91Ó°ÊÓ

Nuclear reaction involves in beta decay of the isotope ${}^{137}\text{Cs}$is given below.

${}_{55}^{137}\text{Cs}→{→}_{56}^{137}\text{Ba}{+}_{−1}^{0}e$

In the each decay, the nucleus ${}^{137}\text{Cs}$decays only one beta particle.

Initial activity of radioactive substance relates with decay rate and initial number of nuclei presented as follows.

$R_0=r{N}_0$

Here, $R_0$is initial activity, $r$is decay rate and $N_0$is initial number of nuclei.

Rewrite this equation for $N_0$,

$N_0=\frac{R_0}{r}$

The relation between decay rate and half-life of a nucleus is,

$r=\frac{\ln 2}{t_{1/2}}$

Here, $t_{1/2}$is half-life.

Substitute $\frac{\ln 2}{t_{1/2}}$for $r$in the equation $N_0=\frac{R_0}{r}$.

$\begin{array}{l}{N}_0=\frac{R_0}{\left(\frac{\ln 2}{t_{1/2}}\right)}\\ =\frac{R_0{t}_{1/2}}{\ln 2}\end{array}$

Convert the unit for time from $y$to $s$.

$\begin{array}{l}{t}_{//2}=30\text{y}\\ =\left(30\text{y}\right)\left(\frac{365\text{days}}{1\text{y}}\right)\left(\frac{24\text{h}}{1\text{day}}\right)\left(\frac{60\text{min}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\\ =9.46×{10}^8\text{s}\end{array}$

Substitute $2.0×{10}^8\text{Bq}$for $R_0$and $9.46×{10}^8\text{s}$for $t_{1/2}$in the equation , $N_0=\frac{R_0{t}_{1/2}}{\ln 2}$and solve for

$N_0$

$\begin{array}{l}{N}_0=\frac{\left(2.0×{10}^8\text{Bq}\right)\left(9.46×{10}^8\text{s}\right)}{\ln 2}\\ =2.729×{10}^{17}\end{array}$

Round off to two significant figures,

$N_0=2.7×{10}^{17}$

Therefore, the number of beta particles emitted is $2.7×{10}^{17}$.