Difference between revisions of "Ionization energy"
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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. | 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:<br> | + | One can calculate the ionization energy of an atom by employing Newton's famous equation:<br><br> |
<math>F = \frac{q e}{4 \pi \epsilon\ r^2} = ma = \frac{m v^2}{r}</math> | <math>F = \frac{q e}{4 \pi \epsilon\ r^2} = ma = \frac{m v^2}{r}</math> | ||
| − | Through algebraic manipulation,<br> | + | Through algebraic manipulation,<br><br> |
<math>\frac{q e}{8 \pi \epsilon\ r} = \frac{1}{2} mv^{2} = KE</math> | <math>\frac{q e}{8 \pi \epsilon\ r} = \frac{1}{2} mv^{2} = KE</math> | ||
| − | To find the [[potential energy]],<br> | + | To find the [[potential energy]],<br><br> |
<math>PE = \int \bold{F} \cdot \mathrm{d}\bold{s} = - \frac{q e}{4 \pi \epsilon\ r}</math> | <math>PE = \int \bold{F} \cdot \mathrm{d}\bold{s} = - \frac{q e}{4 \pi \epsilon\ r}</math> | ||
| − | Then, to find the ionization energy,<br> | + | Then, to find the ionization energy,<br><br> |
<math>E = PE + KE = - \frac{q e}{4 \pi \epsilon\ r} + \frac{q e}{8 \pi \epsilon\ r} = - \frac{q e}{8 \pi \epsilon\ r}</math> | <math>E = PE + KE = - \frac{q e}{4 \pi \epsilon\ r} + \frac{q e}{8 \pi \epsilon\ r} = - \frac{q e}{8 \pi \epsilon\ r}</math> | ||
| − | The last term can be calculated by substituting numerical values for the atomic radius | + | The last term can be calculated by substituting numerical values for the atomic radius <math>r</math>, charge of the atomic nucleus <math>q</math>, and appropriate constants. |
The ionization energy is usually measured with the [[electron volt]] (eV). | The ionization energy is usually measured with the [[electron volt]] (eV). | ||
Revision as of 22:35, 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{q e}{4 \pi \epsilon\ r^2} = ma = \frac{m v^2}{r}</math>
Through algebraic manipulation,
<math>\frac{q e}{8 \pi \epsilon\ r} = \frac{1}{2} mv^{2} = KE</math>
To find the potential energy,
<math>PE = \int \bold{F} \cdot \mathrm{d}\bold{s} = - \frac{q e}{4 \pi \epsilon\ r}</math>
Then, to find the ionization energy,
<math>E = PE + KE = - \frac{q e}{4 \pi \epsilon\ r} + \frac{q e}{8 \pi \epsilon\ r} = - \frac{q e}{8 \pi \epsilon\ r}</math>
The last term can be calculated by substituting numerical values for the atomic radius <math>r</math>, charge of the atomic nucleus <math>q</math>, and appropriate constants.
The ionization energy is usually measured with the electron volt (eV).