Question

Ionisation energy of $\text{ H}{{\text{e}}^{\text{+}}}\text{ }$is $\text{ 19}\text{.6 }\times \text{ 1}{{\text{0}}^{-\text{18}}}\text{ J ato}{{\text{m}}^{-1}}\text{ }$.The energy of the first stationary state $\text{ }\left( \text{ n = 1 } \right)\text{ }$ of $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$is:
A) $\text{ 4}\text{.41}\times {{10}^{-16}}\text{ J ato}{{\text{m}}^{-1}}\text{ }$
B) $\text{ }-\text{4}\text{.41}\times {{10}^{-17}}\text{ J ato}{{\text{m}}^{-1}}$
C) $\text{ }-2.\text{2}\times {{10}^{-15}}\text{ J ato}{{\text{m}}^{-1}}$
D) $\text{ 8}\text{.82}\times {{10}^{-17}}\text{ J ato}{{\text{m}}^{-1}}\text{ }$

Answer 1

the ionization energy is the amount of energy required to knock out the valence shell electron. In simple words, it is the measure of a tendency of an atom to lose its electron. The ionization energy of an atom is related to the atomic number as follows,
$\text{ }{{\text{E}}_{\text{n}}}\text{ = }-{{\text{R}}_{\text{n}}}\dfrac{{{Z}^{2}}\text{hc}}{{{\text{n}}^{\text{2}}}}\text{ }$
Where $\text{ }{{\text{E}}_{\text{n}}}\text{ }$ is the energy of the level n, Z is the atomic number, n is the energy level, h is the Planck's constant, c is the velocity of light, and $\text{ }{{\text{R}}_{\text{n}}}\text{ }$ is Rydberg constant. In a stationary energy state, the value of energy level is equal to 1.

