A sample of ^18 F is used internally as a medical diagnostic tool to look for the effects of the positron decay ( T_(1 2) =110 min ) . How long does…

Question

A sample of ^18 F is used internally as a medical diagnostic tool to look for the effects of the positron decay ( T_(1 2) =110 min ) . How long does it take for 99% of the ^18F to decay?

MARKS Content Team · Accepted Answer

Correct answer: (C) 12.2 h. Radioactive decay equation is N=N_0e^(-λt)=N_0e^(-ln⁡2tT12) After decay of 99% of the initial sample only 1% will be left, and N/N_0=1% . We then have N/N_0=1/100=e^(-ln⁡2tT12) If we take the natural logarithm, we have -ln⁡100=-(ln⁡2×t)/T_12 Which on solving for t yields ∴ t=(ln⁡100)/(ln⁡2)×T_12 =(log⁡100)/(log⁡2)×T_12 =2/0.3010×110 =731min⁡=12.2 h

A sample of $^{18} F$ is used internally as a medical diagnostic tool to look for the effects of the positron decay $(T_{\frac{1}{2}} = 110 min)$ . How long does it take for $99%$ of the $^{18} F$ to decay?

Solution

More Nuclear Physics Questions

A sample of 18 F is used internally as a medical diagnostic tool to look for the effects of the positron decay ( T 1 2 =110 min ) . How long does it take for 99% of the 18F to decay?

Solution

More Nuclear Physics Questions

A sample of $^{18} F$ is used internally as a medical diagnostic tool to look for the effects of the positron decay $(T_{\frac{1}{2}} = 110 min)$ . How long does it take for $99%$ of the $^{18} F$ to decay?