Chemistry
22.07.2020 13:02
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Ahydrogen-like ion is an ion containing only one electron. the energy of the electron

Ahydrogen-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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samueltaye
samueltaye
4,8(23 marks)

Energy required for the ionization of mercury is -13.952\times 10^{-15}kJ

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

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

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:

Hg^{79+}\rightarrow Hg^{80+}+e^-

Mercury has an atomic number = 80

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

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

Putting the values in energy equation, we get

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

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

emj70
emj70
4,6(17 marks)

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I hope this helps you

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