When an unstable or radioactive nucleus disintegrates spontaneously, certain kinds of particles and/or high-energy photons are released. These particles and photons are collectively called “rays.” Three kinds of rays are produced by naturally occurring radioactivity: , , and . They are named according to the first three letters of the Greek alphabet, alpha (), beta (), and gamma (), to indicate the extent of their ability to penetrate matter. rays are the least penetrating, being blocked by a thin sheet of lead, whereas rays penetrate lead to a much greater distance . rays are the most penetrating and can pass through an appreciable thickness of lead.

The three types of radioactivity that occur naturally can be observed in a relatively simple experiment. A piece of radioactive material is placed at the bottom of a narrow hole in a lead cylinder. The cylinder is located within an evacuated chamber, as Figure 31-7 illustrates.

Figure 31-7    and rays are deflected by a magnetic field and, therefore, consist of moving charged particles. rays are not deflected by a magnetic field and, consequently, must be uncharged.

A magnetic field is directed perpendicular to the plane of the paper, and a photographic plate is positioned to the right of the hole. Three spots appear on the developed plate, which are associated with the radioactivity of the nuclei in the material. Since moving particles are deflected by a magnetic field only when they are electrically charged, this experiment reveals that two types of radioactivity ( and rays, as it turns out) consist of charged particles, whereas the third type ( rays) does not.

When a nucleus disintegrates and produces rays, it is said to undergo . Experimental evidence shows that rays consist of positively charged particles, each one being the nucleus of helium. Thus, an particle has a charge of +2e and a nucleon number of . Since the grouping of 2 protons and 2 neutrons in a nucleus is particularly stable, as we have seen in connection with Figure 31-5, it is not surprising that an particle can be ejected as a unit from a more massive unstable nucleus.

Figure 31-8 shows the disintegration process for one example of decay: The original nucleus is referred to as the parent nucleus (P), and the nucleus remaining after disintegration is called the daughter nucleus (D). Upon emission of an particle, the uranium parent is converted into the daughter, which is an isotope of thorium. The parent and daughter nuclei are different, so decay converts one element into another, a process known as transmutation.

Figure 31-8    decay occurs when an unstable parent nucleus emits an particle and in the process is converted into a different, or daughter, nucleus.

Electric charge is conserved during decay. In Figure 31-8, for instance, 90 of the 92 protons in the uranium nucleus end up in the thorium nucleus, and the remaining 2 protons are carried off by the particle. The total number of 92, however, is the same before and after disintegration. decay also conserves the number of nucleons, because the number is the same before (238) and after (234 + 4) disintegration. Consistent with the conservation of electric charge and nucleon number, the general form for decay is

When a nucleus releases an particle, the nucleus also releases energy. In fact, the energy released by radioactive decay is responsible, in part, for keeping the interior of the earth hot and, in some places, even molten. The following example shows how the conservation of mass/energy can be used to determine the amount of energy released in decay.

Example 4   |   α Decay and the Release of Energy

The atomic mass of uranium is 238.0508 u, that of thorium is 234.0436 u, and that of an particle is 4.0026 u. Determine the energy released when decay converts into . Reasoning Since energy is released during the decay, the combined mass of the daughter nucleus and the particle is less than the mass of the parent nucleus. The difference in mass is equivalent to the energy released. We will determine the difference in mass in atomic mass units and then use the fact that 1 u is equivalent to 931.5 MeV. Solution The decay and the masses are shown below: The decrease in mass, or mass defect for the decay process, is . As usual, the masses are atomic masses and include the mass of the orbital electrons. But this causes no error here because the same total number of electrons is included for , on the one hand, and for plus , on the other. Since 1 u is equivalent to 931.5 MeV, the released energy is .

When decay occurs as in Example 4, the energy released appears as kinetic energy of the recoiling nucleus and the particle, except for a small portion carried away as a ray. Conceptual Example 5 discusses how the nucleus and the particle share in the released energy.

