A) \[1.43047\times {{10}^{12}}{{s}^{-1}}\]
B) \[4.844\times {{10}^{-13}}{{s}^{-1}}\]
C) \[4.844\times {{10}^{+13}}s\]
D) \[1.43047\times {{10}^{-12}}{{s}^{-1}}\]
E) \[1.43047\times {{10}^{+12}}{{s}^{-1}}\]
Correct Answer: B
Solution :
We have \[\underset{Radium\,}{\mathop{{{N}_{1}}{{\lambda }_{1}}}}\,=\underset{Uranium}{\mathop{{{N}_{2}}{{\lambda }_{2}}}}\,\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{N}_{2}}}{{{N}_{1}}}=\frac{28}{1}\] \[\frac{{{\left( \frac{0.693}{{{t}_{50}}} \right)}_{1}}}{{{\left( \frac{0.693}{{{t}_{50}}} \right)}_{52}}}=\frac{{{({{t}_{50}})}_{2}}}{{{({{t}_{50}})}_{1}}}=\frac{28}{1}\] \[{{({{t}_{50}})}_{2}}=1620\times 28\,yr=1.43047\times {{10}^{12}}s\] \[{{\lambda }_{2}}=\frac{0.693}{{{t}_{50}}}=4.844\times {{10}^{-13}}{{s}^{-1}}\]
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