Problem 100 - Introductory Chemistry Atoms First

Question

Which has the larger binding energy per mole of nucleons, \({ }_{2}^{4}\) He (molar mass \(4.00150 \mathrm{~g} / \mathrm{mol}\) ) or \({ }_{3}^{6} \mathrm{Li}\) (molar mass \(6.01348 \mathrm{~g} / \mathrm{mol}\) )? [Useful masses: proton, \(1.00730 \mathrm{~g} / \mathrm{mol}\); neutron, \(1.00870 \mathrm{~g} / \mathrm{mol} ;\) electron, \(0.00055 \mathrm{~g} / \mathrm{mol}]\)

Step-by-Step Solution

Answer

The binding energy per mole of nucleons is higher for ${}_{2}^{4}\) He (7.105 MeV/mol of nucleon) compared to ${}_{3}^{6}\mathrm{Li}$ (5.362 MeV/mol of nucleon), making ${}_{2}^{4}\) He more stable as a nucleus.

Step 1: Calculate the number of protons and neutrons for each isotope.

For \({ }_{2}^{4}\) He, Z (Atomic Number) = 2, A (Mass Number) = 4, so the number of protons = 2, and the number of neutrons = 4-2 = 2. For \({ }_{3}^{6}\mathrm{Li}\), Z (Atomic Number) = 3, A (Mass Number) = 6, so the number of protons = 3, and the number of neutrons = 6-3 = 3.

Step 2: Find the total mass of protons and neutrons for each isotope.

Using the given masses of protons and neutrons, Total mass of protons and neutrons in \({ }_{2}^{4}\) He = (2 protons × 1.00730 g/mol) + (2 neutrons × 1.00870 g/mol) = 4.03200 g/mol Total mass of protons and neutrons in \({ }_{3}^{6}\mathrm{Li}\) = (3 protons × 1.00730 g/mol) + (3 neutrons × 1.00870 g/mol) = 6.04800 g/mol

Step 3: Calculate the mass defect for each isotope.

Mass defect (\({ \Delta}m\)) is the difference between the total mass of the protons and neutrons and the actual mass of the isotope. For \({ }_{2}^{4}\) He, \({ \Delta}m\) = total mass of protons and neutrons - actual mass = 4.03200 g/mol - 4.00150 g/mol = 0.03050 g/mol For \({ }_{3}^{6}\mathrm{Li}\), \({ \Delta}m\) = total mass of protons and neutrons - actual mass = 6.04800 g/mol - 6.01348 g/mol = 0.03452 g/mol

Step 4: Use the mass defect to find the total binding energy for each isotope.

Total binding energy (BE) can be found using the famous Einstein equation, BE= \({ \Delta}m\) × \(c^2\). In this case, since we are working with g/mol and not the actual mass, we'll use the conversion factor 931.49 MeV/g to convert the mass defect into MeV. For \({ }_{2}^{4}\) He, BE = 0.03050 g/mol × 931.49 MeV/g = 28.42 MeV/mol For \({ }_{3}^{6}\mathrm{Li}\), BE = 0.03452 g/mol × 931.49 MeV/g = 32.17 MeV/mol

Step 5: Find the binding energy per mole of nucleons for each isotope, and compare them.

Binding energy per mole of nucleons = Total binding energy / Number of nucleons For \({ }_{2}^{4}\) He, binding energy per mole of nucleons = 28.42 MeV/mol / 4 = 7.105 MeV/mol of nucleon For \({ }_{3}^{6}\mathrm{Li}\), binding energy per mole of nucleons = 32.17 MeV/mol / 6 = 5.362 MeV/mol of nucleon Since 7.105 MeV/mol of nucleon is higher than 5.362 MeV/mol of nucleon, \({ }_{2}^{4}\) He has a higher binding energy per mole of nucleons when compared to \({ }_{3}^{6}\mathrm{Li}\). Therefore, \({ }_{2}^{4}\) He is more stable as a nucleus.

Key Concepts

Mass Defect

In nuclear physics, the concept of mass defect is crucial for understanding binding energy and nuclear reactions. The mass defect refers to the difference between the sum of the individual masses of protons and neutrons in a nucleus and the actual mass of the nucleus. Mass defect occurs because when protons and neutrons are bound together in a nucleus, some mass is converted into energy to hold the nucleus together, according to Einstein's famous equation, \(E = mc^2\).

• This "missing mass" is known as the mass defect.
• It is this energy that makes the nucleus stable, and is released during nuclear reactions.

The larger the mass defect, the more energy is needed to disassemble the nucleus, indicating a stronger and more stable nucleus. When calculating mass defects, scientists use the molar masses of the nucleons to determine the theoretical mass and compare it with the actual atomic mass.
In our original problem, \(^{4}_{2}\text{He}\) displayed a mass defect of 0.03050 g/mol, while \(^{6}_{3}\text{Li}\) had a mass defect of 0.03452 g/mol.These differences in mass allow us to calculate the binding energy of each nucleus, giving insight into their stability and nuclear interactions. Understanding mass defect helps researchers explore phenomena such as nuclear fission and fusion.

Nuclear Stability

Nuclear stability is a measure of how tightly nucleons, namely protons and neutrons, are bound within a nucleus. This stability is fundamentally related to the binding energy per nucleon, an indicator of a nucleus's resistance to being broken apart.
If a nucleus has a higher binding energy per nucleon, it generally indicates greater nuclear stability. In the exercise involving \(^{4}_{2}\text{He}\) and \(^{6}_{3}\text{Li}\), the helium isotope had a binding energy of 7.105 MeV per nucleon, compared to lithium's 5.362 MeV per nucleon.

• A higher binding energy per nucleon means more energy is required to separate the nucleons, leading to a more stable configuration.
• Generally, nuclei with intermediate mass numbers, like helium, tend to possess higher nuclear stability, while both very light and very heavy nuclei are less stable and more prone to nuclear reactions.

Understanding nuclear stability is important because it informs us about the likelihood of a nucleus undergoing radioactive decay or participating in nuclear reactions like fission and fusion. Stable nuclei are found at the ends of the spectrum, whereas unstable ones are likely to transform or decay over time.

Isotopes

Isotopes are different forms of the same element, distinguished by the number of neutrons in the nucleus. While they have the same number of protons, the added neutrons lead to variations in their mass numbers and, consequently, their mass.
In our discussed exercise, \(^{4}_{2}\text{He}\) and \(^{6}_{3}\text{Li}\) are isotopes of helium and lithium respectively.

• Helium-4, with two protons and two neutrons, and lithium-6, with three protons and three neutrons, display differences in mass and nuclear properties.
• Isotopes can exhibit different levels of nuclear stability, particularly when the balance of protons to neutrons shifts.

Isotopes are crucial in fields ranging from chemistry to medicine, astrophysics, and archaeology.
For instance, stable isotopes contribute to understanding atomic structures and bonding, whereas radioactive isotopes help in dating geological samples and in medical imaging techniques.Knowing about isotopes helps us understand how elements behave differently in various nuclear conditions and why certain isotopes are more prevalent in nature. This knowledge expands our comprehension of both the practical applications and theoretical principles of atomic structures.