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, \[n=\frac{\text{time(t)}}{\text{half}-\text{life(}{{\text{T}}_{\text{/2}}}\text{)}}\] 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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