Question

The activity of a sample of the cesium isotope 137 Cs, with a half-life of 30 years, is 2.0 * 108 Bq. Many years later, after the sample has fully decayed, how many beta particles will have been emitted?

Short answer

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

the number of beta particles emitted is 2,7x10¹⁷

Convert the unit for time from y to s

t_1/2=30y

=9.46x10⁸s

Step-by-step solution

Step 1

Nuclear reaction involves in beta decay of the isotope ¹³⁷Cs is given below.

${}_{55}{}^{137}Cs\underset{56}{\overset{137}{\to }}Ba{+}_{-1}^{0}e$

In the each decay, the nucleus ¹³⁷Cs 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, R₀ is initial activity, r is decay rate and N₀ is initial number of nuclei.

Rewrite this equation for N₀,

role="math" localid="1649794432288" $N_0=\frac{R_0}{r}$

Step 2

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

$r=\frac{In2}{t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Here t_1/2 is half life

substitute role="math" localid="1649794201361" $$\begin{gathered} \frac{\mathrm{In}2}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\mathrm{for}\mathrm{r}\mathrm{in}\mathrm{the}\mathrm{equation}{\mathrm{N}}_0=\frac{{\mathrm{R}}_0}{\mathrm{r}} \\ {\mathrm{N}}_0=\frac{{\mathrm{R}}_0}{\left({\displaystyle \frac{\mathrm{In}2}{{\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)} \end{gathered}$$

Convert the unit for time from y to s

t_1/2=30y

=9.46x10⁸s

Step 3

Substitute 2.0x10⁴ Bq for R0 and 9.46x10⁸ s for t_1/2 in the equation

$N_0=2.7x{10}^{17}$

Therefore the number of beta particles emitted is 2,7x10¹⁷