A) 2
B) 4
C) between 4 and 6
D) 6
Correct Answer: C
Solution :
Let No be the number of atoms of\[x\]at time\[t=0\]. Then at\[t=4h\], ie, (two half-lives) \[{{N}_{x}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{2}}\] \[{{N}_{x}}=\frac{{{N}_{0}}}{4}\] \[\therefore \] \[{{N}_{y}}={{N}_{0}}-\frac{{{N}_{0}}}{4}=\frac{3{{N}_{0}}}{4}\] \[\therefore \] \[\frac{{{N}_{x}}}{{{N}_{y}}}=\frac{1}{3}\] Now, at\[t=6h\], ie, (three half-lives) \[{{N}_{x}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{3}}=\frac{{{N}_{0}}}{8}\] and \[{{N}_{y}}={{N}_{0}}-{{N}_{x}}\] \[={{N}_{0}}-\frac{{{N}_{0}}}{8}\] \[=\frac{7{{N}_{0}}}{8}\] or \[\frac{{{N}_{x}}}{{{N}_{y}}}=\frac{1}{7}\] The given ratio lies between\[\frac{1}{3}\]and\[\frac{1}{7}\] Therefore, \[t\]lies between\[4h\] and\[6h\].
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