Problem 51 - Chemistry: The Central Science

Question

Which of the following nuclei is likely to have the largest mass defect per nucleon: (a) \({ }^{59} \mathrm{Co}\), (b) \({ }^{11} \mathrm{~B}\), (c) \({ }^{118} \mathrm{Sn},(\mathrm{d}){ }^{243} \mathrm{Cm} ?\) Explain your answer.

Step-by-Step Solution

Answer

Among the given nuclei (Cobalt, Boron, Tin, and Curium), the cobalt nucleus (\(^{59} \mathrm{Co}\)) is likely to have the largest mass defect per nucleon. This is because it is closest to the iron peak elements, which have the highest binding energy per nucleon, resulting in a larger mass defect per nucleon.

Step 1: Compare with 'Iron peak' elements

In the nucleus, the binding energy per nucleon peaks around iron. We can use this fact to determine which nucleus is more likely to have a higher binding energy per nucleon (and thus larger mass defect per nucleon). - Cobalt (\(^{59} \mathrm{Co}\)) is close to the iron peak (\(A \sim 60\)), so it will have a high binding energy per nucleon. - Boron (\(^{11} \mathrm{B}\)) is much lighter than the iron peak elements but is heavier than hydrogen with a mass number of 11. Therefore, it will have a lower binding energy per nucleon compared to the iron peak elements. - Tin (\(^{118} \mathrm{Sn}\)) is heavier than the iron peak elements. Therefore, it will have a lower binding energy per nucleon compared to the iron peak elements. - Curium (\(^{243} \mathrm{Cm}\)) is much heavier than the iron peak elements, so its binding energy per nucleon will be lower than the iron peak elements.

Step 2: Determine the largest mass defect per nucleon

Now that we have analyzed each nucleus relative to the iron peak elements, we can determine which nucleus is likely to have the largest mass defect per nucleon among these four nuclei. Since cobalt is closest to the iron peak, it is most likely to have the highest binding energy per nucleon. A higher binding energy per nucleon implies a larger mass defect per nucleon. Therefore, cobalt is the nucleus with the largest mass defect per nucleon among these four nuclei.

Step 3: Conclusion

Among the given nuclei (Cobalt, Boron, Tin, and Curium), the cobalt nucleus (\(^{59} \mathrm{Co}\)) is likely to have the largest mass defect per nucleon. This is because it is closest to the iron peak elements, which have the highest binding energy per nucleon, resulting in a larger mass defect per nucleon.

Key Concepts

Nuclear Binding Energy

Nuclear binding energy is the energy required to separate an atom's nucleus into its individual protons and neutrons. In essence, it's what keeps the nucleus intact and provides stability to the atom. The larger this energy, the more stable the nucleus is, meaning it's less likely to decay or react.

Atoms strive for stability by optimizing their binding energy. To understand this, imagine you are stacking blocks. If the blocks are loosely stacked, they're more likely to fall apart than if they're tightly packed together. Similarly, in atomic nuclei, the stronger the binding (the 'tighter the blocks are stacked'), the more stable the nucleus.

When calculating the binding energy per nucleon (which is simply the binding energy divided by the number of nucleons), we find that iron and its closest neighbors in the periodic table have the highest values. This is due to the balance between the attractive nuclear force and the repulsive electrostatic force within the nucleus, leading to a stable configuration.

Iron Peak Elements

Iron peak elements are those that exist at the pinnacle of binding energy per nucleon, meaning they have the strongest held nuclei and are the most stable. These include iron (Fe) along with its neighbors, such as cobalt (Co) and nickel (Ni).

The significance of the iron peak is twofold: it serves as a yardstick in nuclear physics for stability and helps us understand stellar processes. Stars produce energy and heavier elements through nuclear fusion but they stop at iron; fusion beyond the iron peak doesn't yield energy, it consumes it.

Comparing Mass Defects

Using the iron peak as a reference, we can deduce that nuclei lighter than iron have less binding energy per nucleon, hence a smaller mass defect per nucleon. Conversely, nuclei heavier than iron also have a lower binding energy per nucleon, which means the 'tight stacking' of protons and neutrons is less efficient in these atoms, hinting at lesser stability and a smaller mass defect per nucleon.

Nuclear Stability

An atom's nucleus stability is an intricate balance of forces. The strong nuclear force attempts to hold the protons and neutrons together, while the electrostatic force due to the protons' positive charge pushes them apart. Stable nuclei are those where these forces are in equilibrium.

However, stability isn't just about forces; it also concerns the ratio of protons to neutrons within the nucleus. Generally, a 1:1 proton-neutron ratio upholds stability in lighter elements. But as atoms get heavier, they require more neutrons to maintain the balance against the growing electrostatic repulsion between protons.

Several factors can thus affect stability: the total number of nucleons, the proton-neutron ratio, and the energy of the individual nucleons. Hence, why nucleon mass defect is a key indicator of nuclear stability: it reflects the binding energy of the nucleus, correlating directly with stability – the greater the mass defect per nucleon, the more tightly bound and stable the nucleus is likely to be.