A) 1 curie
B) 120 micro curie
C) 60 micro curie
D) 8 mili curie
Correct Answer: C
Solution :
1 week » 7 days \[\underline{\tilde{\ }}\] \[7\times 24hrs\underline{\tilde{\ }}\,\ 14\]half lives Number of atoms left \[=\frac{No}{{{(2)}^{14}}}\], Activity =\[N\lambda \] \ Activity left is \[\frac{1}{{{(2)}^{14}}}\] times the initial Þ \[\frac{1}{{{(2)}^{14}}}\times 1\] curie \[=\frac{1}{16384}\times 1\] curie\[\cong \]\[61\times {{10}^{-6}}\] curie \[\approx 60\mu \,\]curie.
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