A) 2
B) 4
C) 8
D) 1
Correct Answer: C
Solution :
[C]For first half sample decay with process A of half lives \[\frac{1}{2}hr.\] |
After first half hrs \[N={{N}_{0}}\frac{1}{2}\] |
For next one hour decay process B of half lives \[\frac{1}{4}\]occur |
So for \[t=\frac{1}{2}\] to \[t=1\frac{1}{2}\] |
Total 4 half lives occur \[N={{N}_{0}}\frac{1}{2}{{\left[ \frac{1}{2} \right]}^{4}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{5}}\] |
For last \[\frac{1}{2}hr\] both A and B process occur simultaneously |
For \[t=1\frac{1}{2}\] to t = 2hr [For both A & B \[\frac{1}{{{t}_{1/2}}}=\frac{1}{1/2}+\frac{1}{1/4}=6]\] |
\[{{t}_{1/2}}=\frac{1}{6}hr\] |
Number of half-life in\[\frac{1}{2}hr\] is 3 |
\[\therefore \,\,\,N={{N}_{0}}{{\left[ \frac{1}{2} \right]}^{5}}\,\,{{\left[ \frac{1}{2} \right]}^{3}}\]\[={{N}_{0}}{{\left[ \frac{1}{2} \right]}^{8}}\] |
a = 8 |
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