Question

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable: (a) \(\frac{92}{44} \mathrm{Ru}\) and \({ }^{102} \mathrm{Ru},(\mathbf{b}){ }_{56}^{138} \mathrm{Ba}\) and \({ }_{56}^{139} \mathrm{Ba},(\mathbf{c})\) tin -109 and tin-120.

Short answer

The radioactive nuclides are \(\frac{92}{44} \mathrm{Ru}\), \({ }_{56}^{138} \mathrm{Ba}\), and Tin-109, while the stable nuclides are \({ }^{102} \mathrm{Ru}\), \({ }_{56}^{139} \mathrm{Ba}\), and Tin-120. This is based on comparing their neutron-to-proton ratios, as higher ratios generally lead to greater stability for medium-sized nuclei.

Step-by-step solution

Calculate the neutron-to-proton ratios for each pair of nuclides

To calculate the neutron-to-proton ratio, we need to find the number of neutrons in each nuclide and divide it by the number of protons. Protons can be found in the bottom number, while the difference between the top and bottom numbers represents the neutrons. (a) \(\frac{92}{44} \mathrm{Ru}\): Neutrons = 92 - 44 = 48, ratio = \( \frac{48}{44} = 1.09 \) \({ }^{102} \mathrm{Ru}\): Neutrons = 102 - 44 = 58, ratio = \( \frac{58}{44} = 1.32 \) (b) \({ }_{56}^{138} \mathrm{Ba}\): Neutrons = 138 - 56 = 82, ratio = \( \frac{82}{56} = 1.46 \) \({ }_{56}^{139} \mathrm{Ba}\): Neutrons = 139 - 56 = 83, ratio = \( \frac{83}{56} = 1.48 \) (c) Tin-109: Nuclide is \({ }_{50}^{109} \mathrm{Sn}\) Neutrons = 109 - 50 = 59, ratio = \( \frac{59}{50} = 1.18 \) Tin-120: Nuclide is \({ }_{50}^{120} \mathrm{Sn}\) Neutrons = 120 - 50 = 70, ratio = \( \frac{70}{50} = 1.4 \)

Predict the stability based on neutron-to-proton ratios

Based on the calculated ratios, we can now predict the stability of the nuclides. Remember that lighter elements with a nuclear ratio close to 1 are stable and as the atomic number increases, stable nuclei have higher neutron-to-proton ratios. (a) \(\frac{92}{44} \mathrm{Ru}\) has a ratio of 1.09 while \({ }^{102} \mathrm{Ru}\) has a ratio of 1.32. Since the atomic number is 44 (medium-sized nucleus), a higher neutron-to-proton ratio will make the nucleus more stable. Therefore, \({ }^{102} \mathrm{Ru}\) is stable and \(\frac{92}{44} \mathrm{Ru}\) is radioactive. (b) \({ }_{56}^{138} \mathrm{Ba}\) has a ratio of 1.46 while \({ }_{56}^{139} \mathrm{Ba}\) has a ratio of 1.48. They both have close ratios, but since the atomic number is 56 (medium-sized nucleus), a slightly higher neutron-to-proton ratio will favor stability. Thus, \({ }_{56}^{139} \mathrm{Ba}\) is stable and \({ }_{56}^{138} \mathrm{Ba}\) is radioactive. (c) Tin-109 (\({ }_{50}^{109} \mathrm{Sn}\)) has a ratio of 1.18 while Tin-120 (\({ }_{50}^{120} \mathrm{Sn}\)) has a ratio of 1.4. Since the atomic number is 50 (medium-sized nucleus), a higher neutron-to-proton ratio will favor stability. Therefore, Tin-120 is stable and Tin-109 is radioactive. In conclusion, the radioactive nuclides are \(\frac{92}{44} \mathrm{Ru}\), \({ }_{56}^{138} \mathrm{Ba}\), and Tin-109, while the stable nuclides are \({ }^{102} \mathrm{Ru}\), \({ }_{56}^{139} \mathrm{Ba}\), and Tin-120.

Key concepts

Neutron-to-Proton Ratio

The neutron-to-proton ratio is a key factor in determining the stability of a nuclide. Neutrons and protons, found in the nucleus of an atom, affect how stable or unstable an atom is. This ratio involves calculating the number of neutrons (mass number minus atomic number) and dividing by the number of protons, which is represented by the atomic number.

For stable lighter elements, a neutron-to-proton ratio close to 1 is typical. However, as you move up the periodic table to heavier elements, this ratio tends to increase. This is because heavier nuclei require more neutrons to balance the repulsive forces between the protons.

This exercise demonstrates that the higher the neutron-to-proton ratio in elements with higher atomic numbers, the more likely the nuclide is stable. For instance, with (a) [ uclide{102}{44}{Ru} uclide{92}{44}{Ru}] , the higher ratio of 1.32 as seen in ( uclide{102}{44}{Ru} uclide{102}{44}{Ru}) indicates greater stability compared to ( uclide{92}{44}{Ru} uclide{92}{44}{Ru}) 's ratio of 1.09.

Nuclear Stability

Nuclear stability is the likelihood of a nuclide to remain consistent without undergoing radioactive decay. It depends predominantly on the neutron-to-proton ratio and the forces within the nucleus.

The strong nuclear force is what holds the protons and neutrons together in a nucleus. When there is an optimal balance, this force is highly efficient at keeping the nucleus intact. If the ratio of neutrons is too high or too low, the nucleus may become unstable, leading to radioactivity.

In this exercise, differentiating between stable and radioactive nuclides is key. (a-c) shows that when nuclides have a balanced or slightly higher neutron presence, as in ([nucide{102}{44}{Ru}],[nucide{139}{56}{Ba}],[nucide{120}{50}{Sn}]), stability is achieved.

Conversely, the fewer neutrons or imbalance gives a radioactive characteristic like in ( (92/44), uclide{138}{56}{Ba}, ).

Atomic Number

The atomic number is the number of protons in an atom's nucleus and it ultimately defines the identity of an element. This number not only tells us about the element but also plays a significant role in nuclear stability.

Each element in the periodic table is represented by its unique atomic number, making it easier to evaluate their properties. When predicting stability, it is crucial to consider the atomic number; larger atomic numbers often mean more complex nucleic interactions and hence a need for more neutrons to stabilize the nucleus.

For example, in the given pairs (a-c), the atomic numbers delineate ([44, 44, 56, 56, 50, 50]). These atomic numbers signal medium-sized nuclei which means more neutrons are needed to maintain stability compared to lighter nuclei such as hydrogen or helium.

Stable Nuclides

A stable nuclide is one that is not radioactive and thus does not spontaneously undergo decomposition to form different nuclides. Stability is primarily about achieving a balance of forces within the nucleus, often correlated with a favorable neutron-to-proton ratio.

In the context of this exercise, the stability of the nuclides ( uclide{102}{44}{Ru}, nuclide{139}{56}{Ba}, nucide{120}{50}{Sn}) can be attributed to their appropriate ratios.

This is contrasted to the other members of their respective pairs, whose neutron-to-proton imbalance makes them susceptible to radioactive decay. Consequently, understanding the needs of a stable nuclide is essential in fields like nuclear chemistry where predicting decay and managing radioactive materials are routine tasks.

• Importance of balanced neutron-to-proton ratios
• Effectiveness of nuclear forces
• Atomic number's role in enabling stability

Overall, stability is achieved by a harmonious relationship between all these factors.