Energy Level - Bohr's Atomic Model

Niels Bohr's atomic model, proposed in 1913, revolutionized our understanding of atomic structure by introducing the concept of quantized energy levels. According to Bohr's model, electrons orbit the nucleus at specific, fixed distances, corresponding to discrete energy levels. This quantization explains why atoms emit or absorb light at characteristic wavelengths, a principle foundational to spectroscopy. In real life, Bohr's model is crucial for technologies like lasers and fluorescent lights, where precise control of electron transitions produces specific colours of light. In this article we will understand that energy levels also underpin the operation of various electronic devices, from semiconductors to imaging systems, demonstrating the profound impact of Bohr's atomic model on both science and technology.

The Energy of an Electron in the nth Orbit

In Bohr's atomic model, the energy levels of an electron in an atom are quantized. This means that electrons can only occupy specific, discrete energy levels and cannot exist in between these levels. The potential energy is always negative, indicating that the electron is bound to the nucleus. The energy levels of the electron are quantized, and the total energy is the sum of kinetic and potential energies.

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. $E=K+U$
$
\begin{aligned}
& \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} \\
& \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 $R=\frac{m e^4}{8 e^2 c^3 h^3}=$ Rydberg's constant $=1.09 \times 10^7 \mathrm{~m}^{-1}$

The Energy Level for Hydrogen

The energy of $\underline{n}^{\text {th }}$ level of hydrogen atom $(z=1)$ is given as :
$
E_n=-\frac{z^2 13.6}{n^2} \mathrm{eV}=-\frac{13.6}{n^2} \mathrm{eV} \quad(\because z=1)
$

Energy of 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 state $(n=4)=E_4=-\frac{13.6}{16} \mathrm{eV}=-0.85 \mathrm{eV}$

Binding Energy(B.E.) of n^th orbit

The binding energy of an electron in the n-th orbit of a hydrogen atom refers to the energy required to remove the electron from that orbit to infinity, essentially ionizing the atom. In the context of the Bohr model, the binding energy of an electron in the n-th orbit can be expressed as:

The energy required to move an electron from $n^{\text {th }}$ orbit to $n=\infty$ is called the Binding energy of $\underline{n}^{\text {th }}$ orbit

OR
The binding energy of $\underline{\underline{t}}^{\text {th }}$ 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

The 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.

The formula for the ionization energy is -

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

On the basis of ionization energy, we can define the ionization potential also -

Ionization Potential

The potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to the ionization energy of the atom is called the ionization potential of the atom.

$V_{\text {ionisation }}=\frac{E_{\text {Ionisation }}}{e}=13.6 Z^2 \mathrm{~V}$

Now let us discuss Excitation energy and Excitation potential

Excitation Energy and Excitation Potential

Excitation Energy and Excitation Potential are related concepts in atomic physics, referring to the energy required to move an electron from a lower energy state to a higher one within an atom.

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

When an atom or molecule absorbs energy, an electron can move from the ground state to a higher energy level. This higher energy level is called an 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
$n=n_0+1, \quad n_0$ th 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.

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

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 state.

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

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Solved Examples Based on Energy Level - Bohr's Atomic Model

Example 1: If the series limit frequency of the Lyman series is ν_L, then the series limit frequency of the Pfund series is :

1) ν_L/25

2) 25 ν_L

3) 16 ν_L

4) ν_L/16

Solution:

P-fund Series

$
\frac{1}{\lambda}=R\left(\frac{1}{5^2}-\frac{1}{n^2}\right)_{\text {(infrared) }}
$
wherein
$
n=6,7,8-\cdots
$

When an electron jumps from a higher orbital to $\mathrm{n}=5$ energy level
Lyman series
$
\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)_{\text {(UV region ) }}
$
wherein
$
n=2,3,4,-----
$

When an electron jumps from a higher orbital to $1^{\text {st }}$ energy level i.e. ground state
For Lyman $\quad h \nu_L=13.6 \mathrm{ev}$

For p-fund
$
h \nu_p=\frac{13.6}{5^2}=\frac{13.6}{25}
$
therefore, $\frac{v_L}{v_P}=25 \quad$ or $\quad v_p=\frac{\nu_L}{25}$

Hence, the answer is the option 1.

Example 2: Arrange the following electromagnetic radiations per quantum in the order of increasing energy :

A: Blue light B: Yellow light
C: X-ray D: Radiowave.

1) D, B, A, C

2) A, B, D, C

3) C, A, B, D

4) B, A, D, C

Solution:

Balmer series

Balmer series
$
\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)
$
(visible light)
wherein
$
n=3,4,5----
$

When an electron jumps from a higher orbital to ^an n=2 energy level

Radio wave < Yellow light < Blue light < X-ray

Hence, the answer is the option 1.

Example 3: The amount of energy required (in eV) to excite an electron from the 3^rd to 4^th orbit of the hydrogen atom:

1) 0.66

2) 0.77

3) 0.88

4) 0.99

Solution:
From Paschen Series,

$
\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{n^2}\right)
$
(infrared)
$
\begin{aligned}
\Delta E= & \frac{h c}{\lambda}=R h c\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\
& =13.6\left(\frac{1}{9}-\frac{1}{16}\right) \mathrm{eV} \\
& =13.6 \frac{7}{144} \mathrm{eV}=0.66 \mathrm{eV}
\end{aligned}
$

Hence, the correct answer is 0.66 eV.

Example 4: Brackett series of lines are produced when electrons excited to a high energy level make the transition to the n=4 level, the wavelength (in nm) corresponding to the transition from n_f =6 to n_i=4 is:

1) 2630

2) 6025

3) 3858

4) 4051

Solution:

From the Brackett Series,

$
\begin{aligned}
& \frac{1}{\lambda}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right) \\
& (\text { infrared })
\end{aligned}
$
wherein
$
n=5,6,7----
$

When an electron jumps from a higher orbital to $n=4$ energy level
$
\begin{aligned}
\frac{1}{\lambda} & =R\left(\frac{1}{4^2}-\frac{1}{n^2}\right)=R\left(\frac{1}{16}-\frac{1}{36}\right) \\
\lambda & =\frac{144}{5 R} \\
\lambda & =\frac{144}{5} \times 91 \mathrm{~nm}=2630 \mathrm{~nm}
\end{aligned}
$

Hence, the answer is option (1).

Example 5: The first member of the Balmer series of hydrogen atoms has a wavelength of 6561 . The wavelength of the second member of the Balmer series (in nm ) is _______.

1) 486

2) 972

3) 986

4) 586

Solution:

$\begin{aligned} \frac{1}{\lambda} & =\mathrm{R} Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ \frac{1}{\lambda_1} & =R(1)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 \mathrm{R}}{36} \\ \frac{1}{\lambda_2} & =\mathrm{R}(1)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 \mathrm{R}}{16} \\ \frac{\lambda_2}{\lambda_1} & =\frac{20}{27} \\ \lambda_2 & =\frac{20}{27} \times 6561 \mathrm{~A}=4860 \mathrm{~A} \\ & =486 \mathrm{~nm}\end{aligned}$

Hence, the answer is option (1).

Summary

Niels Bohr's 1913 atomic model introduced quantized energy levels, explaining why atoms emit or absorb light at specific wavelengths. This model is fundamental to spectroscopy and underpins technologies like lasers and fluorescent lights. The energy of an electron in a given orbit, the binding energy, and the ionization energy are crucial for understanding atomic transitions. Excitation energy refers to the energy required to move an electron from a lower to a higher energy state, while excitation potential is the corresponding voltage needed. Practical examples include determining the frequency of spectral lines and energy transitions in various atomic series.