Monday

December 22, 2014

December 22, 2014

Posted by **Ali** on Monday, August 18, 2014 at 3:23pm.

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 ...

- physics -
**Damon**, Monday, August 18, 2014 at 4:25pmif 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 )

- physics -
**Ali**, Monday, August 18, 2014 at 4:39pmTHANK YOU .....

- physics -
**Damon**, Monday, August 18, 2014 at 4:49pmYou are welcome :)

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