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5.2: Terminology and Some Background Physics

Difficulty Level: At Grade Created by: CK-12
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Lesson Objectives

  • Describe kinetic energy.
  • Describe the unit of energy used to describe small charges.
  • List the mathematical representation of the prefixes used to describe small amounts of charge.
  • Describe the implications of Einstein’s most famous equation.
  • Explain how Einstein’s most famous equation is used.

The amount of charge on a particle is described using a unit called a coulomb. When the electron was believed to carry the smallest size charge, \begin{align*}(1.602 \times 10^{-19} \;\mathrm{C})\end{align*}, physicists created a unit of energy to match the electron’s charge. It is called the electron volt—abbreviated \begin{align*}\;\mathrm{eV}.\end{align*} An \begin{align*}\;\mathrm{eV}\end{align*} is equal to \begin{align*}1.602 \times 10^{-19}\end{align*} joules. Instead of saying a particle carries an energy of \begin{align*}1.602 \times 10^{-19}\;\mathrm{J}\end{align*} or \begin{align*}3.204 \times 10^{-19}\;\mathrm{J}\end{align*}, physicists now can say a particle carries an energy of \begin{align*}1 \;\mathrm{eV}\end{align*} or \begin{align*}2 \;\mathrm{eVs}\end{align*} respectively.

One of the nice aspects of the electron volt is that it also relates the energy gained by an accelerating particle to the potential difference it crosses. This is the mechanism that a linear accelerator uses to accelerate a charged particle. One particle with a charge equal to an electron, changes its kinetic energy by \begin{align*}1 \;\mathrm{eV}\end{align*} when it accelerates between two plates connected to a \begin{align*}1\end{align*} volt potential difference, shown in Figure 1.

LHC Stage 1

A particle with a net electric charge equal to one electron gains one of energy after crossing the metal plates. It gains instead of loses the energy because the plate (on the right) opposite the electron is oppositely charged and attracts the negatively charged electron.

Colliders are large machines designed to smash small charged particles such as protons, electrons, and the nucleus of atoms at extreme speeds. The colliders send particles into each other or into a stationary target. These moving particles have kinetic energy, \begin{align*}KE = \frac{1} {2} mv^2,\end{align*} where \begin{align*}KE\end{align*} is the kinetic energy of the particle, \begin{align*}m\end{align*} is the particle’s mass in kilograms, and \begin{align*}v\end{align*} is the particle’s velocity. An object has kinetic energy as long as it has velocity. One of the ways colliders are classified is by the kinetic energy of the collisions. Because the particles are very small with masses typically in the range of \begin{align*}10^{-28}\;\mathrm{kg}\end{align*} the kinetic energies are measured in \begin{align*}eVs\end{align*}. But the collisions are millions or billions of \begin{align*}eVs\end{align*} not just \begin{align*}10\end{align*} or \begin{align*}20\end{align*}. The collision’s energies are listed using the prefixes listed below.

Prefix Pronunciation Number Math Expressions
\begin{align*}MeV\end{align*} Mega \begin{align*}eVs\end{align*} Millions of \begin{align*}eVs\end{align*} \begin{align*}10^6 \;\mathrm{eVs}\end{align*}
\begin{align*}GeV\end{align*} Giga \begin{align*}eV’s\end{align*} Billions of \begin{align*}eVs\end{align*} \begin{align*}10^9 \;\mathrm{eVs}\end{align*}
\begin{align*}TeV\end{align*} Tera \begin{align*}eVs\end{align*} Trillions of \begin{align*}eVs\end{align*} \begin{align*}10^{12} \;\mathrm{eVs}\end{align*}

Einstein showed that a particle at rest has a rest energy given by \begin{align*}E = mc^2,\end{align*} where \begin{align*}m\end{align*} is the mass of the particle measured in kilograms, \begin{align*}c\end{align*} is the speed of light, \begin{align*}3.00 \times 10^8\;\mathrm{m/s}\end{align*}, and \begin{align*}E\end{align*} is the rest energy. The rest energy is measured in the standard S.I. unit of joules. If an object of mass, \begin{align*}m\end{align*}, was annihilated (destroyed), then this formula would describe how much energy would be released. This equation shows that the mass and energy are equivalent: It allows physicists to quantify the mass of an object in terms of energy.

Example:

The mass of a proton is \begin{align*}1.67 \times 10^{-27}\;\mathrm{kg}\end{align*}. What is the energy associated with proton’s mass in units of joules and \begin{align*}eVs\end{align*}?

Solution:

We use \begin{align*}E= mc^{2}\end{align*} with the speed of light \begin{align*}c = 3.00 \times 10^{8} \;\mathrm{m/s}\end{align*}. Then

\begin{align*} E & = (1.67 \times 10^{-27}\;\mathrm{kg}) (3.00 \times 10^8\;\mathrm{m/s})^2 \\ E & = 1.50 \times 10^{-10} \;\mathrm{J} \\ E & = 938,000,000\;\mathrm{eV} \end{align*}

Using the prefixes shown above, this is typically written as \begin{align*}938 \;\mathrm{MeVs}\end{align*}.

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Feb 23, 2012
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