A) \[3.2\times {{10}^{9}}years\]
B) \[5\times {{10}^{9}}years\]
C) \[1.6\times {{10}^{10}}years\]
D) 5 years
Correct Answer: D
Solution :
[d] \[X\xrightarrow{{{\lambda }_{1}}}Y\xrightarrow{{{\lambda }_{2}}}Z\] At radioactive equilibrium, \[({{\lambda }_{X}})\times ({{N}_{X}})=({{\lambda }_{Y}})\times ({{N}_{Y}})\] \[\frac{{{\lambda }_{X}}}{{{\lambda }_{Y}}}=\frac{{{N}_{Y}}}{{{N}_{X}}}\,\,\frac{{{({{t}_{1/2}})}_{Y}}}{{{({{t}_{1/2}})}_{X}}}=\frac{{{N}_{Y}}}{{{N}_{X}}}\] or \[\frac{{{({{t}_{1/2}})}_{Y}}}{1.6\times {{10}^{10}}}\] \[=\frac{1}{3.2\times {{10}^{9}}}{{({{t}_{1/2}})}_{Y}}=5\,\,years\]
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