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  4. The first excitation potential of a given atom is 10.2V. Then, ionization potential must be

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Q. The first excitation potential of a given atom is 10.2V. Then, ionization potential must be

ManipalManipal 2008Atoms

Solution:

The minimum energy needed to ionize an atom is called ionization energy. The potential difference through which an electron should be accelerated to acquire this much energy is called ionization potential.
$ {{({{E}_{2}})}_{H}}-{{({{E}_{1}})}_{H}}=10.2\,\,eV $
or $ \frac{{{({{E}_{1}})}_{H}}}{4}-{{({{E}_{1}})}_{H}}=10.2\,\,eV $
$ \therefore $ $ {{({{E}_{1}})}_{H}}=-13.6\,\,eV $
Hence, ionization potential energy is
$={{({{E}_{\infty }})}_{H}}-{{({{E}_{1}})}_{H}}=13.6\,\,eV $
$ \therefore $ lonisational potential
$=13.6\,\,eV $