if none were added
dN/dt = k N
number produced per second = 10
number lost per second = k n
so
dN/dt = 10 - k N
N is maximum when dN/dt = 0
0 = 10 - k (1000)
k = .01
so we know that for this fission reaction (without the inflow)
dN/dt = -.01 N
dN/N = -.01 dt
ln N = -.01 t + C
N = Ni e^-(.01 t)
when is N/Ni = .5?
ln .5 = -.01 t
t = 69.3 seconds half life
mean life = 1.443 * half life (see http://www.britannica.com/EBchecked/topic/371549/mean-life )
Radioactive nuclei are produced in an irradiated sample at the rate of 10 s-1. If the number in the sample builds up to a maximum of 1000, calculate the mean life and the half-life of the radioactive nuclei
Using dN / dt = P - λ N
dN / dt = 1000n /10/ 1 = 100
P - λ N = 10 – λ 1000N
Half life is -
(In2) / λ = (In2) / 1000) = 6.93x10^-4 seconds
Not sure of mean life any advice appreciated ...
3 answers
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