24.107 CP - Chemistry: Molecular Nature Of Matter And Change

Question

How many curies are produced by 1.0 mol  of ? How many becquerels ${}^{40} K\left(t_{1/2}=\right.\left.1.25\times {10}^{9}yr\right)$?

Step-by-Step Solution

Answer

• Number of Curie (Ci) (Dose) produced by  $1.0 mo{l}^{40}K$ is $2.9\times {10}^{-4}Ci$ .
• Number of Becquerel ) (Dose) produced by $1.0 mo{l}^{40} K$  is $1.1\times {10}^{7}Bq$ .

Step 1: Find decay constant from half-life

The rate constant, k, is calculated as follows:

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

Step 2: Find the number of Curies produced by 1 . 0   m o l     40 K

No. of  of  ${}^{40}K=1.0 mol$

 

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

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

 

Rate of decay,  A = kN

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

 

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

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

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

                       

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

Step 3: Number of Becquerel formed

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

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

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