Question

The ionisation energy of an electron in the ground state of helium atom is $24.6\, eV$. The energy required to remove both the electron is

• A. 51.8 eV
• B. 79 eV
• C. 38.2 eV
• D. 49.2 eV

Answer 1

The Correct Option is B

lonisation energy in ground state $=24.6 \,eV$
Energy required to remove $2^{n}$ electron from $He ^{2+}$
$=Z^{2}(13.6)\, eV =(2)^{2}(13.6)=54.4\, eV$
So, total energy required $=24.6+54.4=79\, eV$

Answer 2

A Helium atom contains two electrons. The amount of energy needed to remove the outermost electron from an atom is called its ionization energy.

We know that when one electron is removed from helium, the amount of energy needed is given as,

 

E₁ = 24.6eV

 

This is the ionization potential of the helium atom. We also want to remove the other electrons from the helium atom. It looks like when one electron is removed from the helium then it turns into hydrogen-like which means it contains only 1 electron-like hydrogen atom.

Now we can calculate the energy using the Bohr model which is required to eliminate the second electron since now helium becomes hydrogen-like upon the removal of one electron. Therefore, we can use equation (i) where Z = 2 for the helium atom while n = 1 for the ground state of the electron.

E₂ = 13.6×4/1 = 54.4eV

Hence the total amount of energy required to eliminate the 2 electrons from the helium atom is given as 

E = E₁ + E₂

= 24.6+54.4

= 78eV

 

Hence, the correct answer is option B.