91Ó°ÊÓ

The rate constant, k, is calculated as follows:

$$\begin{gathered} t_{1/2}=\frac{\ln 2}{k}k \\ =\frac{\ln 2}{t_{1/2}} \\ =\frac{0.693}{1.25×{10}^{9}yr} \\ =5.544×{10}^{-10}y{r}^{-1} \end{gathered}$$

No. of of ${}^{40}K=1.0mol$

No. of atoms = No. of moles Avogadro's number

$$\begin{gathered} =1.0mol { }^{40}K×6.022×{10}^{23}atoms/mol { }^{40} \\ =6.022×{10}^{23}atoms { }^{40} \end{gathered}$$

Rate of decay, A = kN

$$\begin{gathered} A=\left(5.544×{10}^{-10}y{r}^{-1}\right)×\left(6.022×{10}^{23}atoms\right)×\left(\frac{1yr}{365d}\right)\left(\frac{1d}{24h}\right)\left(\frac{1h}{3600s}\right)\left(\frac{1di\sin tegration}{1atom}\right) \\ A=1.059×{10}^7 \end{gathered}$$

The number of Curie (Ci) (Dose) produced by $1.0mo{l}^{40}K$ is calculated as follows:

$Dose\left(inCi\right)=\left(1.059×{10}^{7}dps\right)\left(\frac{1Ci}{3.70×{10}^{10}dps}\right)$

$$\begin{gathered} =2.86×{10}^{-4}Ci \\ =2.9×{10}^{-4}Ci \end{gathered}$$

Number of Curie (Ci) (Dose) produced bydata-custom-editor="chemistry" $1.0mo{l}^{40}K$ is $2.9×{10}^{-4}Ci$.

Number of Becquerel ( Bq ) (Dose) produced by $1.0mo{l}^{40}K$ is determined as given below,

$$\begin{gathered} Dose\left(Bq\right)=\left(1.059×{10}^{7}dps\right)\left(\frac{1Bq}{1dps}\right) \\ =1.059×{10}^{7}Bq \\ =1.1×{10}^{7}Bq \end{gathered}$$

Therefore, the required number of Becquerel ) (Dose) produced by $1.0mo{l}^{40}K$ is $1.1×{10}^{7}Bq$.