How many electron volt in 1 foot pound?
The answer is 8.4623465101609E+18.
We assume you are converting between electronvolt and foot-pound force.
You can view more details on each measurement unit:
electron volt or foot pound
The SI derived unit for energy is the joule.
1 joule is equal to 6.2415064799632E+18 electron volt, or 0.73756214927727 foot pound.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between electronvolts and foot pounds.
Type in your own numbers in the form to convert the units!
You can do the reverse unit conversion from foot pound to electron volt, or enter any two units below:
electron volt to megaelectron volt
electron volt to joule
electron volt to terajoule
electron volt to zettawatthour
electron volt to yottajoule
electron volt to ton
electron volt to exawatt-hour
electron volt to liter atmosphere
electron volt to gallon
electron volt to pferdestärkenstunde
In physics, the electron volt (eV) is a unit of energy. By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt.
In physics, a foot-pound force (symbol ft·lbf or ft·lbf), usually abbreviated to foot-pound, is an Imperial and U.S. customary unit of mechanical work, or energy, although in scientific fields one commonly uses the equivalent metric unit of the joule (J). There are approximately 1.356 J/ft·lbf, or exactly 1.355 817 948 331 400 4 J/ft·lbf.
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