Products
Rewards
from HOLOOLY

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

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

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

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

HOLOOLY
TEXTBOOKS

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

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 12.12

What is the alpha activity of a 10-kg sample of ${ }^{235} U$ that is used in a nuclear reactor?

Strategy We find the number of radioactive atoms by using Avogadro’s number and the gram-molecular weight. We find in Appendix 8 that ${ }^{235} U$ has a half-life for emitting α particles of $t_{1 / 2}=7.04 \times 10^{8}$ y. Then we use Equation (12.22) to find the activity.

$R=\lambda N(t)$ (12.22)

## Verified Solution

The number of ${ }^{235} U$ atoms in a 10-kg sample is

$N=M \frac{N_{ A }}{M\left({ }^{235} U \right)}$

\begin{aligned}N &=(10 kg )\left(\frac{10^{3} g }{1 kg }\right)\left(\frac{6.02 \times 10^{23} \text { atoms } / mol }{235 g / mol }\right) \\&=2.56 \times 10^{25} \text { atoms }=2.56 \times 10^{25} \text { nuclei }\end{aligned}

The activity is

\begin{aligned}R=\lambda N &=\frac{\ln (2) \cdot N}{t_{1 / 2}} \\&=\frac{\ln (2) \cdot\left(2.56 \times 10^{25} \text { nuclei }\right)}{7.04 \times 10^{8} y } \\&=2.52 \times 10^{16} \text { decays } / y =8.0 \times 10^{8} Bq\end{aligned}