Energy Level - Bohr's Atomic Model MCQ - Practice Questions with Answers

Energy of electron in nth orbit

 Potential energy: An electron possesses some potential energy because it is found in the field of nucleaus potential

energy of electron in $n^{\text{th }}$ orbit of radius $r_n$ is given by:

$$U=k\cdot \frac{(Ze)(-e)}{r_n}=-\frac{kZ{e}^2}{r_n}$$

Kinetic energy : Electron posses kinetic energy because of it's motion. Closer orbits have greater kinetic energy than outer ones. As we know $\frac{m{v}^2}{r_n}=\frac{k\cdot (Ze)(e)}{r_n^2}$

Kinetic energy $K=\frac{kZ{e}^2}{2r_s}=\frac{|U|}{2}$
Total energy : Total energy E is the sum of potential energy and kinetic energy ie. $E=K+U$

$$\begin{array}{rl}& \Rightarrow E=-\frac{kZ{e}^2}{2r_n}\text{ also }{r}_n=\frac{n^2{h}^2{\epsilon }_0}{\pi nz{e}^2}\\ & E=-\left(\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}\right)\frac{z^2}{n^2}=-\left(\frac{m{e}^4}{8{\epsilon }_0^{2}c{h}^3}\right)\mathrm{ch}\frac{z^2}{n^2}\end{array}$$
 

$$=-R\mathrm{ch}\frac{Z^2}{n^2}=-13.6\frac{Z^2}{n^2}\mathrm{eV}$$

where $R=\frac{m{e}^4}{8{\epsilon }_0^2{\mathrm{ch}}^3}=$ Rydberg's constant $=1.09\times {10}^7$ per $m$.

Energy level for Hydrogen

Ionization energy (IE): Total energy zero of a hydrogen atom corresponds to infinite separation between electron and nucleaus. Total positive energy implies that the atom is ionized and electron is in unbound (isolated) state moving

with certain kinetic energy. Minimum energy required to move an electron from ground state to $n=\mathrm{\infty }$ is called ionization energy of the atom or ion.

The formula for the ionisation energy is -

$$E_{\text{ionisition }}=E_{\mathrm{\infty }}-E_n=-E_n=\frac{13.6Z^2}{n^2}\mathrm{eV}$$

On the basis of ionisation energy we can define the ionisation potential also -
Ionization potential (IP): Potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to ionization energy of the atom is called ionization potential of the atom.

$$V_{\text{ionisation }}=\frac{E_n}{e}=\frac{13.6Z^2}{n^2}\mathrm{V}$$
 

Now let us discuss Excitation energy and Excitation potential -

Excitation Energy and Excitation Potential
Now what is Excitation???

The process of absorption of energy by an electron so as to raise it from a lower energy level to some higher energy level is called excitation.

Excited state: The states of an atom other than the ground state are called its excited states. Examples are mentioned below -

$n=2,$ first excited state
$n=3,$ second excited state
$n=4,$ third excited state
$n=n_0+1,n_0$ th excited state

Excitation energy_
Energy required to move an electron from ground state of the atom to any other excited state of the atom is called Excitation energy of that state.

$$E_{\text{excitation }}=E_{\text{higher }}-E_{\text{lower }}$$

Excitation potential can also be defined on the basis of excitation energy. So the excitation potential is the potential difference through which an electron must be accelerated from rest so that its kinetic energy becomes equal to the excitation energy of any state is called excitation potential of that state.

$$V_{\text{excitation }}=\frac{E_{\text{excitation }}}{e}$$