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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 4N0 and that of Q are N0. Half-life of P (for conversion

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 4N0 and that of Q are N0. Half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. 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:

Option: 1

\frac{5\text{N}_{0}}{2}


Option: 2

2N0


Option: 3

3N0


Option: 4

\frac{9\text{N}_{0}}{2}


Answers (1)

\\ \text{Initially, number of nuclei} \ P=4 N_{0} \\ \text{number of nuclei} \ Q =N_{0} \\ \text{Half life of P} , T_{P}=1 min \\ \text{Half life of Q} , T_{Q}=2 min \\ \text{No. of nuclei of P after time t} \ = n_{P}=4 N_{0}\left(\frac{1}{2}\right)^{t / T_{p}}=4 N_{0}\left(\frac{1}{2}\right)^{t / 1} \\ \text{No. of nuclei of Q after time t} \ n_{Q}=N_{0}\left(\frac{1}{2}\right)^{t / T_{Q}}=N_{0}\left(\frac{1}{2}\right)^{t / 2} \\

\text{Suppose number of nuclei P and Q in given sample are equal after time t, then} \\ n_p=n_{Q} \\4N_{0}\left(\frac{1}{2}\right)^{t / 1}=N_{0}\left(\frac{1}{2}\right)^{t / 2} \\ \left(\frac{1}{2}\right)^{t / 2}=\frac{1}{4}=\left(\frac{1}{2}\right)^{2} \\ \ \ or \ \ \frac{t}{2}=2 \\ \\ or \ t=4 m i n \\

\therefore n_p=4 N_{0}\left(\frac{1}{2}\right)^{4 / 1}=\frac{N_{0}}{4} \\ and \ \ n_{Q}=N_{0}\left(\frac{1}{2}\right)^{4 / 2}=\frac{N_{0}}{4} \\ \text{Population of R in the sample is} =\left(4 N_{0}-\frac{N_{0}}{4}\right)+\left(N_{0}-\frac{N_{0}}{4}\right)=\frac{9 N_{0}}{2}

Posted by

Ramraj Saini

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