Question

The barium isotope ¹³¹Ba has a half-life of 12 days. A 250 $\mu$mg

sample of ¹³¹Ba is prepared. What is the mass of ¹³¹Ba after

(a) 1 day, (b) 10 days, and (c) 100 days?

Short answer

Therefore, the mass is

$$\begin{gathered} \text{a.)}N=236\mu \mathrm{g} \\ \text{b.)}\mathrm{N}=140\mathrm{\mu g} \\ \text{c.)}\mathrm{N}=0.775\mathrm{\mu g} \end{gathered}$$

Step-by-step solution

Given information

$$\begin{gathered} N_o=250\mu \mathrm{g} \\ {\mathrm{t}}_{1/2}=12\mathrm{days} \end{gathered}$$

Explanation (a)

(a)After one day, we will determine N's value:

$$\begin{gathered} N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ N=\left(250\mu \mathrm{g}\right){\left(\frac{1}{2}\right)}^{1\text{day}/12\text{days}} \\ N=\left(250\mu \mathrm{g}\right){\left(\frac{1}{2}\right)}^{0.0833} \\ N=236\mu \mathrm{g} \end{gathered}$$

Explanation (b)

(b)After 10 day, we will determine N's value

$$\begin{gathered} N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ N=\left(250\mu \mathrm{g}\right){\left(\frac{1}{2}\right)}^{10\text{days}/12\text{days}} \\ N=\left(250\mu \mathrm{g}\right){\left(\frac{1}{2}\right)}^{0.833} \\ N=140\mu g \end{gathered}$$

Explanation (c)

(c)The N's value after 100 days:

$$\begin{gathered} N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ N=\left(250\mu g\right){\left(\frac{1}{2}\right)}^{100\text{days}/12\text{days}} \\ N=\left(250\mu \mathrm{g}\right){\left(\frac{1}{2}\right)}^{8.33} \\ N=0.775\mu \mathrm{g} \end{gathered}$$