Question

The radium isotope ²²⁶Ra has a half-life of 1600 years. A

sample begins with 1.00x10^{10 226}Ra atoms. How many are left

after (a) 200 years, (b) 2000 years, and (c) 20,000 years?

Short answer

Therefore,

$$\begin{gathered} \text{a.)}N=9.17\times {10}^9\text{atoms} \\ \text{b.)}N=4.20\times {10}^9\text{atoms} \\ \text{c.)}N=1.73\times {10}^6\text{atoms} \end{gathered}$$

Step-by-step solution

Given information

$$\begin{gathered} t_{1/2}=1600\text{years} \\ {N}_o=1.0\times {10}^{10}\text{atoms} \end{gathered}$$

Explanation (a)

(a)After t = 200 years, we must determine the number of nuclei ²²⁶Ra that remain.

$$\begin{gathered} N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right){\left(\frac{1}{2}\right)}^{200\text{years}/1600\text{years}} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right){\left(\frac{1}{2}\right)}^{0.125} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right)\times (0.9170) \\ N=9.17\times {10}^9\text{atoms} \end{gathered}$$

Explanation (b)

(b)After t = 2000 years, we must determine the number of nuclei ²²⁶Ra that remain.

$$\begin{gathered} N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right){\left(\frac{1}{2}\right)}^{2000\text{years}/1600\text{years}} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right){\left(\frac{1}{2}\right)}^{125} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right)\times (0.42044) \\ N=4.20\times {10}^9\text{atoms} \end{gathered}$$

Explanation (c)

(c)After t = 20000 years, we must determine the number of nuclei ²²⁶Ra that remain.

$$\begin{gathered} N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right){\left(\frac{1}{2}\right)}^{20000\text{years}/1600\text{years}} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right){\left(\frac{1}{2}\right)}^{12.5} \\ N=\left(1.0\times {10}^{10}\text{atoms}\right)\times (0.00017) \\ N=1.73\times {10}^6\text{atoms} \end{gathered}$$