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Balmer’s series lies in the region of the spectrum called a visible region. The series of spectral lines produced due to transitions from all energy states to energy states E2 is called Balmer’s series.

For Balmer’s series P=2 and n=3,4,5,6,7………∞

Wavelength in Balmer’s series of H-atom,

1/λ = R [1/p2 – 1/n2 ]

R is Rydberg constant, the value of rydberg constant is 10,973,731.5682 per meter or 1.097×107/m

1/λ = 1.097×107 [1/p2 – 1/n2 ] m-1

For shorter wavelength in Balmer’s series,

p=2 and n=∞

1/λmin = 1.097×107 [1/22 – 1/ ∞ 2 ] m-1

1/λmin = 1.097×107 [1/4] m-1

λmin = 4/ 1.097×107 m

λmin = 3.646×10-7 m

λmin = 364.6×10-9 m

λmin = 364.6 nm

## Bohr’s atomic model:

Basically, Bohr presents his model to understand the atomic structure. Before the Bohr model, the plum pudding model is also present. The Plum pudding model fails to explain the atomic structure.

Rutherford’s model is based upon classical electromagnetic theory. Although, Rutherford’s experiments proved that the Plum pudding model of an atom was not correct. Yet it has the following defects:

### Defect’s of Rutherford’s model:

1. According to the classical theory of radiation, electrons being the charged particles should release or emit energy continuously and they should ultimately fall into the nucleus.
2. If the electron emits energy continuously, it should form a continuous spectrum but in fact, the line spectrum was observed.

According to classical electromagnetic theory, there is a center of positive charge in an atom which is called the nucleus and electrons is revolving around the nucleus. According to classical electromagnetic wave theory, the electrons that revolve in the orbit are accelerated. (every point the direction of velocity is changed)

In electricity and magnetism, when the charged particle is accelerated, they produce electromagnetic radiations. so, according to Rutherford’s the electrons that continuously revolve are accelerated and the energy of electrons is deceased that’s why electrons emit energy continuously and fall in the nucleus.

Keeping in view the defects of Rutherford’s atomic model. Niels Bohr presented another model of the atom in 1931. The Quantum theory of Max Planck was used as the foundation of this model. According to Niel’s Bohr, a revolving electron in an atom does not absorb or emit energy continuously. The energy of revolving electrons is quantized as it revolves only in orbits of fixed energy called energy levels or shells.

### Postulates of Bohr atomic model:

1. The hydrogen atom consists of a tiny nucleus and electrons are revolving in one of the circular orbits of radius ”r” around the nucleus.
2. Each orbit has fixed energy that is quantized.
3. Electrons remain in particular orbits, it does not radiate or emits energy. The energy is emitted or absorbed only when an electron jumps from one orbit to another.
4. When an electron jumps from lower orbit to higher orbit, it absorbs energy and when it jumps from a higher orbit to lower orbit, it released energy.

5. Electron can revolve only in orbits of a fixed angular momentum mvr,

mvr = n×h/2π

where n is the Quantum number having values 1,2,3…so on.

## Derivation of Bohr’s Model:

A hydrogen atom has one proton in the nucleus and only one electron is revolving around it. The ionization energy of hydrogen is 13.6 ev. Suppose electron is revolving with speed ”v” in an orbit of radius ”r”

The column force of attraction between electron and proton provides electron to required centripetal force to revolve around the nucleus. Thus,

Fcoloumb = Fcentripital

• Fcoloumb = k×q1q2 /r2 or k×Ze × e /r2
• Fcentripetal = mv2 /r
• k = 1/4πϵ°

1/4πϵ°. Ze × e /r2 = mv2 /r

1/4πϵ°. Ze2 /r2 = mv2 /r

v2 = Ze2 r / 4πϵ° r2 m

v2 = Ze2 / 4πϵ° mr

v = √ Ze2 / 4πϵ° mr ——– (1)

### Frequency of revolution:

As,

v = rω

v = r(2πf)

v = 2πrf

f = 1/2πr × v

put the value of v from (1)

f = 1/2πr × √ Ze2 / 4πϵ° r m

f = √1/4π2r2 × √ Ze2 / 4πϵ° r m

f = √Ze2/ 16 π3ϵ°mr3 ——– (2)

### Total energy:

E = K.E + P.E ——–(A)

• K.E = 1/2 × mv2

put the value of v from (1)

K.E = 1/2 × m [√ Ze2 / 4πϵ° mr ]2

K.E = 1/2 × m [ Ze2 / 4πϵ° mr ]

