A) 0.014 years
B) 6.66 years
C) 66.6 years
D) 666 years
Correct Answer: C
Solution :
Total amount of \[(A)\text{ }{{N}_{0}}=1\text{ }g\] After\[\beta -\] emission the amount rest \[=0.125\text{ }g\] Total time = 200 years \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}},\frac{0.125}{1}={{\left( \frac{1}{2} \right)}^{n}}\] \[=\frac{125}{1000}={{\left( \frac{1}{2} \right)}^{n}}=\frac{1}{8}={{\left[ \frac{1}{2} \right]}^{n}}\] Or \[{{\left( \frac{1}{2} \right)}^{3}}={{\left( \frac{1}{2} \right)}^{n}}n=3\] \[n=\frac{T}{{{t}_{1/2}}},3=\frac{200}{{{t}_{1/2}}}\] \[{{t}_{1/2}}=\frac{200}{3}\] \[=66.6\text{ }years\]
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