Correct Answer - Option 1 : 9E

_{n}
__Concept:__

According to the Bohr Model of an atom, the electron revolves around the nucleus in certain stable and discrete orbits without radiating energy. Energy is radiated only when an electron jumps from one energy level to another. These stable orbits are known as stationary orbits and are at a fixed distance from the nucleus. There is no other orbit in between these discrete ones. The energy of each orbit is fixed.

The energy of Hydrogen like-atom when its electron is in nth orbit,

\(E_n =-Z^2\times \dfrac{13.6}{n^2}\ eV\)

where n is the orbit in which the electron is revolving.

And Z is the atomic number of the atom.

The doubly ionized Lithium atom contains one electron so Bohr's Model can be applied to it.

__Calculation:__

We know that,

The energy of Hydrogen like-atom when its electron is in nth orbit,

\(E_n =-Z^2\times \dfrac{13.6}{n^2}\ eV\)

For the Hydrogen atom, Z=1

So, \(E_n = -\dfrac{13.6}{n^2}\ eV\)

For doubly ionized Lithium atom, Z=3

So, \(E =-3^2\times \dfrac{13.6}{n^2}\ eV\)

\(E =-9\times \dfrac{13.6}{n^2}\ eV\)

E = 9E_{n }

So, option(1) is the correct answer.

- The ionization energy \(E_\infty\) that is required to remove the electron from the first Bohr orbit is called the Ionization limit of H- atom.
- The formula \(E=E_0 \dfrac{Z^2}{n^2}\), is valid only for H- like atom(one electron).