Resource Lesson Nuclear Reaction
When we speak of atoms, we are speaking of collections of neutrons, protons, and electrons. Electrically, atoms are naturally neutral - that is, there are as many electrons orbiting the nucleus of an atom as there are protons within its nucleus.

• The atomic number, Z, represents the number of protons in the atom and determines its chemical properties.
• The atomic mass number, A, represents the total number of protons and neutrons, collectively known as nucleons.

The number of neutrons can be calculated by subtracting the atomic number from the atomic mass number.

Different isotopes of an element often have great differences in nuclear stability. A nucleus is considered to be stable if it remains together indefinitely. Stable isotopes for light elements have the same number of protons as neutrons; for heavy nuclei, stability arises when the number of neutrons exceeds the number of protons. This can be understood by noting that as the number of protons increases, the strength of the Coulomb repulsive force increases tending to break the nucleus apart. More neutrons are needed to keep the nucleus stable since they only experience attractive nuclear forces provided by the strong nuclear force. When Z = 82, the repulsive force between protons can no longer be buffered resulting in elements that contain more than 82 protons having no stable nuclei. More information about alpha, beta, and positron emitters can be found in this discussion of the Segre Chart.

Refer to the following information for the next four questions.

Let's examine some isotopes of oxygen.
 How many protons are in one atom of 17 8O?

 How many neutrons are in one atom of 17 8O?

 How many neutrons are in one atom of 18 8O?

 99.759% of commonly occurring oxygen atoms are 16 8O, while 0.037% are 17 8O and 0.204% are 18 8O. If the relative atomic masses for each isotope are 15.994915 amu, 16.999133 amu, and 17.999160 amu, then what is the average atomic mass (in amu) of an oxygen atom?

 Natural Radioactive Decay Modes   The three types of natural radioactive decay that can be emitted by a radioactive substance are: alpha radiation, α, beta radiation, β, and gamma rays, γ.   α represents the nucleus of a helium atom. It carries a superscripted mass number of 4 and a subscripted atomic number of 2. It has a charge of +2.   4 2He = 42α   β represents an electron, with a charge of -1, which has been released from the nucleus when a neutron decays into a proton, an electron and an antineutrino by way of the weak force:     In formulas it can also appear as an e with a superscripted mass number of 0 and a subscripted atomic number of -1   e-1 = 0-1e = 0-1β   γ represents radiation released when the nucleus settles to a lower energy level. It has short wavelengths like x-rays [which are produced when excited electrons fall into vacancies in a ground state orbital of atoms with high atomic numbers], is very energetic, and has no mass.
 Detection Methods   These three particles can be distinguished by their levels of penetration and by passing them through magnetic fields. In terms of penetration, alpha particles can be stopped by a sheet of paper, beta particles can make their way through a few millimeters of aluminum, and gamma particles can penetrate several centimeters of lead shielding.   The directions of curvature of charged particles passing through a magnetic field depends on the right-hand rule. In this rule,   your fingers represent the external magnetic field, your thumb represents a positive charge’s velocity, and your palm represents the magnetic force exerted on the positively charged particle by the field.

Balancing Nuclear Equations

Nuclear equations must be balanced by setting the superscripted mass numbers equal to each other on both sides and by setting the subscripted atomic numbers equal to each other on both sides.

Single particles often found in these reactions are:
• neutrons: 10n have a mass equal to 1.6750 x 10-27 kg
• protons: 11H or 11p have a mass equal to 1.6726 x 10-27 kg
• electrons: 0-1e have a mass equal to 9.11 x 10-31 kg
 What is the value of X in the following equation?

 What is the value of X in the following equation?

Refer to the following information for the next three questions.

 Which reaction(s) represents a fusion reaction?

 Which reaction(s) represents a fission-reaction?

 Which reaction(s) represents a chain-reaction?

Refer to the following information for the next two questions.

28 14Si and 2914Si and 3014Si

Silicon-40, 4014Si, lasts for only a few seconds since it has too many neutrons for its 14 protons. One possible decay product for Silicon-40, 4014Si, is calcium-40, 4020Ca.
 Would this decay path most likely involve emission of alpha particles, beta particles, or gamma particles?

 What is the equation for this decay?

Binding Energy

The formula used to calculate the amount of energy released during the complete transformation of mass into energy is

ΔE = Δmc2

This formula can be used during radioactive decay to determine how much kinetic energy is present in either the behavior of the reactants or the products. Customarily, atomic masses are stated in atomic mass units, or amu, when given during nuclear reactions. Energies are also often given in electronvolts, eV, instead of Joules. However, virtually all the formulas require standard SI units of kg and J instead. Subsequently, here are the conversion factors to change amu to kg and eV to J:

1 amu = 1.6606 x 10-27 kg
1 eV = 1.6 x 10-19 J

Suppose you were asked to determine the energy released from the mass deficit in the reaction shown below.

where the atomic masses for the reactants and products are:

The first step in answering this question would be to determine the relative masses of the reactants and then the products.

 reactants products 7 3Li = 7.01601 4 2He = 4.00260 1 1H = 1.00718 4 2He = 4.00260 8.02319 amu 8.00520 amu

The nuclear mass defect (deficit) is the difference in these two values, or 0.01799 amu. Since this additional mass was present in the reactants, it will be released as energy with the two products. This energy can be calculated with the equation E = Δmc2. Notice that the mass unit, amu, was first converted into kilograms before making this final calculation.

 E = Δmc2 E = 0.01799(1.66 x 10-27)(3 x 108)2 E = 2.59056 x 10-11 J, or E = 16.8 MeV

The following graph shows a summary of when nuclear reactions "release energy." Note that both fusion and fission reactions lower the mass per nucleon until the reaction reaches the element iron which has the greatest binding energy per nucleon of 8.8 MeV.