A radioactive sample contains 2.2 mg of pure \[_{6}^{11}C\] which has half-life of 1224 seconds. |
Calculate |
(i) the number of atoms present initially |
(ii) the activity when \[5\mu g\] of the sample will be left. |
Answer:
\[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{1224}=0.000566{{s}^{-1}}\] \[R=\lambda N=0.000566\times \frac{5\times {{10}^{-\,6}}}{11}\times 6.023\times {{10}^{23}}\] \[=1.55\times {{10}^{14}}\,Bq\] (i) \[{{N}_{0}}=1.205\times {{10}^{20}}atoms,\] \[R=1.55\times {{10}^{14}}\,Bq\] (i) \[{{N}_{0}}=\frac{2.2\times {{10}^{-\,3}}}{11}\times 6.023\times {{10}^{23}}=1.205\times {{10}^{20}}\] atoms (ii) \[N=\frac{5\times {{10}^{-\,6}}}{11}\times 6023\times {{10}^{23}}\]
You need to login to perform this action.
You will be redirected in
3 sec