A) 40 year, 0.9173/year
B) 90 year, 9.017/year
C) 80 year, 0.0173/year
D) none of these
Correct Answer: C
Solution :
Here: \[{{T}_{1/2}}=40\] years, \[\frac{N}{{{N}_{0}}}=\frac{1}{4}\] From relation \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{t/40}}\] or \[\left( \frac{1}{4} \right)={{\left( \frac{1}{2} \right)}^{t/40}}\] or \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{t/40}}\] Hence, \[\frac{t}{40}=2\]or t = 80 year Decay constant \[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{40}=0.0173year\]
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