Question #106751
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).
1
Expert's answer
2020-03-27T10:51:32-0400

N=NoeλtN=N_oe^{-\lambda t}

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


At=AoeλtA_t=A_oe^{-\lambda t}

λ=ln2t1/2=0.695.274×365×24×60×60=.690.166×109=4.16×109s1\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}

Ao=λNo=4.16×109×160×6.023×1023A_o=\lambda N_o=4.16\times10^{-9}\times\frac{1}{60}\times6.023\times10^{23} disintegration/sec

Ao=4.17×1014A_o=4.17\times10^{14} disintegration/sec

1 Ci=3.7x1010 disintegration/sec

converting Ao into curie

Ao = 1.13x104 Ci



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