Difference between revisions of "Ionization energy"
(Created page with 'Ionization energy is the amount of energy required to "pluck" an electron from its atom. This value is 13.6 eV for the hydrogen atom. One can calculate the ionization energy of ...') |
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One can calculate the ionization energy of an atom by employing Newton's famous equation: | One can calculate the ionization energy of an atom by employing Newton's famous equation: | ||
| − | <math>F=\frac{1}{4πepsilon_0}\frac{qe}{r^{2}}=ma=\frac{mv^{2}}{r}</math> | + | <math>F = \frac{1}{4πepsilon_0} \frac{qe}{r^{2}} = ma = \frac{mv^{2}}{r}</math> |
Through algebraic manipulation, | Through algebraic manipulation, | ||
| − | <math>\frac{1}{8πepsilon_0}\frac{qe}{r}=\frac{1}{2}mv^{2}=KE</math> | + | <math>\frac{1}{8πepsilon_0} \frac{qe}{r} = \frac{1}{2}mv^{2} = KE</math> |
To find the [[potential energy]], | To find the [[potential energy]], | ||
| − | <math>PE=\int \bold{F} \cdot \mathrm{d}\bold{s}=-\frac{1}{4πepsilon_0}\frac{qe}{r}</math> | + | <math>PE = \int \bold{F} \cdot \mathrm{d}\bold{s} = -\frac{1}{4πepsilon_0}\frac{qe}{r}</math> |
Then, to find the ionization energy, | Then, to find the ionization energy, | ||
| − | <math>E=PE+KE=-\frac{1}{4πepsilon_0}\frac{qe}{r}+\frac{1}{8πepsilon_0}\frac{qe}{r}=-\frac{1}{8πepsilon_0}\frac{qe}{r}</math> | + | <math>E = PE + KE = -\frac{1}{4πepsilon_0} \frac{qe}{r} + \frac{1}{8πepsilon_0}\frac{qe}{r} = -\frac{1}{8πepsilon_0}\frac{qe}{r}</math> |
The last term can be calculated by substituting numerical values for the atomic radius '''r''', charge of the atomic nucleus '''q''', and appropriate constants. | The last term can be calculated by substituting numerical values for the atomic radius '''r''', charge of the atomic nucleus '''q''', and appropriate constants. | ||
| − | + | The ionization energy is usually measured with the [[electron volt]] (eV). | |
Revision as of 22:22, May 9, 2009
Ionization energy is the amount of energy required to "pluck" an electron from its atom. This value is 13.6 eV for the hydrogen atom.
One can calculate the ionization energy of an atom by employing Newton's famous equation: <math>F = \frac{1}{4πepsilon_0} \frac{qe}{r^{2}} = ma = \frac{mv^{2}}{r}</math>
Through algebraic manipulation, <math>\frac{1}{8πepsilon_0} \frac{qe}{r} = \frac{1}{2}mv^{2} = KE</math>
To find the potential energy, <math>PE = \int \bold{F} \cdot \mathrm{d}\bold{s} = -\frac{1}{4πepsilon_0}\frac{qe}{r}</math>
Then, to find the ionization energy, <math>E = PE + KE = -\frac{1}{4πepsilon_0} \frac{qe}{r} + \frac{1}{8πepsilon_0}\frac{qe}{r} = -\frac{1}{8πepsilon_0}\frac{qe}{r}</math>
The last term can be calculated by substituting numerical values for the atomic radius r, charge of the atomic nucleus q, and appropriate constants.
The ionization energy is usually measured with the electron volt (eV).