The question and mark scheme I will write in bold and my own thoughts in normal sized text.

I'm told that:

When a $\bf{_{92}^{235}U} $ nucleus is exposed to free neutrons it can absorb a neutron. The resulting nucleus decays, first to $\bf{_{93}^{239}Np}$ and then to $\bf{_{94}^{239}Pu}$.

I'm then asked a few questions this, one of which I don't understand the answer to. They say that:

The number of $\bf{_{93}^{239}Np}$ nuclei present eventually becomes constant, calculate this constant number of $\bf{_{93}^{239}Np}$ nuclei, given that the half life of $\bf{_{93}^{239}Np} = 2.04 \times 10^{5} s$ and that the number of $\bf{_{93}^{239}Np}$ nuclei produced is at a constant rate of $ \bf{1.80 \times 10^{7} s^{-1}} $

The mark scheme uses the equation:

$ \bf{A = \lambda N}$

and rearranges this to say that:

$\bf{N=\dfrac{A}{\lambda}}$ and we also know that $\bf{\lambda = \dfrac{0.693}{t_\frac{1}{2}}}$

$\bf{\therefore \lambda \approx 3.397 \times 10^{-6}}$

I understand all of this and I understand the maths of the next step, I just don't understand why it gives you they value of the number of nuclei of $_{93}^{239}Np$ when the rate of decay of $_{93}^{239}Np$ equals the rate of formation of $_{93}^{239}Np$.

Substituting this value of lambda into our equation for $\bf{N}$ gives us:

$\bf{N = \dfrac{1.8 \times 10^{7}}{3.397 \times 10^{-6}} \approx 5.3 \times 10^{12}}$

$\bf{{\therefore}}$ the number of $\bf{_{93}^{239}Np}$ nuclei when this number becomes constant is $5.3 \times 10^{12}$


1 Answer 1


The rate of decay of Np is $$-\frac{d}{dt}\left[N e^{-\text{Log}(2)t/\tau}\right]_{t=0}=\frac{n \log (2)}{\tau }$$ where $\tau$ is the Np half-life and $A$ is the number of Np atoms present. Equating this with the Np influx rate $A$ and solving for $N$ yields $$N=\frac{A \tau }{\log (2)}=5.3\times10^{12}.$$ Are you confused why you equate the two? If the influx rate was larger than the decay rate, the population would be increasing, not constant, and vice versa.

  • $\begingroup$ Thank you, is influx rate the same as activity? $\endgroup$
    – Elise
    Commented Feb 4, 2014 at 18:05
  • $\begingroup$ The influx rate is the same as the number of U atoms that decay per second, since each decay produces a Np atom. $\endgroup$ Commented Feb 4, 2014 at 18:39
  • $\begingroup$ Sorry for the late reply, and thank you, I get it now :) $\endgroup$
    – Elise
    Commented Feb 5, 2014 at 15:54

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