Question

Find which of the following is ionization energy of the ionized sodium atom $ N{a^{ + 10}} $ .
(A) $ 13.6eV $
(B) $ 13.6 \times 11eV $
(C) $ \dfrac{{13.6}}{{11}}eV $
(D) $ 13.6 \times {11^2}eV $

Answer 1

Hint :Find the energy of ion $ N{a^{ + 10}} $ at the ground state. The energy of an electron in $ {n^{th}} $ orbit $ {E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV $
Where $ Z $ is the atomic number of the element. $ n $ is positive a integer $ 1,2,3.... $
 $ 1eV = {\text{1}}{\text{.60218}} \times {10^{ - 19}}J $

Complete Step By Step Answer:
We know that the Ionization energy is the minimum energy required for a valence electron in a gaseous atom or ion that has to be absorbed to come out from the atom. It is also sometimes referred to as ionization potential and is usually an endothermic process.
The energy of an electron in $ {n^{th}} $ orbit $ {E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV $
Where $ Z $ is the atomic number of the element.
  $ n $ is positive integer $ 1,2,3.... $ and $ n $ is called principal quantum number
 $ 1eV = {\text{1}}{\text{.60218}} \times {10^{ - 19}}J $
So, if we want to free an electron from it’s $ {n^{th}} $ we have to supply it with $ {E_n} $ amount of energy.
Here, we have given to find the ionization energy of the $ N{a^{ + 10}} $
Now, $ Na $ (sodium) has an atomic number of $ 11 $ .Therefore, $ Na $ has $ 11 $ number of protons and electrons.
Hence, $ N{a^{ + 10}} $ will have $ 10 $ less electrons.
 $ \therefore $ $ N{a^{ + 10}} $ will have only one electron. $ (11 - 10 = 1) $
 $ \therefore $ The electron will be at $ n = 1 $ orbit. This is called the ground state of an electron.
So, removing this electron will further ionize the ion $ N{a^{ + 10}} $ , that energy is our required energy here.
Now, let’s calculate the energy of the ground state electron using the formula of $ {E_n} $ .
Here we have, $ Z = 11 $ , $ n = 1 $
 $ \therefore $ $ {E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV $
Putting the values,
 $ \Rightarrow {E_n} = \dfrac{{ - 13.6 \times {{11}^2}}}{{{1^2}}}eV $
It becomes,
 $ \Rightarrow {E_n} = - 13.6 \times {11^2}eV $
Here, negative sign implies that the energy is stored in the atom.
Which is the energy of an electron at ground state for a $ N{a^{ + 10}} $ ion.
Now, to release this electron the electron needs to absorb this exact amount of energy to come out of the influence of the nucleus hence to come out of the atom.
So, ionization energy of $ N{a^{ + 10}} $ ion will be $ + 13.6 \times {11^2}eV $
Hence, option (D) is correct.

Note :
 $ \bullet $ The ionization energy is the energy given to the electron. Hence, it is always positive.
 $ \bullet $ Energy of an electron in $ H $ -like or single electron atom is $ {E_n} $ , for a multi- electron system energy will be different.