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Chapter 12

Q. 12.10

A sample of { }^{210} Po which α decays with t_{1 / 2}=138 days is observed by a student to have 2000 disintegrations/s (2000 Bq). (a) What is the activity in μCi for this source? (b) What is the mass of the { }^{210} Po sample?

Strategy (a) We already know the activity in disintegrations/s or Bq. We simply use the conversion to μCi. (b) Because we know the activity (2000 decays/s), we can use Equations (12.22) and (12.28) to find the number of radioactive nuclei. From this we can determine the mass of the { }^{210} Po sample.

R=\lambda N(t) (12.22)

\tau=\frac{1}{\lambda}=\frac{t_{1 / 2}}{\ln 2} (12.28)

Step-by-Step

Verified Solution

(a) We multiply the activity of 2000 decays/s by the factor that converts decays/s to Ci.

 

\begin{aligned}2000 \text { decays } / s \left(\frac{1 Ci }{3.7 \times 10^{10} \text { decays } / s }\right) &=0.054 \times 10^{-6} Ci \\&=0.054 \mu Ci\end{aligned}

 

(b) Equations (12.22) and (12.28) give the number of radioactive nuclei.

 

\begin{aligned}N &=\frac{R}{\lambda}=\frac{(R)\left(t_{1 / 2}\right)}{\ln (2)}=\frac{2000 \text { decays } / s }{\ln (2)}(138 \text { days }) \frac{24 h }{1 \text { day }} \frac{3600 s }{1 h } \\&=3.44 \times 10^{10} \text { nuclei }\end{aligned}

 

We use Avogadro’s number to determine the mass from the number of atoms (nuclei).

 

\begin{aligned}\text { Mass } &=3.44 \times 10^{10} \text { atoms } \frac{1 mol }{6.02 \times 10^{23} \text { atoms }} \frac{0.210 kg }{1 mol } \\&=1.2 \times 10^{-14} kg\end{aligned}

 

This is an extremely small mass!