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In an electron gun the electrons move from the negatively charged plate to the positively charged plate. Their kinetic energy will be equal to the potential energy.
A typical electron gun accelerates electrons using a potential difference between two metal plates. The energy of the electron in electron volts is numerically the same as the voltage between the plates. For example, a 5000 V potential difference produces 5000 eV electrons.

Electron volt

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V . In equation form,

1 eV = 1.60 × 10 –19 C 1 V = 1.60 × 10 –19 C 1 J/C = 1.60 × 10 –19 J.

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V is given 50 eV. A potential difference of 100,000 V (100 kV) will give an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V will be given 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron volt a simple and convenient energy unit in such circumstances.

Connections: energy units

The electron volt (eV) is the most common energy unit for submicroscopic processes. This will be particularly noticeable in the chapters on modern physics. Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. There are, for example, calories for food energy, and kilowatt-hours for electrical energy.

The electron volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as 6000 of these molecules ( 30,000 eV ÷ 5 eV per molecule = 6000 molecules ). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can, thus, produce significant biological damage.

Conservation of energy

The total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, KE + PE = constant size 12{"KE"+"PE=constant"} {} . A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as

KE + PE = constant size 12{"KE"+"PE=constant"} {}

or

KE i + PE i = KE f + PE f , size 12{"KE" rSub { size 8{i} } +"PE" rSub { size 8{i} } "=KE" rSub { size 8{f} } +"PE" rSub { size 8{f} } ,} {}

where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

Electrical potential energy converted to kinetic energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)

Strategy

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be KE i = 0, KE f = ½ mv 2 , PE i = qV , and PE f = 0.

Solution

Conservation of energy states that

KE i + PE i = KE f + PE f .

Entering the forms identified above, we obtain

qV = mv 2 2 . size 12{ ital "qV"= { size 8{1} } wideslash { size 8{2} } ital "mv" rSup { size 8{2} } "." } {}

We solve this for v size 12{v} {} :

v = 2 qV m . size 12{v= sqrt { { {2 ital "qV"} over {m} } } "." } {}

Entering values for q , V , and m size 12{q, V", and "m} {} gives

v = 2 –1.60 × 10 –19 C –100 J/C 9.11 × 10 –31 kg = 5.93 × 10 6 m/s.

Discussion

Note that both the charge and the initial voltage are negative, as in [link] . We know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns.

Practice Key Terms 4

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Source:  OpenStax, Concepts of physics. OpenStax CNX. Aug 25, 2015 Download for free at https://legacy.cnx.org/content/col11738/1.5
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