A) \[\frac{1}{4}\]
B) \[\frac{1}{8}\]
C) \[\frac{1}{16}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
From Rutherford Soddy law, the number of atoms left after \[n\] half-lives is \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] where,\[n=\frac{time(t)}{half-time({{T}_{1/2}})}\] Given, \[t=6400\,\,yr,\,\,T=1600\,\,yr\] \[\therefore \] \[n=\frac{6400}{1600}=4\] So, \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{16}\] Hence, \[\frac{1}{16}\] part of element will remain.
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