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?$(5.3)$

Short answer

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

Step-by-step solution

    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 $120Ci$.

Calculate $K-40$activity in becquerel.

     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,

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

    Conclusion

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