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Physics Radioactivity question need help

Been trying to do this question and can't do it, only worth two marks so should be pretty simple if anyone could help that would be great thanks

The daughter nucleus formed in radioactive decay is radioactive . if NP (t) is the number of parent nuclei at time t and ND (t) is the number of daugher nuclei at time t and λP and λ D are the decay constants of them both
show that the number of daughter nuclei at time t is:
dND (t)/dt = λPNP (t) − λD ND (t)

Reply 1

SnoochToTheBooch

Original post by underlyinggirl

For typing out subscripts, use an underscore _, makes it easier to read. Anyway, here's what I reckon:

N_P(t) = number of parent nuclei @ time t
N_D(t) = number of daughter nuclei @ time t
λ_P = decay constant for parents
λ_D = decay constant for daughters

So treating them individually, if you've got an initial number of parent nuclei N_P(t=0), then:

N_P(t) = N_P(0).e^{-λ_P.t}

--> d/dt(N_P(t)) = d/dt(N_P(0).e^{-λ_P.t})
= N_P(0).d/dt(e^{-λ_P.t})
= N_P(0).(-λ_P..e^{-λ_P.t})
= -λ_P.N_P(0)e^{-λ_P.t}
= -λ_P.N_P(t)

So, ignoring for now that any daughter nuclei will also decay, for every parent nucleus you lose, you've gained a daughter nucleus:

N_D(t) = N_D(0) - (N_P(t) - N_P(0))

Therefore d/dt(N_D(t)) = d/dt(N_D(0)) - d/dt(N_P(t)) + d/dt(N_P(0))
= 0 - -λ_P.N_P(t) - 0
= λ_P.N_P(t)

Then you've got to consider the initial number of daughter nuclei you started with that have decayed in time t, so just like the parent nuclei equation N_P(t) = N_P(0).e^{-λ_P.t}, you have:

N_D(t) = N_D(0).e^{-λ_D.t}

meaning that d/dt(N_D(t)) = d/dt(N_D(0).e^{-λ_D.t}) = -λ_D.N_D(t)

so all in all, d/dt(N_D(t)) = λ_P.N_P(t) - λ_D.N_D(t)