Question

How do you calculate the ionization energy $ ( $ in $ kJmo{l^{ - 1}}) $ of the $ H{e^ + } $ ion?

Answer 1

Hint :Ionization energy is the amount of energy required to remove an electron which is present in the outermost shell of an atom. Smaller the size of atom, more tightly will outermost electron have bonded to the nucleus, then higher will the energy required.

Complete Step By Step Answer:
The expression for calculating ionization energy of an atom or ion is given as follows:
 $ E = - \left( {2.18 \times {{10}^{ - 18}}} \right)\left[ {\dfrac{{{Z^2}}}{{{n^2}}}} \right] - - (i) $
The given ion is $ H{e^ + } $ for which the value of $ Z $ is $ 2 $ .
Electronic configuration of $ H{e^ + } = 1{s^1} $ i,e., the outermost electron is present in the first shell of the atom. Therefore, the value of the principal quantum number $ n = 1 $ .
Substituting values in the equation $ (i) $ :
 $ {E_n} = - \left( {2.18 \times {{10}^{ - 18}}} \right)\left[ {\dfrac{{{2^2}}}{{{1^2}}}} \right] $
 $ \Rightarrow - 8.72 \times {10^{ - 18}}J $
Where, $ {E_n} $ is the energy associated with the atom before excitation of electrons.
Energy after transition can be written as,
 $ {E_\infty } = - \left( {2.18 \times {{10}^{ - 18}}} \right)\left[ {\dfrac{{{2^2}}}{{{{(\infty )}^2}}}} \right] $
 $ \Rightarrow 0 $
Now, Ionization energy is the difference of energy associated with the atom after excitation and energy associated with the atom before excitation.
 $ I.E = {E_\infty } - {E_n} $
 $ \Rightarrow 0 - ( - 8.72 \times {10^{ - 18}})J $
We know that $ 1J = {10^{ - 3}}kJ $ . Therefore,
 $ I.E = 8.72 \times {10^{ - 18}} \times {10^{ - 3}} $
 $ \Rightarrow 8.72 \times {10^{ - 21}}kJ $
To calculate ionization energy per mole, we need to multiply the value of ionization energy by Avogadro’s constant as follows:
 $ I.E = 8.72 \times {10^{ - 21}} \times 6.023 \times {10^{23}} $
 $ \Rightarrow 5250kJmo{l^{ - 1}} $
Hence, the ionization energy $ ( $ in $ kJmo{l^{ - 1}}) $ of the $ H{e^ + } $ ion is $ 5250kJmo{l^{ - 1}} $ .

Note :
Trend of ionization energy in modern periodic table is as follows:
On moving top to bottom in a group, the atomic radii of an atom increases due to which effective nuclear charge decreases. Hence, the value of ionization energy decreases down the group.
On moving left to right in a period, the atomic radii of an atom decreases due to which effective nuclear charge increases. Hence, the value of ionization energy increases on moving left to right in a period.