A) \[\frac{1}{4\,\lambda }\]
B) \[4\,\lambda \]
C) \[2\,\lambda \]
D) \[\frac{1}{2\,\lambda }\]
Correct Answer: D
Solution :
Number of nuclei remained after time can be written as \[N={{N}_{0}}{{e}^{-\lambda t}}\] (i) where \[{{N}_{0}}\] is initial number of nuclei of both the substances. \[{{N}_{1}}={{N}_{0}}{{e}^{-5\lambda t}}\] (i) and \[{{N}_{2}}={{N}_{0}}{{e}^{-\lambda t}}\] (ii) Dividing Eq. (i) by Eq. (ii), we obtain \[\frac{{{N}_{1}}}{{{N}_{2}}}={{e}^{(-5\lambda +\lambda )t}}\,={{e}^{-4\lambda t}}\,\,=\frac{1}{{{e}^{4\lambda t}}}\] But, we have given \[\frac{{{N}_{1}}}{{{N}_{2}}}\,=\,\,{{\left( \frac{1}{e} \right)}^{2}}\,\,=\,\,\frac{1}{{{e}^{2}}}\] Hence \[\frac{1}{{{e}^{2}}}\,\,=\,\,\frac{1}{{{e}^{4\lambda t}}}\] Comparing the powers, we get \[2\,\,=\,\,4\lambda t\] \[or\,\,\,{{V}_{s}}={{V}_{p}}\frac{{{N}_{s}}}{{{N}_{p}}}=4\left( \frac{1500}{50} \right)\]
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