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I came across this question when reading about semiconductors, and found out I didn't really get the concept of electron volt with respect to voltage and electric field applied to an electron.

So the question is:

What is the energy in "electron-volts" of an electron that has risen though a potential of 3V?

What is this question asking about?

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  • \$\begingroup\$ the energy gain is 3eV. What the energy is, we don't know, because you never told us the electron's initial state. \$\endgroup\$ – Brian Drummond Aug 25 '17 at 13:22
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3 electron volts

Some definitions:

Energy, usually measured in joules, can be calculated as E=QV, where Q is charge, and V is voltage.

Voltage is a measure of how much energy each electron has. It is usually measured in Volts, which are joules per coulomb.

Charge is measured in coulombs, defined as the charge on 6.2415×10^18 electrons.

These are good units to work with for practical electronics. But if we're working with single electrons, those factors of 10^18 make the maths a pain. So we look for some alternative units which might be easier. For charge, it's obvious - the single electron charge. Volts we keep. So that gives us a new definition for energy, from E=QV, the electron volt. 1eV is 1.60218×10^-19J.

So the question boils down to: "what is the energy gain of an electron which has moved to a potential 3V higher". It doesn't really matter where that energy came from, in this case an electric field. E = QV = (1 electron charge) × (3 volts) = 3eV.

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  • \$\begingroup\$ Imagine a conctant DC voltage applied across two parallel plates which forms a uniform electric filed E. Now for an electron gaining 3eV energy means here(?): The electron moves parallel in the electric field towards the positive plate such that the distance d taken by the electron: d = 3/E. Is this correct? \$\endgroup\$ – atmnt Aug 25 '17 at 11:01
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It's really definition.

If a charge is moved through a potential difference, then the change in energy associated with that move is the charge times the potential difference, when everything is measured in consistent units.

So if the charge is measured in units of charge on an electron e, and the voltage is measured in volts, then the energy is measured in units of eV, electron volts. We don't have another name for this unit.

If the charge is measured in coulombs, and the voltage in volts, then the energy is measured in Joules. You could also measure the energy in coulomb.volts, but we do have a nice name for this unit, Joules.

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  • \$\begingroup\$ Imagine a conctant DC voltage applied across two parallel plates which forms a uniform electric filed E. Now for an electron gaining 3eV energy means here(?): The electron moves parallel in the electric field towards the positive plate such that the distance d taken by the electron: d = 3/E. Is this correct? \$\endgroup\$ – atmnt Aug 25 '17 at 11:01
  • \$\begingroup\$ @user134429 an electric field is measured in volts/unit_distance, so moving a distance d = 3/E, where d is measured in the same distance units, means moving through a voltage of 3v. So as long as your E is defined in volts/distance, yes, it will gain or lose 3eV, depending on direction. \$\endgroup\$ – Neil_UK Aug 25 '17 at 13:14

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