Conceptual Example 5   |

How Energy is Shared During the Decay of In Example 4, the energy released by the decay of is found to be 4.3 MeV. Since this energy is carried away as kinetic energy of the recoiling nucleus and the particle, it follows that . However, and are not equal. Which particle carries away more kinetic energy, the nucleus or the particle? Reasoning and Solution Kinetic energy depends on the mass m and speed v of a particle, since . The nucleus has a much greater mass than the particle, and since the kinetic energy is proportional to the mass, it is tempting to conclude that the nucleus has the greater kinetic energy. This conclusion is not correct, however, since it does not take into account the fact that the nucleus and the particle have different speeds after the decay. In fact, we expect the thorium nucleus to recoil with the smaller speed precisely because it has the greater mass. The decaying is like a father and his young daughter on ice skates, pushing off against one another. The more massive father recoils with much less speed than the daughter. We can use the principle of conservation of linear momentum to verify our expectation. As Section 7.2 discusses, the conservation principle states that the total linear momentum of an isolated system remains constant. An isolated system is one for which the vector sum of the external forces acting on the system is zero, and the decaying nucleus fits this description. It is stationary initially, and since momentum is mass times velocity, its initial momentum is zero. In its final form, the system consists of the nucleus and the particle and has a final total momentum of . According to momentum conservation, the initial and final values of the total momentum of the system must be the same, so that . Solving this equation for the velocity of the thorium nucleus, we find that . Since is much greater than , we can see that the speed of the thorium nucleus is less than the speed of the particle. Moreover, the kinetic energy depends on the square of the speed and only the first power of the mass. As a result of its much greater speed, the particle has the greater kinetic energy. Related Homework: Problem 24

One widely used application of decay is in smoke detectors. Figure 31-9 illustrates how a smoke detector operates. Two small and parallel metal plates are separated by a distance of about one centimeter. A tiny amount of radioactive material at the center of one of the plates emits particles, which collide with air molecules. During the collisions, the air molecules are ionized to form positive and negative ions. The voltage from a battery causes one plate to be positive and the other negative, so that each plate attracts ions of opposite charge. As a result there is a current in the circuit attached to the plates. The presence of smoke particles between the plates reduces the current, since the ions that collide with a smoke particle are usually neutralized. The drop in current that smoke particles cause is used to trigger an alarm.

Figure 31-9   A smoke detector.

The physics of radioactivity and smoke detectors.

The rays in Figure 31-7 are deflected by the magnetic field in a direction opposite to that of the positively charged rays. Consequently, these rays, which are the most common kind, consist of negatively charged particles or particles. Experiment shows that particles are electrons. As an illustration of decay, consider the thorium nucleus, which decays by emitting a particle, as in Figure 31-10: decay, like decay, causes a transmutation of one element into another. In this case, thorium is converted into protactinium . The law of conservation of charge is obeyed, since the net number of positive charges is the same before (90) and after (91 − 1) the emission. The law of conservation of nucleon number is obeyed, since the nucleon number remains at . The general form for decay is The electron emitted in decay does not actually exist within the parent nucleus and is not one of the orbital electrons. Instead, the electron is created when a neutron decays into a proton and an electron; when this occurs, the proton number of the parent nucleus increases from Z to Z + 1 and the nucleon number remains unchanged. The electron is usually fast-moving and escapes from the atom, leaving behind a positively charged atom.

Figure 31-10    decay occurs when a neutron in an unstable parent nucleus decays into a proton and an electron, the electron being emitted as the particle. In the process, the parent nucleus is transformed into the daughter nucleus.

Example 6 illustrates that energy is released during decay, just as it is during decay, and that the conservation of mass/energy applies.

