24.98 CP - Chemistry: Molecular Nature Of Matter And Change

Question

A $5.4-\mu$  gample of ${}^{226}RaC{l}_2$  has a radioactivity of $1.5\times {10}^5 Bq$ . Calculate $t_{1/2}$  of  ${}^{226}Ra$ .

Step-by-Step Solution

Answer

Half-life of ${}^{226}Ra$  is $5.1\times {10}^{10} s$ .

Step 1: Half-life

The half-life of a radioactive substance is defined as the amount of time required for the decay of half the initial amount of a substance. Half-life is denoted by  .

Radioactive decay is a first-order reaction. The expression for half-life is given below.

  

Where, k is the decay constant.

Step 2: Decay constant of R 226 a

Considering the given information:

 $Mass=5.4\mu g=5.4\times {10}^{-6} g$

 

Activity of   ${}^{226}RaC{l}_2=1.5\times {10}^5 Bq$

The following equation can be used to calculate the number of Ra atoms:

No. of   atoms  $=\left(5.4\times {10}^{-6} g\right)\left(\frac{1 molRaC{l}_2}{297 gRaC{l}_2}\right)\left(\frac{1 molRa}{1 molRaCl}\right)\left(\frac{6.022\times {10}^{23}Raatoms}{1 molRa}\right)$

                     $=1.095\times {10}^{16}Ra$                 atoms

The following equation can be used to calculate the rate constant:

 $$\begin{gathered} A=kN \\ \left(1.5\times {10}^5 Bq\right)=k\left(1.095\times {10}^{16}Raatoms\right) \\ k=\left(1.5\times {10}^5 Bq\right)\left(\frac{1 d/s}{Bq}\right)\left(\frac{1}{1.095\times {10}^{16}Raatoms}\right) \\ =1.3699\times {10}^{-11} s^{-1} \end{gathered}$$

Step 3: Half-life of R 226 a

The half-life of    is calculated as follows:

 

 $$\begin{gathered} t_{l/2}=\frac{\ln 2}{k} \\ {t}_{l/2}=\frac{\ln 2}{k} \\ =\frac{0.693}{1.3699\times {10}^{-11} s^{-1}} \\ =5.1\times {10}^{10} s \end{gathered}$$

Therefore, half-life of    is $5.1\times {10}^{10} s$ .