A) \[e\]
B) \[{{e}^{2}}\]
C) \[{{e}^{-1}}\]
D) \[{{e}^{-2}}\]
Correct Answer: D
Solution :
: According to radioactive decay, \[N={{N}_{0}}{{e}^{-\lambda t}}\] where \[{{N}_{0}}\]= Initial number of nuclei at t = 0 N= Number of nuclei left undecayed after time \[\lambda \]= Decay constant For A, \[{{N}_{A}}={{N}_{0A}}{{e}^{^{^{-(15x)(1/6x)}}}}\] \[={{N}_{0A}}{{e}^{-5/2}}\] For B, \[{{N}_{0B}}{{e}^{-(3x)(1/6x)}}\] \[{{N}_{0B}}{{e}^{-1/2}}\] As per question, \[{{N}_{0}}_{A}={{N}_{0B}}\] \[\therefore \] \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{{{e}^{-5/2}}}{{{e}^{-1/2}}}={{e}^{-2}}\]
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