Ionization energy of \[H{e^ + }\]is \[19.6 \times 10 ^{- 18}\;J.atom^{- 1}\]. The energy of the first stationary state (n=1) of is:
A. \[4.41 \times {10^{ - 16}}J.ato{m^{ - 1}}\]
B. \[ - 4.41 \times {10^{ - 17}}J.ato{m^{ - 1}}\]
C. \[ - 2.2 \times {10^{ - 15}}J.ato{m^{ - 1}}\]
D. \[8.82 \times {10^{ - 17}}J.ato{m^{ - 1}}\]

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Hint: Ionisation energy is straightforward terms may be defined as a measurement of the problem in removal of an electron from an atom or ion or the tendency of an atom or ion to give up an electron. The lack of electrons normally happens when it is in a ground state of the chemical species.

Explanation:
Ionization energy is the amount of energy that an isolated, gaseous atom in the ground state should absorb to discharge an electron, resulting in a cation.
When thinking about a first of all neutral atom, expelling the primary electron would require much less energy than expelling the second one, the second one would require much less energy than the third, and so on. Each successive electron calls for extra energy to be released in reaction. This is due to the fact after the primary electron is lost, the net charge of the atom turns into positive, and the negative forces of an electron might be attracted to the positive charge of the newly formed ion. The extra electrons which are lost, the more positive this ion might be, the more difficult it is to split the electrons from the atom.

Complete step by step solution:
Since we know that,
\[\dfrac{{I{E_{L{i^{2 + }}}}}}{{I{E_{H{e^ + }}}}} = {\left( {\dfrac{{{Z_{L{i^{2 + }}}}}}{{{Z_{H{e^ + }}}}}} \right)^2} = \dfrac{9}{4}\]
\[I{E_{L{i^{2 + }}}} = \dfrac{9}{4} \times I{E_{H{e^ + }}}\]
\[I{E_{L{i^{2 + }}}} = \dfrac{9}{4} \times 19.6 \times {10^{ - 18}}\]
\[I{E_{L{i^{2 + }}}} = - 4.41 \times {10^{ - 17}}J/atom\]

Therefore, the correct answer is option B. \[I{E_{L{i^{2 + }}}} = - 4.41 \times {10^{ - 17}}J/atom\]

Note:

Ionisation energies are dependent on the atomic radius. Since, going from right side to left side on the periodic table, the atomic radius has an increase, and the ionization energy also increases from left to right in the periodic table and up side the groups.