Module 8

Module 8—Nuclear Decay, Energy, and the Standard Model of the Atom

 

Try This

 

Complete “Practice Problems” 1 to 3 on page 800 of your physics textbook.

 

Beta Decay (β)

 

The beta decay of carbon-14 can be represented with an equation and animation. Open Nuclear Decay Gizmo and change the “type of decay” to “beta decay.”

 

LAB 3. On the simulation, select “show equation.” Press the “play” button and add the missing information to the equation:

 

 

LAB 4. Vary the “starting element” on the simulation and observe beta decay for several other isotopes.

1. What happened to the neutron?
   
   
2. Is the atomic mass conserved on both sides of the equation?
   
   
3. Is the atomic number conserved on both sides of the equation? Where did the new proton come from?
   
   
4. Which particle is always produced in beta decay?

In LAB 3, you observed carbon-14 as the parent element and nitrogen-14 as the daughter element.

 

In beta-negative decay, a neutron converts into a proton, electron, and antineutrino. The proton is retained by the nucleus, keeping the atomic mass constant while increasing the atomic number and, thus, changing the type of element. The emitted electron is called a beta particle to distinguish it from the electrons around the nucleus. According to the conservation laws, beta decay can be represented by the following equation.

Quantity Symbol SI Unit X – parent element X -- Y - daughter element Y -- β - beta particle β - antineutrino --

atomic mass number (A) and the atomic number (Z) are conserved in nuclear decay reactions

 

The antineutrino listed in the preceding equation was not included in the simulation.

 

Read

 

Read about the neutrino on page 804 of the textbook.

 

Example Problem 2. Amercium-241 is produced by a beta decay of plutonium-241.

1. Write the beta decay reaction for plutonium-241.
   
   
   
   

2. What is the parent element and what are the daughter elements?
   
   

   The parent element is plutonium-241. The daughter element is americium-241. (Electrons and antineutrinos are not elements.)

   
   

3. How does this decay equation obey the law of conservation of charge?
   
   

   The law of conservation of charge is obeyed because there are 94 protons before the transmutation and after there are 95 protons in the americium but negative one on the electron [95+ (−1) = 94]. [A neutron (q = 0) changes into a proton (q = 1) and an electron (q = −1).]
   
   

4. How does this decay equation obey the law of conservation of nucleons?
   
   

   The law of conservation of nucleons is obeyed because there are 241 nucleons before the transmutation and 241 nucleons after the transmutation. (The beta particle and the antineutrino are ejected from the americium 241 atom.)

Try This

 

TR 6. Complete “Practice Problem” 1 with Example 16.8 and “Practice Problem” 1 with Example 16.9 on page 803 of your physics textbook. Remember that each reaction also produces a beta particle and an antineutrino.

 

Read

 

Read about antimatter and the positron on pages 804 and 805 of the textbook. Note in Example 16.10 the extra electron that must be taken into account in the mass defect.

 

Self-Check

 

SC 2. How does the weak nuclear force relate to beta decay (both positive and negative decay)?

 

 

In beta-positive decay, a proton converts into a neutron and a positron. The neutron is retained by the nucleus, keeping the atomic mass constant while decreasing the atomic number and thus changing the type of element. The beta particle emitted is called a positron. According to the conservation laws, beta-positive decay can be represented by the following equation. Note: the parent atom has Z electrons to be electrically neutral, when the proton changes into a neutron and positron there is Z-1 electrons in the daughter nucleus. An electron is released to drift away but this is not shown in the equation. This will affect the mass defect equations you will see later.

Quantity Symbol SI Unit X – parent element X -- Y- daughter element Y -- - beta particle (positron) β - neutrino v --

atomic mass number (A) and the atomic number (Z) are conserved in nuclear decay reactions

 

Example Problem 3. Write the beta-positive decay equation for nickel-56.

 

 

The Neutrino and Antineutrino

 

The neutrino and antineutrino are a matter-antimatter pair. The existence of an antineutrino was hypothesized when the kinetic energy of a beta particle following beta decay was lower than expected. It was predicted that some other particle was carrying this energy away before the particle could be detected. This was later proven to be true.

 

Gamma Decay (γ)

 

Often, the alpha and beta decay processes leave the daughter nucleus in an excited state, with the nucleons spread apart. Similar to that of an electron in an energy level, the nucleons will rearrange to form a more stable ground state and releases very high frequency gamma radiation as a result.

 

According to the EMR spectrum, gamma radiation has extremely high energy, which corresponds to a high frequency and a short wavelength. It has no mass or charge, therefore producing no changes in the atomic number or atomic mass of the nucleus. There is no transmutation with the emission of gamma radiation. Gamma rays are represented by the symbol (γ).

 

The nucleus may be left in an excited state after alpha or beta decay. In the nuclear equation, this excited state is represented with an asterisk (*). The nucleus then experiences gamma decay to return to a ground state.

 

Read

 

Read pages 806 and 807 of your physics textbook for an example of a gamma decay chain and equation as well as a radioactive decay series.

 

Decay Series

 

Many of the daughter nuclei produced by alpha and beta decay are still unstable and, as such, will undergo further transmutation. In such cases a decay series is used to illustrate the successive decays until a stable nucleus is produced. Search the Internet for examples of decay series. In most, each dot in this series represents a new nucleus. Both alpha and beta decay form part of each series as the parent material undergoes successive decay in both forms until a stable nucleus is reached.

 

The Direction of Alpha and Beta Decay in the Diagram

 

 

First, focus on why alpha and beta decay are drawn in the direction shown. Recall that alpha decay (the release of a nucleus) involves a decrease in atomic mass number (by four) and a decrease in protons (by two). On the chart, therefore, you need to move down four and left two to get to the new mass number and atomic number. 

 

Notice on page 807 of your physics textbook that the alpha decays are red arrows. Also notice that the horizontal axis increases by one atomic number and the vertical axis increases by four atomic mass numbers per line.

 

Beta decay results in no change to the atomic mass number, so there is no movement up or down on the grap; but it results in an increase in the atomic number, so there is a movement of one unit to the right. Still on page 807 of the textbook, notice that a horizontal blue arrow that is two units long must, therefore, represent two beta decays in succession.

 

Example Problem 4. Using the decay series on page 807 of your textbook, write the nuclear decay equations that represent the transmutation of protactinium-234 to radium-226 (from the radium decay series).

 

Find the dot for protactinium-234, and read off the chemical symbol and atomic number, . Do the same for its daughter element,  (uranium-234 in this case). Set up the nuclear equation; then use the concepts of conservation of charge and nucleons to balance the equation by adding either an alpha particle or beta particle. (You could just add the particle by referring to the decay type shown in the diagram; but you should still check that the equation is balanced.) If the decay is beta, remember to add the antineutrino.

 

Continue to follow the decay chain in this way until all of the equations have been written.

 

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