Q.5.34 - An Introduction to General, Organic, and Biological Chemistry

Question

Fluorine-18, which has a half-life of 110min, is used in PET scans.

a. If 100 mg of fluorine- 18 is shipped at 8:00 A.M., how many milligrams of the radioisotope are still active after 110min?

b. If 100 mg of fluorine-18 is shipped at 8: 00 A.M., how many milligrams of the radioisotope is still active when the sample arrives at the radiology laboratory at 1:30 PM?

Step-by-Step Solution

Answer

 Part (a).$50$milligrams of the radioisotope is still active

Part (b). $12.5$ milligrams of the radioisotope is still active  

Step 1: Given Information (part a)

We are given$100$ mg of fluorine- which is shipped at $8:00$A.M. Finding the amount of the radioisotope is still active after $110$min. 

Part (a) Step 2: Explanation

Given,

Amount of fluorine = $100mg$

The half-life of fluorine = $110$min

radioisotope that is still active after $110$ mins,

$=\frac{100}{2}=50mg$

Hence $50$ milligrams of the radioisotope is still active.

Part (b) Step 1: Given Information

We are given $50$ mg of fluorine-which is shipped at $8:00$A.M. Finding the amount of the radioisotope which is still active when the sample arrives at the radiology laboratory at $1:30$PM  

Part (b) Step 2: Explanation

We know, $A_o=100mg$ and time t = $330min$

The half-life is, $t_{\frac{1}{2}}=110\min$

Calculating the rate constant,

$k=\frac{0.693}{110}=0.0063/\min$

The first-order rate equation is, $\ln \left[A_t\right]=-kt+\ln \left[A_0\right]$

Substituting the values,

$\begin{array}{r}\ln \left[A_t\right]=-(0.0063\times 330)+\ln \left[100\right]\\ \ln \left[A_t\right]=-2.079+4.605=2.526\end{array}$

Applying antilog we have,

$A_t=12.5mg$

Hence$12.5$milligrams of the radioisotope is still active.