Question

If the ionization energy of hydrogen is $E$, then the ionization energy of $H{e^ + }$would be:
A. $E$
B. $2E$
C. $0.5E$
D. $4E$

Answer 1

Hint: Ionization energy is the least amount of energy required to remove electrons from the loosely bound electrons which is in the outermost orbital of an atom or molecule.
Suppose to remove the first loosely bound electron from atom$A$ the energy required is $I{E_1}$,and once the atom lose it’s outermost electron it becomes${A^ + }$, to remove the second electron the ionization energy is $I{E_2}$and the atom becomes ${A^{2 + }}$and so on.

Complete step by step answer:
To find out the ionization energy
$IE = \dfrac{{E{z^2}}}{{{n^2}}}$
Where $IE = $Ionization energy, $z = $atomic number
$n = $quantum number,
$E = $energy required
For hydrogen,
$
    \\
  z = 1 \\
  IE = E \\
    \\
 $
For $H{e^ + }$,
$
  z = 2 \\
  n = 1 \\
  IE = E{z^2} = E{\left( 2 \right)^2} = 4E \\
 $
So, the correct answer is Option D.

 Note: Ionisation energy is important because we can predict strength of chemical bonds. The higher the ionisation energy the more energy it requires to lose electrons, it indicates reactivity of atom or molecules. Ionisation energy increases from left to right in a periodic table. Ionisation decreases moving top to bottom in the periodic table.