# Using $A = {\lambda}N$ to find when a the amount of a radioactive source becomes constant

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 $5.3 \times 10^{12}$

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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.