A) 250
B) 400
C) 300
D) 200
Correct Answer: D
Solution :
As we know,\[\left[ \frac{N}{{{N}_{0}}} \right]={{\left[ \frac{1}{2} \right]}^{n}}\] n = no. of half life N - no. of atoms left \[{{N}_{0}}-\]initial no.. of atoms By radioactive decay law,\[\frac{dN}{dt}=kN\] k - disintegration constant \[\therefore \]\[\frac{\frac{dN}{dt}}{\frac{d{{N}_{0}}}{dt}}=\frac{N}{{{N}_{0}}}\] ?(ii) From (i) and (ii) we get\[\frac{\frac{dN}{dt}}{\frac{d{{N}_{0}}}{dt}}={{\left[ \frac{1}{2} \right]}^{n}}\]or,\[\left[ \frac{100}{1600} \right]={{\left[ \frac{1}{2} \right]}^{n}}\Rightarrow {{\left[ \frac{1}{2} \right]}^{4}}={{\left[ \frac{1}{2} \right]}^{n}}\] \[\therefore \]w = 4, Therefore, in 8 seconds 4 half life had occurred in which counting rate reduces to 100 counts \[{{\text{s}}^{-1}}\]. \[\therefore \]Half life, \[\frac{{{T}_{1}}}{2}=2\sec \] In 6 sec, 3 half life will occur \[\therefore \]\[\left[ \frac{\frac{dN}{dt}}{1600} \right]={{\left[ \frac{1}{2} \right]}^{3}}\]\[\Rightarrow \]\[\frac{dN}{dt}=200\]counts\[{{\text{s}}^{-1}}\]
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