A) \[2\times {{10}^{10}}\]
B) \[1.5\times {{10}^{10}}\]
C) Zero
D) Infinity
Correct Answer: B
Solution :
| [b]\[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{\frac{t}{{{T}_{1l2}}}}}\] |
| 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, \[{{N}_{2}}=8\times {{10}^{10}}{{\left( \frac{1}{2} \right)}^{\frac{4}{1}}}=\frac{1}{2}\times {{10}^{10}}\] |
| \[\therefore \] 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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