Energy Level - Bohr's Atomic Model - Practice Questions & MCQ

Energy of electron in nth orbit

 Potential energy: An electron possesses some potential energy because it is found in the field of the nucleus potential energy of an electron in  orbit of the radius   is given by

$
U=k \frac{(Z e)(-e)}{r_n}=-\frac{k Z e^2}{r_n}
$

Kinetic energy: Electrons possess kinetic energy because of their motion. Closer orbits have greater kinetic energy than outer ones. As we know $\frac{m v^2}{r_n}=\frac{k(Z e)(e)}{r_n^2}$

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

$
\Rightarrow \quad E=-\frac{k Z e^2}{2 r_n} \quad{ }_{\text {also }} r_n=\frac{n^2 h^2 \varepsilon_0}{z \pi m e^2}
$

$
\begin{aligned}
& \text { Hence } E=-\left(\frac{m e^4}{8 \varepsilon_0^2 h^2}\right) \frac{z^2}{n^2}=-\left(\frac{m e^4}{8 \varepsilon_0^2 c h^3}\right) \operatorname{ch} \frac{z^2}{n^2} \\
& =-R \operatorname{ch} \frac{Z^2}{n^2}=-13.6 \frac{Z^2}{n^2} \mathrm{eV}
\end{aligned}
$

where $R=\frac{m e^4}{8 \varepsilon_0^2 c h^3}=$ Rydberg's constant $=1.09 \times 10^7 \mathrm{~m}^{-1}$

The energy level for Hydrogen
The energy of $\mathrm{n}^{\text {th }}$ level of hydrogen atom ( $\mathrm{z}=1$ ) is given as :

$
E_n=-\frac{z^2 13.6}{n^2} e V=-\frac{13.6}{n^2} e V \quad(\because z=1)
$
Energy of $\operatorname{Ground} \operatorname{state}(n=1)=E_1=-\frac{13.6}{1} \mathrm{eV}=-13.6 \mathrm{eV}$
Energy of first excited state $(n=2)=E_2=-\frac{13.6}{4} \mathrm{eV}=-3.4 \mathrm{eV}$
Energy of second excited state $(n=3)=E_3=-\frac{13.6}{9} \mathrm{eV}=-1.51 \mathrm{eV}$
Energy of third excited $\operatorname{state}(n=4)=E_4=-\frac{13.6}{16} \mathrm{eV}=-0.85 \mathrm{eV}$

Binding Energy(B.E.) of $\mathbf{n}^{\text {th }}$ orbit -Energy required to move an electron from $\boldsymbol{n}^{\text {th }}$ orbit to $n=\infty$ is called the Binding energy of $\underline{n}^{\text {th }}$ orbit

OR
Binding energy of $\mathrm{n}^{\text {th }}$ orbit is the negative of the total energy of that orbit

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

Ionization energy (IE): Total energy of a hydrogen atom corresponds to infinite separation between electron and nucleus. Total positive energy implies that the atom is ionized and the electron is in an unbound (isolated) state moving with certain kinetic energy. The minimum energy required to move an electron from the ground state $(n=1)$ to $n=\infty$ is called the ionization energy of the atom or ion.                                                      

$
E_{\text {ionisition }}=E_{\infty}-E_1=-E_1=13.6 Z^2 \mathrm{eV}
$

On the basis of ionization energy, we can define the ionization 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 the ionization potential of the atom.

$
V_{\text {ionisation }}=\frac{E_{\text {Ionisation }}}{\rho}=13.6 Z^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 -
$n=2, \quad$ first excited state
$n=3, \quad$ second excited state
$n=4, \quad$ third excited state
$n=n_0+1, \quad n_0$ the excited state

Excitation energy -

The energy required to move an electron from the ground state of the atom to any other excited state of the atom is called the 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 the excitation potential of that state.

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