${}^{57}\text{Co}$ decays by electron capture. Its half life is $272$ days. Find the activity left after a year if present activity is $2\mu$ $Ci$

$\lambda$ = $\frac{0.693}{t_{1/2}}$ = $2.95\times 10^{-8}{s}^{-1}$

$N_0$ = $\frac{-dN/dt}{\lambda }$ = $\frac{7.4\times 10^4}{2.95\times 10^{-8}}$ = $2.51\times 10^{12}$ nuclei

$N_t$ $=N_0{e}^{-\lambda t}$ $=2.51\times 10^{12}{e}^{-2.95\times 10^{-8}\times 3.156\times 10^7}$
                      $=0.394(2.52\times 10^{12})$

Activity $=\lambda N(t)$ = $.394(2.5\times 10^{12})\times 2.95\times 10^{-8}$
             $=0.788$ $\mu$ $Ci$

Alternative method:
$\frac{dN(t)}{dt}$ = $\frac{dN(0)}{dt}{e}^{\lambda t}$$=(2\mu Ci)$ $(e^{-2.95\times 10-8\times 3.156\times 10^{-7}})$$=2(.394)$ = $0.788$ $\mu$ $Ci$