K.E = m/2 × [ Ze2 / 4πϵ° mr ]

K.E = Ze2 / 8πϵ° r ——– (B)

• P.E = Potential difference × charge of e

∴ v = ω/q

ω = vq

∴ v = 1/4πϵ° × q/r

P.E = 1/4πϵ° . Ze/r × (−e)

P.E = −1/4πϵ° . Ze2/r ——– (C)

Put (B) and (C) in (A)

E = Ze2 / 8πϵ° r × ( −1/4πϵ° . Ze2/r )

E = Ze2 − 2Ze2 / 8πϵ° r

E =−Ze2 / 8πϵ° r ——– (3)

### Angular momentum:

L = r × p

L = r × (mv)

L = mvr sinθ

∴ θ = 90°

L = mvr

put the value of v from (1)

L = mr × √ Ze2 / 4πϵ° r m

L = √ m2r2 × √ Ze2 / 4πϵ° mr

L = √ Ze2 m2r2 / 4πϵ° mr

L = √ Ze2 mr / 4πϵ° ——–(4)

### Relation between frequency and energy:

From eq. (3)

E =−Ze2 / 8πϵ° r

r =−Ze2 / 8πϵ° E

From eq. (2)

f = √Ze2/ 16 π3ϵ°mr3

put the value of ”r”

f = √Ze2/ 16 π3ϵ°m ( −Ze2 / 8πϵ° E )3

f = √Ze2/ 16 π3ϵ°m ( −Z3e6 / 512π3ϵ°3 E3 )

f = √512 π3ϵ°3 E3 . Ze2/ 16 π3ϵ°m. −Z3e6

f = √−32 ϵ°2 E3 / Z2 m e4

eq. (5) derived classically so,

fcm = (−32 ϵ°2 E3 / Z2 m e4 )1/2 ——– (5)

For large Quantum number,

put E = −hcR/n2 in (5)

fcm = (−32 ϵ°2 / Z2 m e4 × ( −hcR/n2 )3 )1/2

fcm = (32 ϵ°2 / Z2 m e4 × (h3c3R3) )1/2 × ((1/n2)3 )1/2

fcm = (32 ϵ°2 / Z2 m e4 × (h3c3R3) )1/2 × 1/n3 ——– (6)

As,

1/λ = R (1/m2 − 1/n2 )

put m=n−1

1/λ = R (1/( n−1)2 − 1/n2 )

Multiply both side by ”c”

c/λ = cR (n2 − (n−1)2 / n2 (n−1)2 )

c/λ = cR (n2 − n2(1−2n) / n2 (n−1)2 )

c/λ = cR (2n−1 / n2 (n−1)2 )

fqm= cR (2n−1 / n2 (n−1)2 ) ——– (7)

for n>>>1 so, eq, (7) becomes

fqm≅ cR (2n) / n2 (n)2

fqm≅ 2cRn / n4

fqm= 2cR / n3 ——– (8)

### Expression for Rydberg constant:

According to the correspondence principle, for large quantum numbers, quantum frequency and classical frequency should be equal.

fqm = fcm

2cR / n3 = (32 ϵ°2 / Z2 m e4 × (h3c3R3) )1/2 × 1/n3

simplify, we get,

R = m Z2 e4 / 8 ϵ°2 h3c ——– (9)

### Quantum expression for energy:

En = −hcR/n2

En = −hc/n2 × m Z2 e4 / 8 ϵ°2 h3c

En = −m Z2 e4 / 8 ϵ°2 h2 × 1/n2 ——– (10)

This is a quantum expression for the energy of stationary state of H-atom.

E =−Ze2 / 8πϵ° r ——–from Eq.(3)

comparing with eq. (10)

−Ze2 / 8πϵ° r = −m Z2 e4 / 8 ϵ°2 h2 × 1/n2

r = ϵ° h2 / Ze2 πm × n2 OR

r = 90 n2

where,

• 90 = ϵ° h2 /Z e2 πm is called Bohr’s radius. For n=1 where, 90 = 0.53 angstrum

## Atomic line spectra:

First we undersrand the concept of spectrum,

### Spectrum:

The arrangement is according to the wavelength of visible, UV, and infrared light.

### Types of the spectrum:

• An instrument designed for visual observation of spectrum is spectroscopy.
• An instrument that maps spectrum is spectroscopy.
• The study of the interaction between matter and electromagnetic radiation is called spectroscopy.

