InterviewSolution
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InterviewSolution
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Two radioactive nuclei `P` and `Q`, in a given sample decay into a stable nucleus `R`. At time `t = 0`, number of `P` species are `4 N_0` and that of `Q` are `N_0`. Half-life of `P` (for conversation to `R`) is `1mm` whereas that of `Q` is `2 min`. Initially there are no nuclei of `R` present in the sample. When number of nuclei of `P` and `Q` are equal, the number of nuclei of `R` present in the sample would be :A. `(5N_(0))/(2)`B. `2N_(0)`C. `3N_(0)`D. `(9N_(0))/(2)` |
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Answer» Correct Answer - D Initeally `Pto4N_(0),QtoN_(0)` Half-life, `T_(P)=1min,T_(Q)=2min` Let after time t, number of nuclei of P and Q equal `i.e.,(4N_(0))/(2^(t//1))=(N_(0))/(2^(t//2))" "(becauseN=(N_(0))/(2^(t//T_(1)//2)))` `or(4)/(2^(t//2))=1or t=4min` So, at t= 4 min `N_(P)=((4N_(0)))/(2^(4//1))=(N_(0))/(4)` and at `t=4 min,N_(Q)=(N_(0))/(2^(4//2))=(N_(0))/(4)` So, number of nuclei of R in teh sample `=(4N_(0)-(N_(0))/(4))+(N_(0)-(N_(0))/(4))=(9N_(0))/(2)` |
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