A) \[2\times {{10}^{10}}\]
B) \[1.5\times {{10}^{10}}\]
C) Zero
D) Infinity
Correct Answer: B
Solution :
\[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{\frac{t}{{{T}_{1l2}}}}}\] No of atoms at t = 2hr, \[{{N}_{1}}=8\times {{10}^{10}}{{\left( \frac{1}{2} \right)}^{\frac{2}{1}}}=2\times {{10}^{10}}\] No. of atoms at t = 4hr, \[{{N}_{2}}=8\times {{10}^{10}}{{\left( \frac{1}{2} \right)}^{\frac{4}{1}}}=\frac{1}{2}\times {{10}^{10}}\] \ No. of atoms decayed in given duration \[=\left( 2-\frac{1}{2} \right)\times {{10}^{10}}=1.5\times {{10}^{10}}\]
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