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 ${}^{226}Ra$ .

Short answer

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

Step-by-step solution

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.

 Decay constant of  R226a

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$=\left(5.4\times {10}^{-6}g\right)\left(\frac{1molRaC{l}_2}{297gRaC{l}_2}\right)\left(\frac{1molRa}{1molRaCl}\right)\left(\frac{6.022\times {10}^{23}Raatoms}{1molRa}\right)$

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

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

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

 Half-life of  R226a

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 $5.1\times {10}^{10}s$.