 half lives for your 2.8 kg to decay to 0.1 kg.  This is 4.8 times 2.44 x 10^5 = 1.1712 x 10^6 years.  

Answer 3

The basic relationship is A = Ao e^-kt.

where A is the activity at time t

Ao is the original acitivey at t=0

k is the decay constant which can be determined from the half-life

e is the base of the natural logarithms

This can be rewritten as ln (A/Ao) = -kt

k = 0.693 / t(1/2) = 0.693 / (2.44x10^5 yrs) = 2.84 x 10^-6 1/yr

We can substitute the mass for the activity, since they are proportional.

Solve for t

t = ln(A/Ao) / -k

t = ln(0.10 kg / 2.80 kg) / -2.84 x 10^-6 1/yr

t = 1,170,000 yr

Answer 4

infinity, effectively .... a long long time.  (I despise these so called scientific calculations that have to be a joke - just to see if inteligent being will barf back up what is forced fed to them in a classroom.... )  Does it really matter if it is 20000 thousand years or 150 thousand?  I don't think so.