Example 6   |

Decay and the Release of Energy The atomic mass of thorium is 234.043 59 u, and the atomic mass of protactinium is 234.043 30 u. Find the energy released when decay changes into . Reasoning To find the energy released, we follow the usual procedure of determining how much the mass has decreased because of the decay and then calculating the equivalent energy. Solution The decay and the masses are shown below: When the nucleus of a thorium atom is converted into a nucleus, the number of orbital electrons remains the same, so the resulting protactinium atom is missing one orbital electron. However, the given mass includes all 91 electrons of a neutral protactinium atom. In effect, then, the value of 234.043 30 u for already includes the mass of the particle. The mass decrease that accompanies the decay is The equivalent energy is . This is the maximum kinetic energy that the emitted electron can have. Problem solving insight In decay, be careful not to include the mass of the electron twice. As discussed here for the daughter atom , the atomic mass already includes the mass of the emitted electron.

A second kind of decay sometimes occurs.* In this process the particle emitted by the nucleus is a positron rather than an electron. A positron, also called a particle, has the same mass as an electron but carries a charge of +e instead of −e. The disintegration process for decay is The emitted positron does not exist within the nucleus but, rather, is created when a nuclear proton is transformed into a neutron. In the process, the proton number of the parent nucleus decreases from Z to Z − 1, and the nucleon number remains the same. As with decay, the laws of conservation of charge and nucleon number are obeyed, and there is a transmutation of one element into another.

The nucleus, like the orbital electrons, exists only in discrete energy states or levels. When a nucleus changes from an excited energy state (denoted by an asterisk *) to a lower energy state, a photon is emitted. The process is similar to the one discussed in Section 30.3 for the photon emission that leads to the hydrogen atom line spectrum. With nuclear energy levels, however, the photon has a much greater energy and is called a ray. The decay process is written as follows: decay does not cause a transmutation of one element into another. In the next example the wavelength of one particular -ray photon is determined.

  Need more practice?

	Interactive LearningWare31.1 Sodium (atomic mass = 23.99 u) emits a ray that has an energy of 0.423 MeV. Assuming that the nucleus is initially at rest, find the speed with which the nucleus recoils. Ignore relativistic effects. Related Homework: Problem 28

A N A L Y Z I N G    M U L T I P L E - C O N C E P T    P R O B L E M S

Example 7   |   The Wavelength of a Photon Emitted During γ Decay What is the wavelength (in vacuum) of the 0.186-MeV -ray photon emitted by radium ? Reasoning The wavelength of the photon is related to the speed of light and the frequency of the photon. The frequency is not given, but it can be obtained from the 0.186-MeV energy of the photon. The photon is emitted with this energy when the nucleus changes from one energy state to a lower energy state. The energy is the difference between the two nuclear energy levels, in a way very similar to that discussed in Section 30.3 for the energy levels of the electron in the hydrogen atom. In that section, we saw that the energy difference is related to the frequency f and Planck’s constant h, so that we will be able to obtain the frequency from the given energy value. Knowns and Unknowns The following table summarizes the available data: Description Symbol Value Comment  Energy of -ray photon 0.186 MeV Will be converted into joules Unknown Variable        Wavelength of -ray photon ?

Problem solving insight The energy of a-ray photon, like that of photons in other regions of the electromagnetic spectrum (visible, infrared, microwave, etc.), is equal to the product of Planck’s constant h and the frequency f of the photon: .

	Step 1 The Relation of Wavelength to Frequency
	The photon wavelength is related to the photon frequency f and the speed c of light in a vacuum according to Equation 16.1, as shown at the right. We have no value for the frequency, so we turn to Step 2 to evaluate it. (16.1)

Step 2 Photon Frequency and Photon Energy

Section 30.3 discusses the fact that the photon emitted when the electron in a hydrogen atom changes from a higher to a lower energy level has an energy , which is the difference between the energy levels. A similar situation exists here when the nucleus changes from a higher to a lower energy level. The -ray photon that is emitted has an energy given by (Equation 30.4). Solving for the frequency, we obtain which we can substitute into Equation 16.1, as indicated at the right. Solution Combining the results of each step algebraically, we find that The wavelength of the -ray photon is Note that we have converted the value of into joules by using the fact that . Related Homework: Problem 22

