Q.The isotope 57Co decays by electron capture to 57Fe with a half-life of 272 d. The 57Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.

[!TLDR] $\tau=T_{1/2}/0.693=392.5$ d $\approx3.39\times10^7$ s, $\lambda=1/\tau\approx2.95\times10^{-8}$ s$^{-1}$; N from 2.0 $\mu$Ci activity $\approx2.51\times10^{12}$; after 1 year, $A\approx0.789\,\mu$Ci. [!ANSWER] $\tau\approx3.39\times10^7$ s, $\lam …

1. Mean lifetime $\tau=T_{1/2}/0.693=272\text{ d}/0.693\approx392.5$ days, which in seconds is $392.5\times86400\approx3.392\times10^7$ s. The decay constant is $\lambda=1/\tau\approx2.948\times10^{-8}$ s$^{-1}$.
2. Converting the given activity to becquerel, $2.0\,\mu\text{Ci}=2.0\times10^{-6}\times3.7\times10^{10}=7.4\times10^4$ decays/s; the number of nuclei present is $N=A/\lambda=\frac{7.4\times10^4}{2.948\times10^{-8}}\approx2.51\times10^{12}$. …

Find $\tau$ and $\lambda$ from the half-life, use $N=A/\lambda$ (after converting the given activity from $\mu$Ci to Bq), and apply $A …

Using the half-life directly as if it were the decay constant, or forgetting to convert both the half-life and the elapsed time consistently i …