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NEET 2011
Exam
NEET
Year
2011
Subject
Physics
Chapter
Nuclear Physics

NEET 2011 · Physics · Nuclear Physics

Two radioactive nuclei P and Q , in a given sample decay into a...

Two radioactive nuclei PP and QQ, in a given sample decay into a stable nucleous RR. At time t=0t=0, number of PP species are 4N04 N_0 and that of QQ are N0N_0. Half-life of PP (for conversion to RR ) is 1 min where as that of QQ is 2 min2 \mathrm{~min}. Initially there are no nuclei of RR present in the sample. When number of nuclei of PP and QQ are equal, the number of nuclei of RR present in the sample would be

Options

  1. A3N03 N_0
  2. B9N02\frac{9 N_0}{2}
  3. C5N02\frac{5 N_0}{2}
  4. D2N02 N_0

Correct answer

B. 9N02\frac{9 N_0}{2}

Step-by-step solution

Initially P→4N0P \rightarrow 4 N_0 Q → N_0 Half life Tp→1 minT_p \rightarrow 1 \mathrm{~min} T_Q → 2 \min Let after time tt number of nuclei of PP and QQ are equalie, 4N02t/1=N02t/2\quad \frac{4 N_0}{2^{t / 1}}=\frac{N_0}{2^{t / 2}} 4=2t/222=2t/2t2=2t=4 min\begin{aligned}4 & =2^{t / 2} \\ 2^2 & =2^{t / 2} \\ \frac{t}{2}& =2 \\ t & =4 \mathrm{~min} \end{aligned}Disactive nucleus or Nuclei of RR=(4N0−4N024)+(N0−N022)=4N0−N04+N0−N04=5N0−N02=92N0\begin{aligned}& =\left(4 N_0-\frac{4 N_0}{2^4}\right)+\left(N_0-\frac{N_0}{2^2}\right)\\ & =4 N_0-\frac{N_0}{4}+N_0-\frac{N_0}{4}\\ & =5 N_0-\frac{N_0}{2}\\ & =\frac{9}{2}N_0 \end{aligned}$
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