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NEET Sample Paper NEET Sample Test Paper-81

  • question_answer
    If a mixture contains two radioactive material \[{{\operatorname{X}}_{1}}\,\,and\,\,{{X}_{2}}\] with half-lives of 20 s and 10 s respectively. If at \[\operatorname{t}=0\] the mixture has 40 g of \[{{\operatorname{X}}_{1}}\] and 160 g of\[{{X}_{2}}\]. The amount of the two in the mixture will become equal after t

    A)  80 s                            

    B)  40 s

    C)  60 s                            

    D)  20 s

    Correct Answer: B

    Solution :

    \[\because \,\,N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\,[n=number\,\,of\,\,half-lives]\] \[For\,\,{{X}_{1}},\,\,\,\,\,\,{{N}_{1}}={{({{N}_{0}})}_{1}}\,\,{{\left( \frac{1}{2} \right)}^{t/10}}\] \[For\,\,{{X}_{2}},\,\,\,\,\,\,{{N}_{2}}={{({{N}_{0}})}_{2}}\,\,{{\left( \frac{1}{2} \right)}^{t/10}}\] According to question, At \[\operatorname{t}= 0, {{\left( {{N}_{0}} \right)}_{1}} = 40 g, {{\left( {{N}_{0}} \right)}_{2}} =160 g\] And after t sec, \[{{N}_{1}}={{N}_{2}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{40}{{{2}^{t/20}}}=\frac{160}{{{2}^{t/10}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{2}^{t/10}}\,\,=\,\,4({{2}^{t/20}})\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{2}^{t/10}}={{2}^{(t/20\,+\,2)}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{t}{10}=\frac{t}{20}+2\,\,\,\Rightarrow \,\,t=40\,\,s\]


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