# beta decay of a growing sample

This is an exercise about neutron capture. I want the remaining $$^{198}$$Au after a time t and I know $$\lambda$$.

$$^{197}$$Au absorps a neutron and transforms to $$^{198}$$Au which then decays. The sample absorps 10$$^{20}$$ neutrons every second, but how do I include the decay for the atoms the sample gains every second?

Because for N(t) = $$C*e^{-t\lambda}+10^{20}*t$$ wouldn't C = 0 because of N(0) = 0?

Edit: So I guess I have to solve the differential $$N'(t) + \lambda N(t) = 10^{20}$$

• Search term: "secular equilibrium." Also, $10^{20}\rm\,s^{-1}$ is a lot of neutrons. – rob May 28 '19 at 0:49
• For the differential eq. I found $N(t) = N(0)e^{-\lambda t} + 10^{20}/\lambda$ Thank you for the tip, I keep thinking about it. – quielywhis May 28 '19 at 1:05
• @rob that isn’t just a lot of neutrons. That’s 10x or more of the neutron dose for structural materials across the lifetime of a modern nuclear reactor... – Jon Custer May 28 '19 at 2:15
• The exercise was posed in class, maybe I didn't write it down correctly.:) – quielywhis May 28 '19 at 2:17
• Speaking of lots of neutrons, check out physics.stackexchange.com/questions/437245/… – PM 2Ring May 28 '19 at 5:01