Chapter 25, Problem 10STP

Interpretation Introduction

Interpretation:

The radioactive isotope is to be identified.

Concept interpretation:

The stability of a nucleus and isotope depends on the ratio of neutrons to protons (neutron-to-proton ratio). When an isotope has too many neutrons relative to the number of protons, it becomes unstable and undergoes decay.

Answer to Problem 10STP

Carbon- $\text{14}$ is the radioactive isotope.

Explanation of Solution

Isotope $\text{C}$ has $6$ protons and $8$ neutrons and the neutron-to-proton ratio is

  $\text{ratio=}\frac{\text{8}}{\text{6}}\text{=1}\text{.33}$ .

Isotope $\text{N}$ has $7$ protons and $7$ neutrons and the neutron-to-proton ratio is

  $\text{ratio=}\frac{7}{7}\text{=1}\text{.00}$

Isotope $\text{O}$ has $8$ protons and $8$ neutrons and the neutron-to-proton ratio is

  $\text{ratio=}\frac{10}{8}\text{=1}\text{.25}$

Conclusion

From these results, it can be concluded that the isotope Carbon- $\text{14}$ is the least stable and radioactive because it has a higher neutron-to-proton ratio $\left(\text{1}\text{.33}\right)$ . Also, isotope oxygen- $18$ is unstable because of its neutron-to-proton ratio is $1.25$ .

Only isotope nitrogen- $14$ is stable because it has a neutron-to-proton ratio equivalent to $1$ .