A) \[1.82\times {{10}^{18}}\]
B) \[0.67\times {{10}^{18}}\]
C) \[1.34\times {{10}^{18}}\]
D) \[2.01\times {{10}^{11}}\]
Correct Answer: D
Solution :
\[Number\text{ }of\text{ }half-life=\frac{time\text{ }taken}{half-life}\] \[Number\text{ }of\text{ }half-life=\frac{3240}{1620}=2\] From Rutherford-Soddy law, fraction of original nuclei left is \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[\Rightarrow \] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}\] Hence, number of nuclei that disintegrate \[=\frac{3}{4}\times 2.68\times {{10}^{18}}=2.01\times {{10}^{18}}\]
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