Nuclear Physics And Radioactivity Question 458
Question: The ratio of number of atoms of $ ^{14}C $ to $ ^{12}C $ in living matter is measured to be $ 1.3\times {{10}^{-12}} $ at the present time. A 12 g sample of carbon produces 180 decays/min due to the small amount of $ ^{14}C $ in it. The half-life of $ ^{14}C $ is nearly [1 year $ =3.15\times 10^{7}s $ ]
Options:
A) 574 years
B) 5740 years
C) 2870 years
D) 287 years
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{14_{C}}{12_{C}}=1.3\times {{10}^{-12}} $ $ 12g $ contain $ 6.022\times 10^{23} $ atoms No. atoms of $ 14_{C}=6.022\times 10^{23}\times 1.3\times {{10}^{-12}} $ $ =7.8286\times 10^{11}\Rightarrow N=N_{0}{{e}^{-\lambda t}} $ $ -\frac{dN}{dt}=N_{0}{{e}^{-\lambda t}}\times \lambda \Rightarrow -\frac{dN}{dt}=N\times \lambda $ $ \frac{180}{60}=7.8286\times 10^{11}\times \lambda $
$ \Rightarrow \frac{1}{\lambda }=\frac{7.8286\times 10^{11}}{3}\Rightarrow \lambda =0.3832\times {{10}^{-11}} $ $ {t_{1/2}}=\frac{0691}{\lambda }=\frac{0.692}{0.3832\times {{10}^{-11}}}=1.80\times 10^{11}\sec $ Half life $ =1.8032\times 10^{11}\sec =0.5740\times 10^{4} $ year=5740 years.
BETA
