A) \[\frac{1}{2}g\]
B) \[\frac{1}{4}g\]
C) \[\frac{1}{8}g\]
D) \[\frac{1}{16}g\]
Correct Answer: D
Solution :
\[T=n\times {{t}_{1/2}}\] Where, \[{{t}_{1/2}}=\]half-life period, \[n=\frac{T}{{{t}_{1/2}}}=\frac{560}{140}=4\] Now, \[{{N}_{t}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[=1\times {{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{16}g\]
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