Answer to Question #106751 in Atomic and Nuclear Physics for Ambika

Question

The half-life of cobalt-60 is 5.274 years. Calculate the activity of a sample of
cobalt-60 weighing 1 g in units of Curie (Ci).

Answer 1

$N=N_oe^{-\lambda t}$

activity is $\frac{dN}{dt}$=A

$A_t=A_oe^{-\lambda t}$

$\lambda=\frac{\ln2}{t_{1/2}}=\frac{0.69}{5.274\times365\times24\times60\times60}=\frac{.69}{0.166\times10^9}=4.16\times10^{-9}s^{-1}$

$A_o=\lambda N_o=4.16\times10^{-9}\times\frac{1}{60}\times6.023\times10^{23}$ disintegration/sec

$A_o=4.17\times10^{14}$ disintegration/sec

1 Ci=3.7x10¹⁰ disintegration/sec

converting A_o into curie

A_o = 1.13x10⁴ Ci