### History of spectroscopy:

The History of spectroscopy began with Newton’s optics experiments (1666-1672)

• Newton applied the word ”spectrum” to describe the rainbow of colors that combine to form white light and that are revealed when the white light is passed through a prism.
• In 1802, William Hyde Wallaston built an improved spectrometer that included a lens to focus the sun’s spectrum on a screen. Wallaston realized that color was not spread uniformly, but instead had missing patches of color, which appeared as dark bands in the spectrum.
• Later in 1815, German physicist Joseph Fraunhofer also examined the solar spectrum and found about 600 such dark lines (missing colors) which are known as Fraunhofer lines.

### Importance of spectroscopy:

Spectroscopy is a fundamental tool in the field of physics, chemistry, and astronomy. knowing the composition of atoms (different atoms show different spectrum), physical structure, and electronic structure of matter to be investigated at the atomic scale, molecule scale, macro scale, and over astronomical distances.

• It provides information not only about the arrangement and motion of the outer electrons (optical spectroscopy), but also the inner electrons( X-rays spectroscopy) and about the angular momentum, magnmeticquantum, distribution of charge and magnetism of nucleus.
• Atoms and molecules have unique spctra. As a result these spectra can be used to detect, identify and quantify information about the atoms and molecules.

### Things that are used to make spectrum:

1. Ligth source
2. Material
3. Screen

## Atomic line spectra:

”The spectrum of emitted radiations having specific wavelengths when current is passed through atomic gas at low pressure”

### Spectral series:

The spectrum of any element has some regularities. These are classified into certain groups called spectral series.

The first series → In 1885 → Balmer series → lies in the visible region

There are three regions of the atomic spectrum of hydrogen:

1. Visible region
2. UV region
3. Infrared region

### Balmer formula:

λ = 364.4 n2 /n2 −4 nm

where, n= 3,4,5….

Then, the Rydberg formula arranges this formula by reciprocal it:

1/λ = 1/364.4 × n2 − 4 /n2

### Rydberg formula:

1/λ = 27×106 [n2 /n2 − 4/n2]

1/λ = 27×106 [1 − 4/n2]

1/λ = 4 × 27×106 /4 [1 − 4/n2]

1/λ = 1.09 × 107 [1/4 − 1/n2]

1/λ = R [1/4 − 1/n2]

where R is the Rydberg constant ( 1.09 × 107 m-1 )

Re-write the above formula,

1/λ = R [1/p2 − 1/n2]

where always p<n

## Energy Level Diagram for Hydrogen:

In a hydrogen atom, when an electron is move in different energy levels (from lower energy level to higher energy) it emits or absorbs energy in the form of electromagnetic radiation. (energy is quantized)

En = − 13.6/n2 ev

### UV- region:

#### Series: Lyman series

The series of spectral lines produced due to transitions from all energy states to the ground state is called the Lyman series.

1/λ = R [1/p2 − 1/n2]

where p=1 and n=2,3,4,5….

##### when the longest wavelength is released?

The longest wavelength is released when n=2 and p=1

##### when a shorter wavelength is released?

The shorter wavelength is released when n=∞ and p=1

### Visible region:

#### Series: Balmer series

1/λ = R [1/p2 − 1/n2]

where p=2 and n=3,4,5,6….

##### when the longest wavelength is released?

The longest wavelength is released when n=3 and p=2

##### when a shorter wavelength is released?

The shorter wavelength is released when n=∞ and p=2

### Infrared region:

#### Series: Paschen series, Bracket series, Pfund series

• Paschen series:

The series of spectral lines produced due to transitions from all energy states to energy states E3 is called the Paschen series.

p=3 and n=4,5,6,7…..

Longest wavelength: when n=4 and p=3

Shortest wavelength: when n=∞ and p=3

• Bracket series:

The series of spectral lines produced due to transitions from all energy states to energy states E4 is called the Bracket series.

p=4 and n=5,6,7,8…..

Longest wavelength: when n=5 and p=4

Shortest wavelength: when n=∞ and p=4

• Pfund series:

The series of spectral lines produced due to transitions from all energy states to energy states E5 is called the Pfund series.

p=5 and n=6,7,8,9…..

Longest wavelength: when n=6 and p=5

Shortest wavelength: when n=∞ and p=5

#### Bilal kamboh

A pioneer in the Chemistry space, Bilal is the Content writer at UO Chemists. Driven by a mission to Success, Bilal is best known for inspiring speaking skills to the passion for delivering his best. He loves running and taking fitness classes, and he is doing strength training also loves outings.