A) \[\frac{1}{8}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
The decay constant \[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{24000}=2.89\times {{10}^{-5}}\] Hence, the fraction of plutonium left is \[\frac{N}{{{N}_{o}}}={{e}^{-\lambda t}}={{e}^{(-2.89\times {{10}^{-5}})\times 72000}}\] \[={{e}^{-2.08}}=\frac{1}{{{e}^{2.08}}}=\frac{1}{8}\]
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