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 ...
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 )
THANK YOU .....
You are welcome :)
To calculate the mean life of the radioactive nuclei, we can use the relationship between the mean life and the decay constant (λ).
The mean life (τ) is the average time it takes for a radioactive nucleus to decay. It can be calculated using the formula:
τ = 1 / λ
In your case, the decay constant (λ) is not given directly. However, we can calculate it using the rate of decay (P) and the current number of radioactive nuclei in the sample (N).
Given that the rate of decay (P) is 10 s^(-1) and the maximum number of nuclei in the sample (N) is 1000, we can rewrite the equation:
P - λN = 10 - 1000λ = 0
Solving for λ, we find:
λ = 10 / 1000 = 0.01 s^(-1)
Now, we can calculate the mean life using the formula mentioned earlier:
τ = 1 / λ = 1 / 0.01 = 100 seconds
Hence, the mean life of the radioactive nuclei is 100 seconds.
To summarize:
- Mean life (τ) = 100 seconds
- Half-life = 6.93x10^(-4) seconds