Problem 42 - Fundamentals of Physics

Question

What is the minimum energy that is required to break a nucleus of \({ }^{12} \mathrm{C}\) (of mass \(11.99671 \mathrm {u}\) ) into three nuclei of "He (of mass 4.00151 u each \() ?\)

Step-by-Step Solution

Answer

7.28 MeV is required.

Step 1: Identify the Initial and Final State

The initial state is the carbon-12 nucleus with a mass of 11.99671 atomic mass units (u). The final state consists of three helium nuclei, each with a mass of 4.00151 u.

Step 2: Calculate Total Mass of the Products

Calculate the total mass of the three helium nuclei. Since each helium nucleus has a mass of 4.00151 u, the total mass of the products is:\[3 \times 4.00151 = 12.00453 \text{ u}.\]

Step 3: Apply Mass-Energy Equivalence

Use the mass-energy equivalence principle, expressed by Einstein's equation, to find the energy difference related to the mass difference:\[\Delta m = m_{final} - m_{initial} = 12.00453 \text{ u} - 11.99671 \text{ u} = 0.00782 \text{ u}.\]

Step 4: Convert Mass Difference to Energy

Convert the mass difference into energy using the conversion factor 1 atomic mass unit (u) = 931.5 MeV. The energy required is given by:\[E = \Delta m \times c^2 = 0.00782 \times 931.5 \text{ MeV} = 7.28 \text{ MeV}.\]

Step 5: Conclusion

The minimum energy required to break a nucleus of \(^{12}\text{C}\) into three \(^4\text{He}\) nuclei is 7.28 MeV.

Key Concepts

Mass-Energy Equivalence

The concept of mass-energy equivalence is fundamental in nuclear physics. It states that mass and energy are interchangeable, and this relationship is expressed by Einstein's famous equation: \[ E = mc^2 \]

• E stands for energy
• m represents mass
• c is the speed of light in a vacuum (~3 x 10⁸ meters per second)

This equation reveals that a small amount of mass can be converted into a significant amount of energy. It plays a crucial role in nuclear reactions, where even tiny differences in mass can result in large energy changes.
The exercise demonstrates this concept by showing how the "missing" mass when a carbon nucleus splits into helium nuclei is converted into energy. So, any change in mass during nuclear reactions leads to a corresponding amount of energy being released or absorbed.

Atomic Mass Unit

In nuclear physics, the atomic mass unit (amu or simply u) is used to express the masses of atoms and their subatomic particles. One atomic mass unit is equal to one twelfth of the mass of a carbon-12 atom. The mass of one atomic mass unit is approximately:\[1 ext{ u } = 1.66053906660 \times 10^{-27} ext{ kg}\]This measurement is crucial because it provides a convenient way to discuss the extremely small masses of atoms and nuclei without resorting to cumbersome scientific notations.
In our problem, we detail the atomic mass of carbon-12 and helium-4 in atomic mass units. Using these units helps simplify the calculation of mass differences between reactants and products in a nuclear reaction. When combined with mass-energy equivalence, these differences can be easily converted into energy values.

Nuclear Reactions

Nuclear reactions involve changes in the nucleus of an atom, resulting in the transformation of elements and the release or absorption of energy. These reactions are distinct from chemical reactions, which only involve electron reconfigurations around atoms. Key aspects of nuclear reactions include:

• Involvement of the atomic nucleus
• Transformation of elements
• Release or absorption of substantial energy

In the given exercise, a nuclear reaction is shown when the carbon-12 nucleus is split into three helium nuclei. The energy required to do this is the amount needed to overcome the nuclear forces holding the particles together.
The calculation of this energy incorporates concepts such as mass defects, the difference in mass between reactants and products, and their conversion into energy. This kind of reaction is fundamental in processes such as nuclear fission and fusion.