Gamma Knife radiosurgery is becoming a very promising medical procedure for treating certain problems of the brain, including benign and cancerous tumors, as well as blood vessel malformations. The procedure, which involves no knife at all, uses powerful, highly focused beams of rays aimed at the tumor or malformation. The rays are emitted by a radioactive cobalt-60 source. As Figure 31-11a illustrates, the patient wears a protective metal helmet that is perforated with many small holes. Part b of the figure shows that the holes focus the rays to a single tiny target within the brain. The target tissue thus receives a very intense dose of radiation and is destroyed, while the surrounding healthy tissue is undamaged. Gamma Knife surgery is a noninvasive, painless, and bloodless procedure that is often performed under local anesthesia. Hospital stays are 70 to 90% shorter than with conventional surgery, and patients often return to work within a few days.

Figure 31-11   (a) In Gamma Knife radiosurgery, a protective metal helmet containing many small holes is placed over the patient’s head. (© Custom Medical Stock Photo) (b) The holes focus the beams of rays to a tiny target within the brain.

The physics of Gamma Knife radiosurgery.

An exercise thallium heart scan is a test that uses radioactive thallium to produce images of the heart muscle. When combined with an exercise test, such as walking on a treadmill, the thallium scan helps identify regions of the heart that do not receive enough blood. The scan is especially useful in diagnosing the presence of blockages in the coronary arteries, which supply oxygen-rich blood to the heart muscle. During the test, a small amount of thallium is injected into a vein while the patient walks on a treadmill. The thallium attaches to the red blood cells and is carried throughout the body. The thallium enters the heart muscle by way of the coronary arteries and collects in heart-muscle cells that come into contact with the blood. The thallium isotope used, , emits rays, which a special camera records. Since the thallium reaches those regions of the heart that have an adequate blood supply, lesser amounts show up in areas where the blood flow has been reduced due to arterial blockages (see Figure 31-12). A second set of images is taken several hours later, while the patient is resting. These images help differentiate between regions of the heart that temporarily do not receive enough blood (the blood flow returns to normal after the exercise) and regions that are permanently damaged due to, for example, a previous heart attack (the blood flow does not return to normal).

Figure 31-12   An exercise thallium heart scan indicates regions of the heart that receive insufficient blood during exercise.

The physics of an exercise thallium heart scan.

The physics of brachytherapy implants. The use of radioactive isotopes to eliver radiation to specific targets in the body is an important medical technique. In treating cancer, for example, the method of delivery should ideally apply a high dose of radiation to a malignant tumor in order to kill it, while applying only a small (non-damaging) dose to healthy surrounding tissue. Brachytherapy implants offer such a delivery method. In this type of treatment radioactive isotopes are formed into small seeds and implanted directly in the tumor according to a predesigned pattern. The energy and type of radiation emitted by the isotopes can be exploited to optimize a treatment design and minimize damage to healthy tissue. Seeds containing iridium are used to treat many cancers, and seeds containing iodine and palladium are used for prostate cancer. Research has also indicated that brachytherapy implants may have an important role to play in the treatment of atherosclerosis, in which blood vessels become blocked with plaque. Such blockages are often treated using the technique of balloon angioplasty. With the aid of a catheter inserted into an occluded coronary artery, a balloon is inflated to open the artery and place a stent (a metallic mesh that provides support for the arterial wall) at the site of the blockage. Sometimes the arterial wall is damaged in this process, and as it heals, the artery often becomes blocked again. Brachytherapy implants (using iridium or phosphorus , for instance) have been found to inhibit repeat blockages following angioplasty.

Check Your Understanding 3

Polonium undergoes decay to produce a daughter nucleus that itself undergoes decay. Which one of the following nuclei is the one that ultimately results: (a) , (b) , (c) , (d) , (e) ? (The answer is given at the end of the book.) Background: During a nuclear disintegration, the electric charge and the nucleon number are conserved, meaning that these quantities remain unchanged when a nucleus disintegrates into nuclear fragments and the accompanying and particles. For similar questions (including calculational counterparts), consult Self-Assessment Test 31.1, which is described at the end of Section 31.5.