Question

a drug prepared for a patient is tagged with $_{43}^{99}tc$, which has a half - life of 6.05 h. how many $_{43}^{99}tc$ nuclei are required to give an activity of 1.50 $mu$ci?

a. $3.5\times10^{7}$ nuclei
b. $4.1\times10^{4}$ nuclei
c. $6.2\times10^{7}$ nuclei
d. $8.6\times10^{6}$ nuclei
e. $1.7\times10^{9}$ nuclei

Explanation:

Step1: Convert half - life to seconds

The half - life $T_{1/2}=6.05\ h$. Since $1\ h = 3600\ s$, then $T_{1/2}=6.05\times3600\ s = 21780\ s$.

Step2: Calculate the decay constant $\lambda$

The formula for the decay constant is $\lambda=\frac{\ln 2}{T_{1/2}}$. So, $\lambda=\frac{\ln 2}{21780\ s}\approx3.18\times 10^{-5}\ s^{-1}$.

Step3: Convert the activity to decays per second

The activity $A = 1.50\ \mu Ci$. Since $1\ Ci=3.7\times 10^{10}\ decays/s$ and $1\ \mu Ci = 10^{- 6}\ Ci$, then $A = 1.50\times10^{-6}\times3.7\times 10^{10}\ decays/s=5.55\times 10^{4}\ decays/s$.

Step4: Use the activity formula $A=\lambda N$ to find the number of nuclei $N$

We know that $A = \lambda N$, so $N=\frac{A}{\lambda}$. Substituting the values of $A$ and $\lambda$, we get $N=\frac{5.55\times 10^{4}\ s^{-1}}{3.18\times 10^{-5}\ s^{-1}}\approx1.7\times 10^{9}$ nuclei.

Answer:

e. $1.7\times 10^{9}$ nuclei