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difference in number of nuclei over difference in time ...

p = 10, n = 1000 nuclei

dN / dt = 10 /1000 = 0.01t - the rate at which nuclei are produced minus the rate at which they decay P - is the production rate = 10 s-1 N - number of nuclei = 1000 So, N(t) = N0 exp(-λt), Where N0 = number of radioactive nuclei at t = 0. N(t) = N0 exp (.01t) N0 = N0 /2...

In the capture reaction 115In + n 116In, 116In nuclei are produced at the rate of 10 s-1. Calculate the maximum number of 116In nuclei (number in equilibrium) in the sample if the half-life of 116In is 54 m. The half life is 54 minutes, converted to seconds = 54m x 60 s-1 = 3240 s-1 (In2)...

Homework Statement Can someone check this please ? Calculate the maximum number of 116In nuclei (number in equilibrium) in the sample if the half-life of 116In is 54 m. Homework Equations The Attempt at a Solution The half life is 54 minutes, converted to seconds = 54m...

Please don't be sorry - I have been struggling with this for a while. I am going to sleep on this, I am sure I am not too far off the mark. I am so grateful to you for steering me along .. thank you very much. I must work through these issues and find my own way and solutions. Thank you once...

dN / dt = 1000n /10/ 1 = 100 P - λ N = 10 – λ 1000N (In2) / λ = (In2) / 1000) = 6.93x10^-4 seconds

dn/dt = 1000 - 10 = 990 p = 10 n = 1000 mind block is here ...

dn / dt is the total number divided by time ... P is the production rate N is the total of dn/dt - production rate

dN is the total number of nuclei 1000 dt is the time the rate of production = 10s-1 Production rate 10 s-1 N is the total number of nuclei after I have deducted 10 nuclei from the total amount 990 nuclei

There is a total of 1000 nuclei in the solution, the rate of increase (decrease) is 10s-1 If there are 1000 nuclei and the rate of decay is 10 per second, then the difference is 990 nuclei Divided by by the decay constant 0.693 ..

1000 nuclei - 10s-1 = 990 nuclei s-1 (In2) / λ = (In2) / (990s-1) = 7.00 x 10^-4 s-1

So the steady state is the 1000 nuclei So I would just need to use that amount and apply the decay constant equation ?? I am confused and going round in a circle ...

Is the steady state is the difference between the rate of production 10s-1 and the rate of decay ?

Homework Statement 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 Can anyone advise on this one not sure how mean life is found...