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Question

Physics Question on deccay rate

Two radioactive nuclei PP and QQ, in a given sample decay into a stable nucleus 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 11 minute where as that of QQ is 22 minutes. 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

A

2N02\,N_0

B

3N03\,N_0

C

9N02\frac{9\,N_0}{2}

D

5N02\frac{5\,N_0}{2}

Answer

9N02\frac{9\,N_0}{2}

Explanation

Solution

T1/2T _{1 / 2}
NP=NQN _{ P }= N _{ Q }
4N02t/1=N02t/2\frac{4 N _{0}}{2^{ t / 1}}=\frac{ N _{0}}{2^{ t / 2}}
4=2t/24=2^{ t / 2}
22=2t/22^{2}=2^{ t / 2}
t2=2\frac{ t }{2}=2
t=4min\Rightarrow t =4 \,min
Disactive nucleus or Nuclei of RR
=(4N04N024)+(N0N022)=\left(4 N _{0}-\frac{4 N _{0}}{2^{4}}\right)+\left( N _{0}-\frac{ N _{0}}{2^{2}}\right)
=4N0N04+N0N04=5N0N02=4 N _{0}-\frac{ N _{0}}{4}+ N _{0}-\frac{ N _{0}}{4}=5 N _{0}-\frac{ N _{0}}{2}
=92N0=\frac{9}{2} N _{0}