We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

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)


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!