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Question: Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, ...

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 electrons and therefore would exist as Hg80+H{{g}^{80+}} . The energy required for the step is: Hg79+(g)→Hg80++e−H{{g}^{79+}}(g)\to H{{g}^{80+}}+{{e}^{-}}
[A] 4.095×106KJ/mol4.095\times {{10}^{6}}KJ/mol
[B] 6.25×106KJ/mol6.25\times {{10}^{6}}KJ/mol
[C] 8.39×106KJ/mol8.39\times {{10}^{6}}KJ/mol
[D] 4.43×106KJ/mol4.43\times {{10}^{6}}KJ/mol

Explanation

Solution

To solve this, calculate the energy at ground state and at infinite excited state (use the energy equation given for a hydrogen like system).Their difference will give us the energy absorbed by the electron which is the ionisation energy. Multiply it by Avogadro’s number and divide it by 1000 to get the answer in KJ/mol.

Complete step by step answer:
In the question it is given that mercury will exist as Hg80+H{{g}^{80+}} in plasma state if we remove all of its 80 electrons. We are asked about the energy required for the last ionisation step i.e. energy required for conversion of Hg79+H{{g}^{79+}} to Hg80+H{{g}^{80+}}

Here we have Hg79+H{{g}^{79+}}. We know that the atomic number of mercury is 80. Therefore, Hg79+H{{g}^{79+}}acts as an iso-electronic system. Therefore, we can write the energy equation as-
En=−2.18×10−18J(Z2)(1n2){{E}_{n}}=-2.18\times {{10}^{-18}}J\left( {{Z}^{2}} \right)\left( \dfrac{1}{{{n}^{2}}} \right)
Where, n is the principal quantum number (n = 1, 2, 3…), Z is the atomic number and E corresponds to the energy level of the corresponding state.
When n = 1, it is the ground state energy and when n is infinity, (1n2)\left( \dfrac{1}{{{n}^{2}}} \right) tends to 0, which means the electron is completely ionised.

Therefore, for n = 1, energy will be, E1=−2.18×10−18J(802)(112)=−1.3952×10−14J{{E}_{1}}=-2.18\times {{10}^{-18}}J\left( {{80}^{2}} \right)\left( \dfrac{1}{{{1}^{2}}} \right)=-1.3952\times {{10}^{-14}}J
When completely ionised, (1n2)\left( \dfrac{1}{{{n}^{2}}} \right) has to be zero therefore n will be infinity. Therefore, energy at the corresponding state will be - E∞=−2.18×10−18J(802)(1∞)=0J{{E}_{\infty }}=-2.18\times {{10}^{-18}}J\left( {{80}^{2}} \right)\left( \dfrac{1}{\infty } \right)=0J
Therefore, energy absorbed by the electrons from n = 1 to n = ∞\infty will be-
ΔE=E∞−E1=0J−(−1.3952×10−14J)=1.3952×10−14J\Delta E={{E}_{\infty }}-{{E}_{1}}=0J-\left( -1.3952\times {{10}^{-14}}J \right)=1.3952\times {{10}^{-14}}J

In terms of moles we can write that energy required = 1.3952×10−14×6.023×10231000KJ of Hg79+\dfrac{1.3952\times {{10}^{-14}}\times 6.023\times {{10}^{23}}}{1000}KJ\text{ of H}{{\text{g}}^{79+}}
Therefore, energy required for the step Hg79+(g)→Hg80++e−H{{g}^{79+}}(g)\to H{{g}^{80+}}+{{e}^{-}} is
1.3952×10−14×6.023×10231000KJ/mol = 8.39×106KJ/mol\dfrac{1.3952\times {{10}^{-14}}\times 6.023\times {{10}^{23}}}{1000}KJ/mol\text{ = 8}\text{.39}\times \text{1}{{\text{0}}^{6}}KJ/mol
So, the correct answer is “Option C”.

Note: It is important to remember here that the energy equation that we used here is only for hydrogen like, iso-electronic systems. We can also solve this question without calculating the energy absorbed and directly applying n = 1 to the equation because from 79 to 80, one electron is ionised. We can directly write the ionisation energy equation as-
2.18×10−18×(80)2×6.023×10231000KJ/mol \dfrac{2.18\times {{10}^{-18}}\times {{\left( 80 \right)}^{2}}\times 6.023\times {{10}^{23}}}{1000}KJ/mol\text{ }