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# Electric energies

An electric field differs from gravitational field in that it requires an electric potential (in volt or V) over a distance of s. A charge q coulomb accelerated by a voltage V V acquires an energy E J ( = 1 C V). E = V q Note that the amount of energy is proportional to the charge and the voltage.

Here, we will introduce some common units used in nuclear technology. We will discuss the discovery of electron, its charge and mass later, but for the moment, just accept the fact that an electron has a charge of 1.60221x10-19 C. The energy acquired by an electron when accelerated by 1.0 V is 1.0 eV.

1.0 eV = 1.60221x10-19 C * 1.0 V
= 1.60221x10-19 J
1.0 keV = 1000 eV
1.0 MeV = 1,000,000 eV

Example 1

Convert 1.0 J into units of eV or its multiple.

Solution

For one electron to acquire 1.0 J energy requires a very high voltage, but when 1018 electrons are accelerated, the a 6.2414-V potential does the job.

1.0 J / 1.60221x10-19 J eV-1 = 6.2414x1018 eV.

Example 2

Assume that 2.0 F charge (F = 96485 C mol-1) and a voltage of 1.30 V is required to electrolyze 18.0 mL of water. Estimate the amount of energy required to decompose 18.0 mL or 1.0 mol water.

Solution

The amount of energy to decompose each mole of water is estimated as follows:

E = 2.0 * 96485 C mol-1 * 1.30 V
= 250861 J mol-1
= 251 kJ mol-1
Theoretically, a minimum of 1.23 V is required for the reaction, H2O ® H2 + 0.5 O2 but some over-voltage is required to excute the electrolysis. This problem illustrates the conversion of electric energy into chemical energy.
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E-mail: cchieh@uwaterloo.ca