Question

Consider the following fission reaction. $$_{0}^{1} n+_{5}^{10} \mathrm{B} \rightarrow_{3}^{7} \mathrm{Li}+_{2}^{4} \mathrm{He}$$ $$1.0087 \quad 10.0129 \quad 7.0160 \quad 4.0026$$ The masses of the species involved are given in atomic mass units below each species, and 1 amu can create 932 MeV of energy. What is the energy liberated due to transformation of mass into energy during this reaction? A. \(0.003 \mathrm{MeV}\) B. \(1.4 \mathrm{MeV}\) C. \(2.8 \mathrm{MeV}\) D. \(5.6 \mathrm{MeV}\)

Short answer

C. 2.8 MeV

Step-by-step solution

Calculate the total mass of reactants

The reactants are and _{5}^{10} \text{B}. Using the atomic masses provided, sum the masses of the reactants: \[1.0087 + 10.0129 = 11.0216 \text{ amu} \]

Calculate the total mass of products

The products are _{3}^{7} \text{Li} and _{2}^{4} \text{He}. Using the atomic masses provided, sum the masses of the products: \[ 7.0160 + 4.0026 = 11.0186 \text{ amu} \]

Find the mass defect

The mass defect is the difference between the mass of reactants and the mass of products: \[ 11.0216 - 11.0186 = 0.003 \text{ amu} \]

Convert mass defect to energy

Convert the mass defect into energy using the formula \( E = \text{Mass Defect} \times 932 \text{ MeV/amu} \): \[ 0.003 \times 932 = 2.796 \text{ MeV} \]

Approximate to fit the given options

Round 2.796 MeV to 2.8 MeV to match the choices given.

Key concepts

mass-energy equivalence

In nuclear reactions, the mass-energy equivalence principle is crucial. It tells us that mass can be converted into energy and vice versa. This principle is explained by Einstein's famous equation, \( E = mc^2 \). This means that even a tiny amount of mass can be converted into a huge amount of energy because the speed of light, \( c \), is a very large number. This equation is key to understanding why nuclear reactions, like the one in the exercise, release so much energy. The mass lost during the reaction doesn't just disappear; it transforms into energy.

atomic mass unit (amu)

An atomic mass unit (amu) is a measurement unit for atomic and molecular weights. One atomic mass unit is defined as one twelfth of the mass of a carbon-12 atom. This unit is extremely useful in chemistry and physics for calculating the mass of atoms and molecules. In nuclear reactions, we often use amu to measure the mass of the participating species. For example, the neutron (_{0}^{1} n) in the given exercise has a mass of 1.0087 amu. Knowing these precise masses allows us to calculate the mass defect and energy released during nuclear reactions.

mass defect

The mass defect in a nuclear reaction is the difference between the mass of the reactants and the products. It represents the mass that has been converted into energy during the reaction. To find the mass defect, we subtract the total mass of the products from the total mass of the reactants. In the exercise, the mass defect calculation is as follows:
Reactants: 1.0087 amu (neutron) + 10.0129 amu (Boron-10) = 11.0216 amu
Products: 7.0160 amu (Lithium-7) + 4.0026 amu (Helium-4) = 11.0186 amu
Mass defect = 11.0216 amu - 11.0186 amu = 0.003 amu.
This tiny mass loss contributes to a significant amount of energy.

energy conversion in nuclear reactions

Converting the mass defect into energy is an essential step in understanding nuclear reactions. Using the conversion factor, 1 amu = 932 MeV, we can determine how much energy a nuclear reaction releases. In the exercise, the mass defect is 0.003 amu. By multiplying this defect by 932 MeV/amu, we get:
\(0.003 \times 932 = 2.796 \text{ MeV}\)
For practical purposes and to match the provided options, this value is approximated to 2.8 MeV. This energy is what liberates during the reaction and illustrates the power of nuclear fission. It's incredible how a small mass can turn into such a large amount of energy, showcasing nature's efficiency in energy conversion.