Answer:The correct answer is option C.
Explanation:
Initial mass of the sample = [tex]N_o[/tex]
Sample left after t years ,=[tex]N=\frac{N_o}{8}[/tex]
Half-life of sample of cobalt-60 [tex]t_{\frac{1}{2}}[/tex] = 5 years
[tex]\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5 years}=0.1386 year^{-1}[/tex]
[tex]N=N_o\times e^{-\lambda t}[/tex]
[tex]ln[N]=ln[N_o]-\lambda t[/tex]
[tex]\log[\frac{[N_o]}{8}]=\log[N_o]-\frac{\lambda t}{2.303}[/tex]
[tex]\log[\frac{1}{8}]=-\frac{0.1386 year^{-1}\times t}{2.303}[/tex]
[tex]t=15.33 years\approx 15 years[/tex]
Hence ,the correct answer is option C.
