Question

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

• A. $0.788$ $\mu$ $Ci$
• B. $0.431$ $\mu$ $Ci$
• C. $0.39$ $\mu$ $Ci$
• D. none of these

Answer 1

Answer: (A)

$0.788$ $\mu$ $Ci$

$\lambda$ = $\frac{0.693}{t_{1/2}}$ = $2.95\times {10}^{-8}{s}^{-1}$
$N_0$ = $\frac{-dN/dt}{\lambda }$ = $\frac{7.4\times {10}^4}{2.95\times {10}^{-8}}$ = $2.51\times {10}^{12}$ 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 $=\lambda N\left(t\right)$ = $.394(2.5\times {10}^{12})\times 2.95\times {10}^{-8}$
$=0.788$ $\mu$ $Ci$

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