Homework Helpers: Physics

10 Nuclear Physics

Perhaps there is something about the model of the atom that has bothering you? In Lesson 10–1 you learned that each proton has a positive charge, and most atoms have more than one proton in their nucleus. However, when we studied Chapter 5, we learned that particles with like charges repel each other. We also learned that the electrostatic force of repulsion between like charges is greatest when the objects are close together, and a particle can”t get much closer to another object than when they occupy the same nucleus. How is it that multiple protons can exist in the nucleus, when they must exert a relatively strong electrostatic force of repulsion on each other?

Based on your knowledge of Newton”s second law, you can assume that the answer must be that there is another strong force involved that holds the nucleus together. This force is called the strong force, or the strong nuclear force.

The fundamental force of nature that exists between nucleons that holds them together against the electrostatic force of repulsion between protons.

The strong force holding the nucleus together is related to the binding energy of the atom. Would it surprise you to learn that if you were to add up all of the individual masses of all of the subatomic particles that make up the atom, they would be greater than the mass of the actual atom? You may have heard it said that something is “greater than the sum of its parts.” An atom is literally less than the sum of its parts, at least if you only focus on mass. The difference between the masses of the individual particles and the total mass of the atom is called the mass defect.

Where did the missing mass go? Does this violate the law of conservation of mass? Einstein”s answer to these questions was E = mC². One of the most famous equations in physics shows us that mass and energy are actually equivalent, and it is the sum of all of the mass and energy in a system that is conserved. Einstein combined the laws of conservation of mass and conservation of energy into the law of conservation of mass-energy.

How does this apply to what we are discussing now? Some of the mass from the individual subatomic particles, the so-called mass defect, is converted into energy, called binding energy, which is used to hold the nucleons together. This is the energy that is released when an atom is split.

To calculate the binding energy of a particular isotope, we must first calculate the mass defect of the atom, and then we calculate the equivalent energy using the conversion factor: 1u = 931 MeV

Example 1

The nucleus of an atom of tritium with an atomic mass of 3.016049 u consists of two neutrons and one proton. Calculate the mass defect (Δm) and the binding energy of this isotope of hydrogen.

Note that the mass of the electron is not taken into account in these calculations.

Given: m_p = 1.007 825 u m_n = 1.008 665 u m_H-3 = 3.016049 u

Find: Δm and E_b

Solution:

Lesson 10–3 Review

1. _______________ is the difference between the masses of the individual particles and the total mass of the atom.

2. The nucleus of an atom of deuterium with an atomic mass of 2.014102 u consists of one neutron and one proton. Calculate the mass defect (Δm) and the binding energy of this isotope of hydrogen.

3. Protons are all positively charged, and like charges repel. What prevents the protons found in the nucleus of all atoms from repelling each other, destroying the atom?

Radioactivity

The strong force doesn”t only exist between proton pairs, it exists between pairs of neutrons and between neutrons and protons. In unstable isotopes, the strong force is unable to compensate for the total electrostatic force of repulsion between protons, and such atoms experience naturalradioactivity. Radioactivity is the spontaneous release of energy, which is sometimes accompanied by particles, from an atom. Several major types of radioactivity have been classified.

Alpha decay (α): Alpha decay involves the release of an alpha particle, which is identical to the nucleus of a helium atom. They consist of two protons and two neutrons, and are often represented with the notation , which indicates a mass of four atomic mass units and an atomic charge of two.

Beta-minus decay (β): There are two forms of beta decay. Beta-minus decay, which is often simply referred to as beta decay, involves the release of an electron from the nucleus of an atom. When an electron is ejected from the nucleus of an atom, one of the neutrons is transformed into a proton, which increases the atomic number of the atom by one and changes its identity.

Beta-plus decay (β⁺) involves the release of a positron from the nucleus. A positron is the antiparticle of the electron, having the same mass but the opposite charge. When a positron is released from the nucleus of an atom, a proton is transformed into a neutron, decreasing the atomic number of the atom by one and changing its identity.

Gamma decay (γ): Gamma decay involves the release of high-energy photons, commonly called gamma rays. Of the three forms of radiation, gamma rays have the most penetrating power by far. Alpha particles can be blocked by clothing or paper. A thin sheet of metal can stop beta particles. Gamma rays are capable of passing through several centimeters of lead.

Nuclear Reactions

There are many different types of nuclear reactions, but they fall into two main categories: natural radioactivity and artificial radioactivity. Natural radioactivity includes all of the nuclear reactions that take place in nature. When you look at the elements on the periodic table with an atomic number of 83 or higher, you should be aware that all of the isotopes of these elements are unstable. This means that, over time, their nuclei undergo natural radioactive decay, transforming them into other elements. You may have heard of a radioactive isotope of carbon called carbon-14. Carbon-14 is naturally radioactive, undergoing beta-decay according to the following reaction:

Notice that carbon (C) is being transformed into nitrogen (N) in this reaction. Also note that both the mass numbers (top numbers) and atomic numbers (bottom numbers) are conserved between each side of the equation. This will be the key to determining the missing isotope or particle in our practice problems.

Artificial radioactivity includes scientific efforts to produce previously undiscovered elements and isotopes by forcing known isotopes to combine or fuse with other particles, as shown in this reaction:

In case you don”t recognize it, the particle on the far right side of the equation is a neutron. You can see that it has a mass of one and an atomic number (charge) of zero. This reaction is an example of nuclear fusion, where two or more smaller nuclei combine to make a larger one.

Nuclear fission is the process in which a heavy nucleus splits into two or more parts. Nuclear reactors release energy using controlled nuclear fission reactions. An example of a fission reaction is shown here:

Notice that there are neutrons on both sides of the equation. The neutron on the left side of the equation is used to split the atom of uranium-235. The three neutrons on the right side of the equation can then, under the proper conditions, go on to split more atoms of uranium-235. This, in turn, will release more neutrons and will set up a process called a chain reaction.

Example 1

What does X represent in the following nuclear equation?

We can tell that this reaction represents aluminum (Al) combining with a particle (X) and then releasing a neutron to form phosphorus (P). How do we determine what the particle (X) is? We must make sure that both the atomic number and mass number are conserved. That is, they show the same total on each side of the equation.

Starting with the mass number, we have 27 on the left side and a total of (30 + 1) 31 on the right side. Therefore, the particle X must have a mass number of four.

As for the atomic number, we have 13 on the left and a total of (15 + 0) 15 on the right. The atomic number for the missing particle must be two, in order to make both sides equal.

What is the identity of our missing particle? Check the periodic table and see which element has a mass of four and an atomic number of two. The answer is helium (He).

Be aware that this is not an example of alpha decay. It is a fusion reaction, because two lighter nuclei were combined to form a heavier one.

Example 2

What does X represent in the following nuclear equation?

Whatever our mystery particle is, it has a mass of zero. We already have a mass of 30 on each side of the equation, and we need to conserve it. So:

We read about two particles that have a mass of zero, a positron (beta-plus) and an electron (beta-minus). The atomic number of the particle represented by X must be equal to one, because we have 15 on the left side of the equation and only 14 on the right side.

The fact that the atomic number of the mystery element is a +1 as opposed to a –1 means that it must be a positron.

So, this reaction is an example of beta-plus decay.

Lesson 10–4 Review

1. ________________ is the process in which two or more smaller nuclei combine to make a larger one.

2. What does X represent in the following nuclear equation?

3. What does X represent in the following nuclear equation?