A) \[2\times {{10}^{2}}\] year
B) \[3\times {{10}^{3}}\] year
C) \[3.3\times {{10}^{4}}\] year
D) \[4\times {{10}^{3}}\] year
Correct Answer: B
Solution :
| [b] Radioactive decay is first order reaction. |
| So, \[\lambda =\frac{0.693}{{{T}_{1/2}}}\] |
| \[\therefore \] \[{{T}_{1/2}}=\frac{0.693}{2.31\times {{10}^{-4}}}\,year\] |
| \[=0.3\times {{10}^{4}}=3\times {{10}^{3}}\,year\] |
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