Foot-pounds to Electronvolts Converter

To convert a measurement in foot-pounds to electronvolts, multiply the value by 8.462350e+18. The conversion factor comes from the relationship between both units and the joule.

For example, to convert 5 ft·lb to electronvolts: 5 × 8.462350e+18 = 4.231175e+19 eV.

A foot-pound is an imperial unit of energy equal to the work done when a force of one pound-force acts through a distance of one foot, about 1.356 joules. The foot-pound comes from imperial mechanics and is still used in the US for torque and energy, notably to describe the muzzle energy of firearms and the output of engines.

Common uses of foot-pounds include: Firearm muzzle energy; Torque specifications (as ft·lbf); Mechanical work in imperial units; Engine and tool energy ratings.

An electronvolt is the energy gained by one electron moving through a potential difference of one volt, equal to about 1.602 × 10⁻¹⁹ joules. The electronvolt became essential in atomic and particle physics, where joules are impractically large. Particle accelerator energies are expressed in mega-, giga-, and tera-electronvolts.

Common uses of electronvolts include: Atomic and particle physics; Semiconductor band-gap energies; Photon and X-ray energy; Chemistry ionization energies.

Here are some common values converted:

1 ft·lb = 8.462350e+18 eV 5 ft·lb = 4.231175e+19 eV 10 ft·lb = 8.462350e+19 eV 50 ft·lb = 4.231175e+20 eV 100 ft·lb = 8.462350e+20 eV