A) 574 years
B) 5740 years
C) 2870 years
D) 287 years
Correct Answer: B
Solution :
[b] \[\frac{{{14}_{C}}}{{{12}_{C}}}=1.3\times {{10}^{-12}}\] \[12\text{ }g\] 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.
You need to login to perform this action.
You will be redirected in
3 sec
