A) \[2{{N}_{0}}\]
B) \[3{{N}_{0}}\]
C) \[\frac{9{{N}_{0}}}{2}\]
D) \[\frac{5{{N}_{0}}}{2}\]
Correct Answer: C
Solution :
[c] : P Q No. of nuclei, at t = 0 \[4{{N}_{0}}\] \[{{N}_{0}}\] Half- life 1 min 2 min No. of nuclei after time t \[{{N}_{p}}\] \[{{N}_{Q}}\] Let after t min the number of nuclei of P and Q are equal. \[\therefore \]\[{{N}_{P}}=4{{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/1}}\]and\[{{N}_{Q}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/2}}\] As\[{{N}_{P}}={{N}_{Q}}\] \[\therefore \]\[4{{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/1}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/2}},\frac{4}{{{2}^{t/1}}}=\frac{1}{{{2}^{t/2}}},4=\frac{{{2}^{t}}}{{{2}^{t/2}}}\] \[4={{2}^{t/2}},{{2}^{2}}={{2}^{t/2}}\]\[\frac{t}{2}=2\]or\[t=4\]min After 4 minutes, both P and Q have equal number of nuclei. \[\therefore \]Number of nuclei of R \[=\left( 4{{N}_{0}}-\frac{{{N}_{4}}}{4} \right)+\left( {{N}_{0}}-\frac{{{N}_{0}}}{4} \right)\] \[=\frac{15{{N}_{0}}}{4}+\frac{3{{N}_{0}}}{4}=\frac{9{{N}_{0}}}{2}\]
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