Question

How much of this isotope ${(}_6^{11}C)$ is left after 24 h of its preparation?

• A. $2.28\times {10}^{-21}g{(}_6^{11}C)$.
• B. $3.28\times {10}^{-21}g{(}_6^{11}C)$.
• C. $4.58\times {10}^{-21}g{(}_6^{11}C)$.
• D. None of these

Answer 1

Answer: (A)

$2.28\times {10}^{-21}g{(}_6^{11}C)$.

Using equation : $C=C_0{\left(\frac{1}{2}\right)}^y$

where, $y=\frac{totaltime}{T_{50}}=\frac{24\times 60min}{21min}=68.57$

$C_0=initialamount=1g$

$C=$amount of radioactive substance left after y half-lives

$\therefore$ $C=1{\left(\frac{1}{2}\right)}^{68.57}$

Taking log and then antilog, $C=2.28\times {10}^{-21}g{(}_6^{11}C)$.