PhysicsLAB: 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

Linked here are three EXCELLENT video lessons by Khan Academy on these decay modes: alpha decay, beta decay, and gamma decay. Please take time to view them!

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. Alpha decay usually occurs when an unstable nucleus has too many protons. It has a very low penetrability; in fact, it can be stopped by a single sheet of paper.

⁴ ₂He = ⁴₂α

β represents an electron, with a charge of -1 and a mass of zero which is released from an unstable nucleus that has a high ratio of protons to neutrons. In beta-minus decay a neutron decays into a proton by way of the weak nuclear force producing a beta-minus particle and an antielectron neutrino.

This decay mode has the following equation:

⁰₀n → ⁰_-1e + ¹₁p + ν_e

In formulas beta particles usually appear as an e with a superscripted mass number of 0 and a subscripted atomic number of -1: ⁰_-1e, or ⁰_-1β Beta particles have a greater penetrability than that of alpha particle requiring several sheets of paper or a sheet of aluminum to be stopped.

A second type of beta decay is called a beta-plus decay. In this decay mode, a proton is converted into a neutron, a positively charged electon (called a positron), and an electron neutrino. The equation is

¹₁p → ⁰₀n + ⁰₊₁e + ν_e.

γ represents radiation released when an excited nucleus settles to a lower energy level. Gamma rays are very energetic and have 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]. Gamma rays are electromagnetic in nature and have no mass. They are usually stopped by using a thick piece of lead.

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.

There are three stable isotopes of silicon:

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?

Nuclear 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.