Complete step by step solution:
Ionization energy is the amount of energy required by an atom to lose its electron. The atom loses its valence shell electron and acquires the positive charge. The general representation is as follows,
$\text{ A + IE }\to \text{ }{{\text{A}}^{\text{+}}}\text{ + }{{\text{e}}^{-}}\text{ }$
Where, $\text{ IE }$ is the ionization energy applied to atom A.
We have given the following data:
The helium atom loses its electrons and forms $\text{ H}{{\text{e}}^{\text{+}}}\text{ }$.
The ionization energy of the $\text{ H}{{\text{e}}^{\text{+}}}\text{ }$is equal to the $\text{ 19}\text{.6 }\times \text{ 1}{{\text{0}}^{-\text{18}}}\text{ J ato}{{\text{m}}^{-1}}\text{ }$
We are interested to determine the energy of the first stationary state i.e. $\text{ n = 1 }$ of $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$ ion.
The ionization energy of an atom is related to the atomic number as follows,
$\text{ }{{\text{E}}_{\text{n}}}\text{ = }-{{\text{R}}_{\text{n}}}\dfrac{{{Z}^{2}}\text{hc}}{{{\text{n}}^{\text{2}}}}\text{ }$
Where $\text{ }{{\text{E}}_{\text{n}}}\text{ }$ is the energy of the level n, Z is the atomic number, n is the energy level, h is the Planck's constant, c is the velocity of light, and $\text{ }{{\text{R}}_{\text{n}}}\text{ }$ is Rydberg constant.
Since, h, c and $\text{ }{{\text{R}}_{\text{n}}}\text{ }$are constant, the ionization energy is directly related to the atomic number and inversely related to the number of energy levels.
For the first stationary state i.e. when $\text{ n = 1 }$the ionization energy of the $\text{ H}{{\text{e}}^{\text{+}}}\text{ }$written as,
$\text{ IE}{{\text{ }}_{\text{H}{{\text{e}}^{\text{+}}}}}\text{ }\propto \text{ }\dfrac{{{\left( {{Z}_{\text{H}{{\text{e}}^{\text{+}}}}} \right)}^{2}}}{{{\left( {{\text{n}}_{\text{H}{{\text{e}}^{\text{+}}}}} \right)}^{2}}}\text{ }$ (1)
Similarly, for$\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$, the ionization energy can be written as,
$\text{ IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}\text{ }\propto \text{ }\dfrac{{{\left( {{Z}_{_{\text{L}{{\text{i}}^{\text{2+}}}}}} \right)}^{2}}}{{{\left( {{\text{n}}_{_{\text{L}{{\text{i}}^{\text{2+}}}}}} \right)}^{2}}}\text{ }$ (2)
Take the ratio of (1) and (2), we have,
$\text{ }\dfrac{\text{IE}{{\text{ }}_{\text{H}{{\text{e}}^{\text{+}}}}}}{\text{IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}}\text{ = }\dfrac{{{\left( {{Z}_{\text{H}{{\text{e}}^{\text{+}}}}} \right)}^{2}}}{{{\left( {{\text{n}}_{\text{H}{{\text{e}}^{\text{+}}}}} \right)}^{2}}}\text{ }\times \text{ }\dfrac{{{\left( {{\text{n}}_{_{\text{L}{{\text{i}}^{\text{2+}}}}}} \right)}^{2}}}{{{\left( {{Z}_{_{\text{L}{{\text{i}}^{\text{2+}}}}}} \right)}^{2}}}\text{ }$
Since we are interested in the ionization energy at the stationary level .i.e. at $\text{ n = 1 }$. The equation becomes,
$\text{ }\dfrac{\text{IE}{{\text{ }}_{\text{H}{{\text{e}}^{\text{+}}}}}}{\text{IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}}\text{ = }\dfrac{{{\left( {{Z}_{\text{H}{{\text{e}}^{\text{+}}}}} \right)}^{2}}}{{{\left( {{Z}_{_{\text{L}{{\text{i}}^{\text{2+}}}}}} \right)}^{2}}}\text{ }$
We know that the atomic number of $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$is 3 and for $\text{ H}{{\text{e}}^{\text{+}}}\text{ }$is is 2 and ionization energy of $\text{ H}{{\text{e}}^{\text{+}}}\text{ }$. Thus on substituting the values we have,
$\begin{aligned} & \text{ }\dfrac{\text{IE}{{\text{ }}_{\text{H}{{\text{e}}^{\text{+}}}}}}{\text{IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}}\text{ = }\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( 3 \right)}^{2}}}\text{ = }\dfrac{4}{9} \\\ & \Rightarrow \text{IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}\text{ = IE}{{\text{ }}_{\text{H}{{\text{e}}^{\text{+}}}}}\text{ }\times \text{ }\dfrac{9}{4} \\\ & \Rightarrow \text{IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}\text{ = }\left( 19.6\times {{10}^{-18}} \right)\text{ }\times \text{ }\dfrac{9}{4} \\\ & \therefore \text{IE}{{\text{ }}_{\text{L}{{\text{i}}^{\text{2+}}}}}\text{ = 4}\text{.41}\times {{10}^{-17}}\text{ J ato}{{\text{m}}^{-1}}\text{ } \\\ \end{aligned}$
Therefore, the ionization energy of $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$ the ion stationary state is equal to.
$\text{ Energy in stationary state = }-\text{IE = }-\text{ 4}\text{.41}\times {{10}^{-17}}\text{ J ato}{{\text{m}}^{-1}}\text{ }$. As, the energy is released during the ionization hence it is always negative.
Hence, (B) is the correct option.

Note: The ionization energy is related to the atomic number ‘Z’ and the energy level ‘n’.It is clear that the atomic number is directly related to the atomic number, thus with increases in the Z, the ionization energy increases and decreases with the increase in the energy level or with the addition of energy shell.
$\text{ Ionisation energy }\propto \text{ }\dfrac{\text{Atomic number (}Z)}{\text{energy level(n)}}\text{ }$
Therefore, the ionization energy is found to increase as we move from left to right and decreases as we move top to bottom in the periodic table.