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# Atomic and Nuclear Physics for AP Physics B

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By McGraw-Hill Professional
Updated on Feb 12, 2011

Practice problems for these concepts can be found at:

Atomic and Nuclear Physics Practice Problems for AP Physics B

### Subatomic Particles

Atoms, as you know, are the fundamental units that make up matter. An atom is composed of a small, dense core, called the nucleus, along with a larger, diffuse "cloud" of electrons that surrounds the nucleus. The nucleus of an atom is made up of two types of subatomic particles: protons and neutrons. Protons are positively charged particles, neutrons carry no charge, and electrons are negatively charged. For an atom to be electrically neutral, therefore, it must contain an equal number of protons and electrons. When an atom is not electrically neutral, it is called an ion; a positive ion has more protons than electrons, and a negative ion has more electrons than protons.

One particle we didn't introduce in Chapter 20 is the photon. A photon is a particle of light.1 Photons have no mass, but they can transfer energy to or from electrons.

Table 25.1 summarizes the properties of the four subatomic particles. Note that the mass of a proton (and, similarly, the mass of a neutron) equals one atomic mass unit, abbreviated "1 amu." This is a convenient unit of mass to use when discussing atomic physics.

### The Electron-Volt

The electron-volt, abbreviated eV, is a unit of energy that's particularly useful for problems involving subatomic particles. One eV is equal to the amount of energy needed to change the potential of an electron by one volt. For example, imagine an electron nearby a positively charged particle, such that the electron's potential is 4 V. If you were to push the electron away from the positively charged particle until its potential was 5 V, you would need to use 1 eV of energy.

The conversion to joules shows that an eV is an itty-bitty unit of energy. However, such things as electrons and protons are itty-bitty particles. So the electron-volt is actually a perfectly-sized unit to use when talking about the energy of a subatomic particle.

### Momentum of a Photon

Given that photons can "knock" an electron away from a nucleus, it might seem reasonable to talk about the momentum of a photon. Photons don't have mass, of course, but the momentum of a photon can nonetheless be found by this equation:

The momentum of a photon equals Planck's constant divided by the photon's wavelength. You can rearrange this equation a bit to show that the energy of a photon, which equals hc/λ, also equals the photon's momentum times c.

### de Broglie Wavelength

Light acts like a wave, also has the characteristics of particles. As it turns out, things with mass, which we generally treat like particles, also have the characteristics of waves. According to de Broglie, a moving mass behaves like a wave. The wavelength of a moving massive particle, called its de Broglie wavelength, is found by this formula:

So an object's de Broglie wavelength equals Planck's constant divided by the object's momentum. For a massive particle, momentum p = mv. Note that the formula above is valid even for massless photons— it's just a rearrangement of the definition of the photon's momentum. What's interesting here is that the formula can assign a "wavelength" even to massive particles.

Okay, sure, massive particles behave like waves… so how come when you go for a jog, you don't start diffracting all over the place?

Let's figure out your de Broglie wavelength. We'll assume that your mass is approximately 60 kg and that your velocity when jogging is about 5 m/s. Plugging these values into de Broglie's equation, we find that your wavelength when jogging is about 10–36 m, which is way too small to detect.3

Only subatomic particles have de Broglie wavelengths large enough to actually be detected. Moving neutrons can be diffracted, but you can't.

### E = MC2

We can imagine that the nuclei in the last few examples are at rest before emitting alpha, beta, or gamma particles. But when the nuclei decay, the alpha, beta, or gamma particles ome whizzing out at an appreciable fraction of light speed.6 So, the daughter nucleus must recoil after decay in order to conserve momentum.

But now consider energy conservation. Before decay, no kinetic energy existed. After the decay, both the daughter nucleus and the emitted particle have gobs of kinetic energy.7 Where did this energy come from? Amazingly, it comes from mass.

The total mass present before the decay is very slightly greater than the total mass present in both the nucleus and the decay particle after the decay. That slight difference in mass is called the mass defect, often labeled Δm. This mass is destroyed and converted into kinetic energy.

How much kinetic energy? Use Einstein's famous equation to find out. Multiply the mass defect by the speed of light squared:

E = (Δm)c2

And that is how much energy is produced.

Practice problems for these concepts can be found at:

Atomic and Nuclear Physics Practice Problems for AP Physics B

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