Problem 26 - Chemistry: The Central Science in SI Units, Global Edition

Question

Which of the following nuclides would you expect to be radioactive: \({ }_{26}^{58} \mathrm{Fe},{ }_{27}^{60} \mathrm{Co},{ }_{41}^{92} \mathrm{Nb},\) mercury- \(202,\) radium \(-226 ?\) Justify your choices.

Step-by-Step Solution

Answer

Among the given nuclides, mercury-202 and radium-226 are expected to be radioactive due to their higher neutron-to-proton ratios (1.53 and 1.57, respectively) and lack of magic numbers. The other nuclides, \({}_{26}^{58} \mathrm{Fe}, { }_{27}^{60} \mathrm{Co},\) and \({}_{41}^{92} \mathrm{Nb}\), have neutron-to-proton ratios close to 1 (1.23, 1.22, and 1.24, respectively), suggesting that they are likely to be stable and not radioactive.

Step 1: Calculate Neutron-to-Proton Ratios

For each nuclide, calculate the neutron-to-proton (n/p) ratio. The neutron number (N) can be found by subtracting the atomic number Z (number of protons) from the mass number A (sum of protons and neutrons). The calculation of the n/p ratio for each nuclide is as follows: For \({}_{26}^{58} \mathrm{Fe}\), \(N = A - Z = 58 - 26 = 32\). The n/p ratio = \(\frac{32}{26} \approx 1.23\). For \({}_{27}^{60} \mathrm{Co}\), \(N = 60 - 27 = 33\). The n/p ratio = \(\frac{33}{27} \approx 1.22\). For \({}_{41}^{92} \mathrm{Nb}\), \(N = 92 - 41 = 51\). The n/p ratio = \(\frac{51}{41} \approx 1.24\). For mercury-202, \(A=202\) and \(Z=80\), \(N = 202 - 80 = 122\). The n/p ratio = \(\frac{122}{80} \approx 1.53\). For radium-226, \(A=226\) and \(Z=88\), \(N = 226 - 88 = 138\). The n/p ratio = \(\frac{138}{88} \approx 1.57\).

Step 2: Analyze Neutron-to-Proton Ratios and Magic Numbers for Stability

Using the neutron-to-proton ratios calculated in Step 1, along with the concept of magic numbers, analyze each nuclide for stability: For \({}_{26}^{58} \mathrm{Fe}\), the n/p ratio is close to 1, suggesting it may be stable. Neither the proton number (26) nor neutron number (32) corresponds to magic numbers. For \({}_{27}^{60} \mathrm{Co}\), the n/p ratio is close to 1, suggesting it may be stable. The neutron number (33) is not a magic number, but the proton number (27) is close to the magic number 28. For \({}_{41}^{92} \mathrm{Nb}\), the n/p ratio is close to 1, suggesting it may be stable. Neither the proton number (41) nor neutron number (51) corresponds to magic numbers. For mercury-202, the n/p ratio is higher, suggesting it may be unstable (i.e., radioactive). Neither the proton number (80) nor neutron number (122) corresponds to magic numbers. For radium-226, the n/p ratio is higher, suggesting it may be unstable (i.e., radioactive). Neither the proton number (88) nor neutron number (138) corresponds to magic numbers.

Step 3: Determine the Radioactive Nuclides

From the analysis in Step 2, it is expected that mercury-202 and radium-226 would be radioactive due to their higher neutron-to-proton ratios and lack of magic numbers. The other nuclides (\({_26}^{58} \mathrm{Fe}\), \({_27}^{60} \mathrm{Co}\), and \({_41}^{92} \mathrm{Nb}\)) have neutron-to-proton ratios close to 1 and are not expected to be radioactive.

Key Concepts

Neutron-to-Proton Ratio

The neutron-to-proton (n/p) ratio is a key factor in determining the stability of a nucleus. This ratio represents the balance between the number of neutrons and protons in an atom's nucleus. To find the n/p ratio, subtract the atomic number, Z (number of protons), from the mass number, A, which gives the number of neutrons, N. Then, the neutron-to-proton ratio is calculated as \(n/p = \frac{N}{Z}\). For nuclear stability:

• A ratio close to 1 is typically ideal for lighter elements (up to calcium with proton number 20).
• For heavier elements, a higher n/p ratio is necessary due to the increased demand for neutron-induced nuclear binding.

A higher n/p ratio beyond a certain limit generally indicates instability, which often leads to radioactivity. This is particularly evident in elements with higher atomic numbers, where the repulsive forces between protons increase, thereby requiring more neutrons to stabilize the nucleus.

Magic Numbers

The concept of magic numbers is crucial in understanding nuclear stability, as certain numbers of protons or neutrons result in more stable nuclei. Magic numbers are: 2, 8, 20, 28, 50, 82, and 126. When either the number of protons or the number of neutrons equals one of these magic numbers, the nucleus is considered highly stable. Magic numbers arise because of complete energy levels within a nucleus, leading to a more energetically favorable and stable configuration. Some key points include:

• Nuclei with magic numbers have extra binding energy, making them less likely to undergo radioactive decay.
• Examples of stable nuclei include ⁴He, ¹⁶O, and ²⁰⁸Pb, which are characterized by magic numbers.

However, if a nuclide has neither neutron nor proton numbers that correspond to magic numbers, like mercury-202 and radium-226, it may be less stable and prone to radioactivity.

Nuclear Stability

Nuclear stability is a fundamental aspect of atomic nuclei, influenced significantly by the neutron-to-proton ratio and the presence of magic numbers. A stable nucleus tends to avoid decay and remains unchanged over time, while unstable ones exhibit radioactivity.
Factors affecting nuclear stability include:

• Energy balance: Nuclei with a lower overall energy state are more stable.
• Nuclear forces: The strong nuclear force acts between neighboring nucleons (protons and neutrons) to hold the nucleus together.
• Neutron excess: An increased number of neutrons helps counterbalance the repulsive electrostatic force between positively charged protons.

Instability can occur when the delicate balance of forces is disrupted. In heavier elements, an excess of neutrons is needed to maintain this balance, explaining the higher n/p ratios required for stability. Disruptions in these factors can lead to decay processes like alpha, beta, or gamma decay, through which the nuclide strives to achieve a more stable condition.