Q. 5.73 - An Introduction to General, Organic, and Biological Chemistry

Question

The activity of $\mathrm{K}-40$ in a$70.-\mathrm{kg}$ human body is estimated to be $120\mathrm{nCi}$. What is this activity in becquerels? $(5.3)$

Step-by-Step Solution

Answer

As a result,$K-40\text{'}s$ activity in a$70-kg$ human body is$4.4\times {10}^3 \mathrm{Bq}$

Step 1: Introduction

      The number of nuclear disintegrations per second is used to determine a sample's activity. Curie $Ci$ is a typical unit for measuring activity.

     The $Sl$ unit of activity is the Becquerel $Bq$.

     In a $70-kg$ human body, the activity of$K-40$ is $120 Ci$.

     Calculate $K-40$ activity in becquerel.

Step 2: The following relationships

Curie and Becquerel have the following relationships:

$$\begin{gathered} 1\mathrm{Ci}=3.7\times {10}^{10} \mathrm{Bq} \\ lnCi=37 \mathrm{Bq} \end{gathered}$$

Therefore,

$120\mathrm{nCi}\times \frac{37 \mathrm{Bq}}{1\mathrm{nCi}}=4.4\times {10}^3 \mathrm{Bq}$

Step 3: Conclusion

        As a result,$K-40\text{'}s$ activity in a $70 - Kg$ human body is $4.4\times {10}^3 \mathrm{Bq}$.