Problem 109 - Chemistry: An Atoms-Focused Approach

Question

In Section 21.6 we state that "no energy would be released if two \(^{4} \mathrm{He}\) nuclei were to fuse together to form \(^{8} \mathrm{Be}\). Similarly, \(^{8} \mathrm{Be}\) nuclei require no energy to spontaneously decompose into \(^{4}\) He nuclei, so they would immediately do so." Verify this statement by calculating the binding energy of \(^{8} \mathrm{Be}\) and comparing it to that of \(^{4} \mathrm{He}\).

Step-by-Step Solution

Answer

Based on the calculations, we have verified that the binding energy of two \(^{4}\mathrm{He}\) nuclei is almost equal to the binding energy of one \(^{8}\mathrm{Be}\) nucleus. Therefore, no energy would be released if two \(^{4}\mathrm{He}\) nuclei were to fuse together to form \(^{8}\mathrm{Be}\), and \(^{8}\mathrm{Be}\) would require no energy to decompose into two \(^{4}\mathrm{He}\) nuclei.

Step 1: Recall the binding energy formula

The binding energy (\(B\)) of a nucleus can be calculated using the formula: \(B = \Delta m c^2\) Where \(\Delta m\) is the mass defect of the nucleus and \(c\) is the speed of light.

Step 2: Calculate the binding energy of \(^{4}\mathrm{He}\)

We know that the binding energy per nucleon for \(^{4}\mathrm{He}\) is approximately 7.073 MeV. We can calculate the total binding energy of \(^{4}\mathrm{He}\) nucleus by multiplying the binding energy per nucleon with the total number of nucleons: \(B_{^{4}\mathrm{He}} = 4 \times 7.073 \ \text{MeV}\) \(B_{^{4}\mathrm{He}} = 28.292 \ \text{MeV}\) The binding energy of one \(^{4}\mathrm{He}\) nucleus is 28.292 MeV.

Step 3: Calculate the binding energy of \(^{8}\mathrm{Be}\)

We know that the binding energy per nucleon for \(^{8}\mathrm{Be}\) is approximately 7.062 MeV. We can calculate the total binding energy of \(^{8}\mathrm{Be}\) nucleus by multiplying the binding energy per nucleon with the total number of nucleons: \(B_{^{8}\mathrm{Be}} = 8 \times 7.062 \ \text{MeV}\) \(B_{^{8}\mathrm{Be}} = 56.496 \ \text{MeV}\) The binding energy of one \(^{8}\mathrm{Be}\) nucleus is 56.496 MeV.

Step 4: Compare the binding energies

Now, let's compare the binding energies of two \(^{4}\mathrm{He}\) nuclei and one \(^{8}\mathrm{Be}\) nucleus: \(2 \times B_{^{4}\mathrm{He}} = 2 \times 28.292 \ \text{MeV} = 56.584 \ \text{MeV}\) We can see that the binding energy of two \(^{4}\mathrm{He}\) nuclei (56.584 MeV) is very close to the binding energy of one \(^{8}\mathrm{Be}\) nucleus (56.496 MeV).

Step 5: Verify the statement

Since the binding energy of two \(^{4}\mathrm{He}\) nuclei is almost equal to the binding energy of one \(^{8}\mathrm{Be}\) nucleus, we can conclude that no energy would be released if two \(^{4}\mathrm{He}\) nuclei were to fuse together to form \(^{8}\mathrm{Be}\). Similarly, the \(^{8}\mathrm{Be}\) nucleus requires no energy to spontaneously decompose into \(^{4}\mathrm{He}\) nuclei, so it would immediately do so. This confirms the statement given in the exercise.

Key Concepts

Nuclear Fusion

Nuclear fusion is an important process in which two lighter atomic nuclei come together to form a heavier nucleus, releasing a large amount of energy. It occurs naturally in stars, including our sun, where hydrogen nuclei fuse to form helium, powering the star's light and heat. Another example is when two helium-4 nuclei fuse to form beryllium-8. In the case of helium-4 fusion, however, the energy dynamics are such that forming beryllium-8 doesn’t release any energy.
Usually, during nuclear fusion:

• New elements are formed.
• Energy is released in tremendous quantities.
• It requires a significant amount of initial energy to overcome the electrostatic forces between the positively charged nuclei.

Understanding fusion is crucial as it holds potential for being a cleaner, more powerful source of energy compared to current methods like fossil fuels or even nuclear fission.

Mass Defect

The mass defect is the difference between the mass of an atom and the sum of the masses of its protons, neutrons, and electrons. This concept is foundational in understanding why energy is released in nuclear reactions, like fusion and fission.
Here's why mass defect occurs:

• The binding energy required to hold the nucleus together leads to a loss in mass, as some of the original mass is converted to energy.
• This binding energy equates to the mass defect according to Einstein's equation: \( E = \Delta m c^2 \) where \( \Delta m \) is the mass defect and \( c \) is the speed of light.

When looking at helium-4 and beryllium-8 fusion, the mass defect comes into play in calculating the binding energy, which shows why fusing helium-4 to form beryllium-8 does not result in a net energy release.

Helium-4

Helium-4 is a stable isotope of helium, consisting of two protons and two neutrons. It is a product of stellar nucleosynthesis and plays a pivotal role in both nuclear fusion reactions and helium-4 fusion scenarios. In nuclear fusion:

• Helium-4 is often used to illustrate nuclear reactions because of its stability and the high binding energy per nucleon.
• Its fusion into other elements usually results in significant energy releases.

Present in the core of stars, helium-4 nuclei contribute to nuclear processes that sustain stars for billions of years. However, when two helium-4 nuclei attempt to combine into beryllium-8, the resulting compound is unstable and quickly reverts, indicating energy is not favorably released.

Beryllium-8

Beryllium-8 is an isotope composed of 4 protons and 4 neutrons. Interestingly, it is a very unstable nucleus that forms transiently during specific fusion processes in stars. Here's what happens in fusion involving beryllium-8:

• It is formed when two helium-4 nuclei combine.
• Due to its instability, it immediately decomposes back into two helium-4 nuclei, releasing no significant energy in the process.
• This rapid decay illustrates why beryllium-8 formation doesn't contribute to net energy output in fusion reactions.

Understanding beryllium-8's instability helps in explaining why its formation is not favored in terms of energy release, unlike other fusion reactions.