Problem 83 - Chemistry The Science in Context

Question

How much energy is released per nucleus of tritium produced during the following reactions? a. \(_{0}^{1} n+\frac{6}{3} L i \rightarrow \frac{4}{2} H e+\frac{3}{1} H\) b. \(_{0}^{1} n+_{3}^{7} L_{1} \rightarrow_{2}^{4} \mathrm{He}+_{1}^{3} \mathrm{H}+_{0}^{1} \mathrm{n}\)

Step-by-Step Solution

Answer

Answer: The energy released per nucleus of tritium produced during the two nuclear reactions is: a. 5.82 MeV b. 9.32 MeV

Step 1: Determine the masses of the reactants and products

We will need to find the atomic masses of the particles involved in the reaction: - mass of neutron (m_n) = 1.008665 u - mass of lithium-6 (m_Li-6) = 6.015122 u - mass of helium-4 (m_He-4) = 4.001506 u - mass of tritium (m_H-3) = 3.016049 u

Step 2: Calculate the mass difference (∆m) of the reaction

The mass difference can be found as the difference in mass between the reactants and products: ∆m = (m_n + m_Li-6) - (m_He-4 + m_H-3)= (1.008665 + 6.015122) - (4.001506 + 3.016049) = 0.006232 u

Step 3: Convert the mass difference to energy

To find the energy released, we can use Einstein's mass-energy equivalence formula, E = ∆m * c^2, where c = 299,792,458 m/s (speed of light). First, we will need to convert ∆m from atomic mass units (u) to kilograms (kg). 1 u = 1.66053906660 × 10^-27 kg ∆m (kg) = 0.006232 u * 1.66053906660 × 10^-27 kg/u = 1.03453070299 × 10^-29 kg Now, we can find E (in joules): E = (1.03453070299 × 10^-29 kg) * (299,792,458 m/s)^2 = 9.318960283 × 10^-13 J Finally, we will convert the energy to electronvolt (eV) using the conversion factor: 1 eV = 1.602176634 × 10^-19 J E = (9.318960283 × 10^-13 J) / (1.602176634 × 10^-19 J/eV) ≈ 5.817485 × 10^6 eV = 5.82 MeV b. \(_{0}^{1} n+_{3}^{7} L_{1} \rightarrow_{2}^{4} \mathrm{He}+_{1}^{3} \mathrm{H}+_{0}^{1} \mathrm{n}\)

Step 4: Determine the masses of the reactants and products for reaction b

We will need to find the atomic masses of the particles involved in the reaction: - mass of lithium-7 (m_Li-7) = 7.016004 u

Step 5: Calculate the mass difference (∆m) of the reaction b

The mass difference for reaction b can be found as the difference in mass between the reactants and products: ∆m = (m_n + m_Li-7) - (m_He-4 + m_H-3 + m_n)= (1.008665 + 7.016004) - (4.001506 + 3.016049 + 1.008665) = 0.998449 u

Step 6: Convert the mass difference to energy for reaction b

First, convert ∆m to kg: ∆m (kg) = 0.998449 u * 1.66053906660 × 10^-27 kg/u = 1.65859185795 × 10^-27 kg Now, we can find E (in joules): E = (1.65859185795 × 10^-27 kg) * (299,792,458 m/s)^2 = 1.492284009 × 10^-12 J Finally, we will convert the energy to electronvolt (eV) using the conversion factor: E = (1.492284009 × 10^-12 J) / (1.602176634 × 10^-19 J/eV) ≈ 9.321759 × 10^6 eV = 9.32 MeV So, the energy released per nucleus of tritium produced during the two reactions is: a. 5.82 MeV b. 9.32 MeV

Key Concepts

Mass-Energy Equivalence

One of the most fascinating insights of modern physics is the concept of mass-energy equivalence, introduced by Albert Einstein's famous equation: \[ E = mc^2 \] This equation tells us that mass can be converted into energy, and vice versa. The symbol \( E \) stands for energy, \( m \) is the mass, and \( c \) is the speed of light, which is a constant value \( c = 299,792,458 \ m/s \). This principle shows that mass and energy are two forms of the same thing, interconvertible and incredibly powerful. Understanding this relation helps explain how nuclear reactions release so much energy. When a nuclear reaction occurs, a small amount of mass is lost and converted into a large amount of energy because of the multiplication with \( c^2 \). In simpler terms, even tiny amounts of mass can produce tremendous energy, which is why nuclear reactions are so energetic compared to chemical reactions. This is also how energies in nuclear reactions are calculated, which you'll find is essential in understanding nuclear chemistry.

Atomic Mass Units

To measure atomic-scale masses, scientists use atomic mass units (AMU), which conveniently reflect the mass of atoms and subatomic particles. The atomic mass unit is defined as one twelfth of the mass of a carbon-12 atom, the most common isotope of carbon. This unit simplifies calculations involving the very tiny masses of atoms and atomic particles. For example, in nuclear chemistry, we deal with particles like neutrons, protons, and other isotopes, each with their specific atomic masses listed in AMUs.

• A neutron has a mass of 1.008665 u.
• Lithium-6 has a mass of 6.015122 u.
• Helium-4 is 4.001506 u.
• Tritium, or hydrogen-3, weighs in at 3.016049 u.
• Lithium-7 is slightly heavier at 7.016004 u.

Using these atomic masses, we can find mass differences in a nuclear reaction to calculate how much energy it releases.

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus and result in the transformation of elements and the release or absorption of energy. They are different from chemical reactions, which only involve interactions between electrons in the shells surrounding an atom's nucleus. In the given reactions:

• The first reaction involves a neutron and lithium-6, producing helium-4 and tritium (hydrogen-3).
• The second reaction involves a neutron and lithium-7, forming helium-4, tritium, and an additional neutron.

These reactions not only transform the elements involved but also release significant energy due to the change in their atomic nuclei. This is possible because of the small mass differences between reactants and products, which, though minuscule, result in significant energy output when converted using the mass-energy equivalence formula. Understanding nuclear reactions helps us harness energy for various applications, such as in nuclear power plants or in understanding stellar processes.

Energy Conversion

Energy conversion is crucial in understanding and calculating the energy released in nuclear reactions. Once the mass difference \( \Delta m \) from the reactants and products is determined in atomic mass units, we convert it into energy using Einstein's \( E = mc^2 \) equation. Because atomic mass units are more intuitive for these calculations, we must first convert \( \Delta m \) into kilograms: \[ 1 \ u = 1.66053906660 \times 10^{-27} \ kg \] Then, we use this mass to calculate energy in joules, multiplying by the speed of light squared. This energy is often converted to electronvolts (eV) for convenience, using \[ 1 \ eV = 1.602176634 \times 10^{-19} \ J \]. This entire process helps to better illustrate how mass differences lead to energy release:

• Reaction \( a \) releases \( 5.82 \rm{MeV} \) of energy per nucleus of tritium produced.
• Reaction \( b \) releases \( 9.32 \rm{MeV} \) per tritium nucleus.

Understanding this conversion process is vital for evaluating how nuclear reactions can be utilized for energy production and various scientific applications.