A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by En = ?(2.18 × 10?18J) Z2 ( 1 n2 ) where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step: Hg79+(g) ? Hg80+(g)+ e?

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RomeliaThurston RomeliaThurston · Answer 1 · 2018-11-22T08:09:43+01:00

Answer: Energy required for the ionization of mercury is [tex]-13.952\times 10^{-15}kJ[/tex]

Explanation: Energy of the electron in a hydrogen-like ion is given by the equation:

[tex]E_n=\frac{(-2.18\times 10^{-18})\times Z^2}{n^2}J[/tex]

where, Z = atomic number

n = principle quantum number

Ionization energy is the energy required to release the outermost electron from an isolated gaseous atom.

For the ionization of mercury, the equation follows:

[tex]Hg^{79+}\rightarrow Hg^{80+}+e^-[/tex]

Mercury has an atomic number = 80

As, in this element 79 protons are already released, which means that the electronic configuration for [tex]Hg^{79+}[/tex] is [tex]1s^1[/tex]

and the principle quantum number for the last ionization step = 1

Putting the values in energy equation, we get

[tex]E_1=\frac{(-2.18\times 10^{-18})\times (80)^2}{(1)^2}J=-13952\times 10^{-18}J[/tex]

[tex]E_1=-13.952\times 10^{-15}kJ[/tex] (Conversion Factor: 1kJ = 1000J)