Question

$1gram$ of $A{u}_{79}^{198}(t_{1/2}=65hrs)$ decays by beta emission and produces stable mercury. How much mercury will be present after $260$ hours?

Answer 1

Since $\frac{260}{65}=4$, 260 hours corresponds to 4 half lives

Using the relation $N=N_o{e}^{-kt}$ where $k=\frac{ln\left(2\right)}{t_{1/2}}$and $t=4t_{1/2}$,

$N=N_o{e}^{-4ln\left(2\right)}=\frac{1}{16}{N}_o$

$\frac{N}{N_o}=\frac{1}{16}$