Question

Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are \(^{1}_{1}{H}, 1.00782 \mathrm{u} ; \frac{2}{1} \mathrm{H}, 2.01410 \mathrm{u} ;\) and an electron, \(5.4858 \times\) \(10^{-4}\) u. (Hint: Think carefully about how to account for the electron mass.)$$\mathrm{i} \mathrm{H}+\mathrm{i} \mathrm{H} \longrightarrow_{\mathrm{i}}^{2} \mathrm{H}+_{+\mathrm{i}}^{0}$$

Short answer

The energy released per gram of hydrogen nuclei reacted for the given reaction can be calculated by first finding the mass difference between the initial reactants and final products, then converting the mass difference into energy using Einstein's equation, and finally converting the energy per reaction to energy per gram of hydrogen nuclei reacted. Following these steps, we find that the energy released per gram of hydrogen nuclei reacted is: Energy released per gram = Energy released per reaction × Number of reactions in 1 gram of hydrogen nuclei

Step-by-step solution

Write down the given reaction and identify the initial reactants and final products

The given reaction can be written as: \(^{1}_{1}\mathrm{H} + ^{1}_{1}\mathrm{H} \longrightarrow ^{2}_{1}\mathrm{H} + e^-\) In this reaction: - Initial reactants are: \(^{1}_{1}\mathrm{H}\) and \(^{1}_{1}\mathrm{H}\) - Final products are: \(_{1}^{2}\mathrm{H}\) and \(e^-\)

Find the mass difference between the initial reactants and final products

To calculate the mass difference, we need to subtract the total mass of the final products from the total mass of the initial reactants. Mass of initial reactants = \(2 \times 1.00782 \mathrm{u}\) (since two \(^{1}_{1}\mathrm{H}\) are reacted) Mass of final products = \(1 \times 2.01410 \mathrm{u} + 1 \times 5.4858 \times 10^{-4} \mathrm{u}\) (one \(_{1}^{2}\mathrm{H}\) and one \(e^-\)) Now, let's calculate the mass difference: Mass difference = Mass of initial reactants - Mass of final products Mass difference = \((2 \times 1.00782) - (2.01410 + 5.4858 \times 10^{-4}) \mathrm{u}\)

Convert the mass difference into energy

Now we will use Einstein's expression, \(E = mc^2\), to convert the mass difference into energy. Here, \(c\) is the speed of light, which is approximately \(3 \times 10^8 \mathrm{m.s^{-1}}\). First, let's convert the mass difference from unified atomic mass units (u) to kilograms (kg), using the conversion factor \(1.66054 \times 10^{-27}\mathrm{kg.u^{-1}}\): Mass difference = Mass difference in u × \(1.66054 \times 10^{-27}\mathrm{kg.u^{-1}}\) Now we can calculate the energy using \(E = mc^2\): Energy released = Mass difference × \((3 \times 10^8)^2 \mathrm{m^2.s^{-2}}\)

Calculate the energy released per gram of hydrogen nuclei reacted

We were asked to find the amount of energy released per gram of reacting hydrogen nuclei. Since we have calculated the energy released per reaction, we can now convert it into energy per gram. First, we need to find the mass of 1 gram of hydrogen nuclei in terms of the number of hydrogen nuclei: 1 gram of hydrogen nuclei = \(\frac{1 \mathrm{g}}{1.00782 \mathrm{u} \times 1.66054 \times 10^{-27} \mathrm{kg.u^{-1}}}\) Now, we multiply the energy released per reaction by the number of reactions in 1 gram of hydrogen nuclei: Energy released per gram = Energy released per reaction × Number of reactions in 1 gram of hydrogen nuclei

Key concepts

Mass-Energy Equivalence

The principle of mass-energy equivalence is beautifully expressed through Einstein's famous equation, \( E = mc^2 \). This equation reveals that mass and energy are interchangeable. This means that when a certain amount of mass is converted in a nuclear reaction, it becomes energy. In the context of the problem, the mass difference between the reactants and the products is transformed into energy.

• The 'm' in the equation is the mass difference between the initial reactants and final products, usually very small but significant in nuclear reactions.
• 'c' represents the speed of light, a universal constant, which is approximately \(3 \times 10^8 \mathrm{m.s^{-1}}\). Its square is an enormous number, indicating that even tiny masses can convert to immense energy.

Thus, any tiny mass defect in the reaction among hydrogen nuclei results in a substantial energy release. This principle underpins many nuclear processes, including how stars like the sun produce energy.

Nuclear Reaction

A nuclear reaction involves changes in an atom's nucleus and leads to the transmutation of elements. Unlike chemical reactions, nuclear reactions can release or absorb significant amounts of energy due to transformations within the nucleus. In the given reaction, two hydrogen nuclei (protons) fuse to create a heavier form of hydrogen, known as deuterium, alongside an electron.

• The original reactants in this process are \(^{1}_{1}\mathrm{H}\) and \(^{1}_{1}\mathrm{H}\), which are simply two protons.
• The deuterium atom produced \((^{2}_{1}\mathrm{H})\) contains one proton and one neutron, showing that a new subatomic structure results from the fusion process.
• Electrons are also involved, meaning there isn't just a transformation in nuclear structure, but also a net change in particle mass.

Understanding nuclear reactions is essential for harnessing nuclear energy and for understanding natural processes like how stars evolve and emit energy over billions of years.

Atomic Mass

Atomic mass is fundamental to understanding how nuclear reactions liberate energy and is represented by the symbol 'u', a unit based on one-twelfth the mass of a carbon-12 atom. It helps quantify how kinetic energy converts to mass or vice versa.

• In the exercise, knowing the precise atomic masses of the individual hydrogen isotopes and the electron provides the information to calculate the mass difference, crucial for determining energy release.
• For the reaction, the original masses were \(1.00782 \mathrm{u}\) for \(^{1}_{1}{H}\) and \(2.01410 \mathrm{u}\) for \(^{2}_{1}\mathrm{H}\), with an additional mass for the electron.
• Atomic mass is linked with isotopes, differing only in the number of neutrons within the nucleus, which impacts the mass defect observed in nuclear reactions.

The mass difference, though minuscule, explains the energy release in nuclear processes. This concept not only aids in theoretical physics but is also crucial in practical applications such as nuclear power generation.