A) 2 : 1
B) 1 : 2
C) 1 : 4
D) 1 : 16
E) 4 : 1
Correct Answer: A
Solution :
For X radioactive substances \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{x}}}}\] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{x}}}}\] \[\frac{1}{16}={{\left( \frac{1}{2} \right)}^{8/{{T}_{x}}}}\] \[\Rightarrow \] \[{{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{x}}}}\] \[{{T}_{X}}=2\] For Y radioactive substances \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{Y}}}}\] \[\frac{1}{256}={{\left( \frac{1}{2} \right)}^{8/{{T}_{Y}}}}\] \[{{\left( \frac{1}{2} \right)}^{8}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{Y}}}}\] Or \[{{T}_{Y}}=1\] \[\therefore \] \[\frac{{{T}_{X}}}{{{T}_{Y}}}=\frac{2}{1}\]You need to login to perform this action.
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