Question

The ionization energy of hydrogen atom is 13.6 $eV$.What will be the ionization energy of $H{{e}^{+}}$ and $L{{i}^{+2}}$ ions?

Answer 1

As $H{{e}^{+}}$ and $L{{i}^{+2}}$ both ions contain one electron similar to hydrogen atom. The ionization energy of such ions also known as hydrogen-like species can be calculated using the following formula: ionization energy = $\dfrac{13.6{{Z}^{2}}}{{{n}^{2}}}eV$ where Z is the atomic number and n is shell number. Value of Z for helium is 2 and for lithium is 3.

Complete step by step solution:
-Bohr’s atomic model is based on the following postulates:
-The electron in a hydrogen atom can move in a circular path around the nucleus. These paths are called orbits.
-These orbits are arranged around the nucleus concentrically.
-the energy of the electron in orbit does not change with time.
-when the electron moves from a lower stationary state to higher stationary state when the required amount of energy is absorbed.
-when an electron moves from a higher stationary state to lower stationary state, energy is emitted.
-the energy change does not take place in a continuous manner.
-The frequency of radiation absorbed or emitted when the transition occurs between two stationary states that differ in energy $\Delta E$is given by:
$\nu =\dfrac{\Delta \varepsilon }{h}=\dfrac{{{E}_{2}}-{{E}_{1}}}{h}$
Where ${{E}_{1}}$and ${{E}_{2}}$ are energies of lower and higher energy states.
-The stationary states for electrons are numbered n=1,2,3…
-The radii of stationary states are expressed as:
${{r}_{n}}={{n}^{2}}{{a}_{0}}$where ${{a}_{0}}$=52.9pm. the radius of the first stationary state called Bohr orbit.
-Energy associated with the electron in its stationary state is given by
${{E}_{n}}=-{{R}_{H}}\dfrac{1}{{{n}^{2}}}$
Where ${{R}_{H}}$is Rydberg constant and value for this constant is $-2.18\times {{10}^{-18}}J$
The energy of the lowest state also called ground state is
${{E}_{1}}$=$-2.18\times {{10}^{-18}}\dfrac{1}{{{(1)}^{2}}}$ =$-2.18\times {{10}^{-18}}J$
Ionization energy is defined as the amount of energy required to remove an electron resulting in the formation of the cation.
$2.18\times {{10}^{-18}}J=13.6eV$
-Bohr’s theory can also be applied to ions containing only one electron. So, for $H{{e}^{+}}$ and $L{{i}^{+2}}$, ionization energy can be calculated as
Ionization energy= $\dfrac{13.6{{Z}^{2}}}{{{n}^{2}}}eV$ Where Z is atomic number and n is number of shells
Ionization energy for $H{{e}^{+}}$= $\dfrac{13.6\times {{2}^{2}}}{1}=54.4eV$
Ionization energy for $L{{i}^{+2}}$= $\dfrac{13.6\times {{3}^{2}}}{1}=122.4eV$

Note: Ionization energy is defined as the amount of energy that must be absorbed to remove an electron resulting in the formation of the cation. For hydrogen atom energy of $13.6eV$must be absorbed to remove an electron. The ionization energy of ions containing one electron can be calculated using the following formula:
ionization energy = $\dfrac{13.6{{Z}^{2}}}{{{n}^{2}}}eV$ where Z is atomic number and n is shell number, n is considered as 1.
Also, $2.18\times {{10}^{-18}}J=13.6eV$