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__Rydberg Equation:__

The __Rydberg Equation:__

__emission spectrum of hydrogen atom was determined experimentally and it was possible to relate the frequencies of different spectral lines by simple equation__,

**⋎ = R [1/nı² - 1/nıı²]**

Where

**n₁**and**n₂**__are integers__and**R**__is a constant, called Rydberg constant__after the name of the discoverer.
The existence of discrete spectral emission could hardly be explained by any model at Rydberg's time.

The Energy associated with the permitted orbits is given by ,

**En**= - (1/2)(e²/r) where r = n²h²/4𝜋²me²

Thus,

**En = - (2π²me⁴/n²h²)**
Again E₁ = - (2π²me⁴/h²)

∴

**En = E₁/n²**
Thus the values of E₂, E₃, E₄, E₅ etc. in terms of E₁(

**) are given as,**__- 13.6 eV__
E₂ = -13.6/2² = - 3.4 eV

E₃ = -13.6/3² = - 1.51 eV

E₃ = -13.6/3² = - 1.51 eV

E₄ = -13.6/4² = - 0.85 eV

E₅ = -13.6/5² = - 0.54 eV

__As n increases the energy becomes less negative and hence the system becomes less stable.__

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**Explanation of the Rydberg Equation:**

**Explanation of the Rydberg Equation:**

The explanation of Rydberg equation is now simple.

__The most stable orbit must be the one with the lowest energy (ground state).__
When n = 1

__the energy of an electron is lowest__and it__the lowest energy state__and the states with n 〉 1 are the__higher energy exited states__.__Provided the right amount of energy is supplied the electron may jump from n = 1 to another higher level__.
Conversely energy will be released in the form of light of

__definite frequency__when the exited electron returns to its ground state. Since energy and frequency of the emitted light are connected by__Plank Relation__,**E = hν**

**or, ν = E/h**

__The energy corresponding to a particular line in the emission spectrum of hydrogen atom is the energy difference between the initial state I and the final state II__.

So that, E = Eıı - Eı

= - (2π²me⁴/nıı²) - {-(2π²me⁴/nı²)}

=

**(2π²me⁴/h²){(1/nı²) - (1/nıı²)}**
Then

**ν**corresponding to the energy**E**is given by,**ν = E/h = (2π²me⁴/h³)[1/nı² - 1/nıı²] = R[1/nı² - 1/nıı²]**

__Where R is the Rydberg Constant.__

__This relation tells us that as we go to higher and still higher orbits the energy gap between two immediate orbits continues to decreases.__

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__Value of the Rydberg Constant:__

__Value of the Rydberg Constant:__

Using the value of 9.108 × 10⁻²⁸ gm for the

__mass of an electron__, the__Bohr Theory__Predicts the following value of__Rydberg Constant__.
R = 2π² me⁴/h³

= {2 × (3.1416)² × (9.108 × 10⁻²⁸ gm) × (4.8 × 10⁻¹⁰ esu)}/(6.627 × 10⁻²⁷ erg sec)³

__= 3.2898 × 10¹⁵ cycles/sec__
Evaluate R in terms of wave number (

*instead of frequency. Wave number and frequency are connected with,***⊽**)
λν = c and ν = c/λ = c

**⊽**
Thus

*⊽*= ν/c
where c

__is the velocity of light__. Recall the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second. Wave number (**) on the other hand stands for the number of waves connected in unit length that is per centimeter (***⊽***cm⁻¹**) or per metre (**m⁻¹**).
Thus

####

Putting n = 1, n = 2, n = 3, etc in

The

Hydrogen atom contains only one electron but its spectrum gives a large number of lines - Explain.

This is because any given

####

Transition to the

The

**R**= 3.2898 × 10¹⁵ sec⁻¹/2.9979 × 10¹⁰ cm sec⁻¹ =**=**__109737 cm⁻¹____10973700 m⁻¹____The experimental values of R is__.**109677 cm⁻¹**(**10967700 m⁻¹**) showing a remarkable agreement between experiment and Bohr's Theory####
__Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom__:

Putting n = 1, n = 2, n = 3, etc in __Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom__:

__Rydberg Equation we get the energies of the different stationary states for the hydrogen electron. The transition energies can be calculated from the Rydberg Equation.__The

__experimental spectra of hydrogen atom also exhibited several series of lines__.Energy Level Diagram for the Hydrogen Spectrum |

__Problem:__Hydrogen atom contains only one electron but its spectrum gives a large number of lines - Explain.

__Answer:__This is because any given

__sample of hydrogen contains almost infinite number of atoms__. Under the normal conditions the electron of the__each hydrogen atom remains in the ground state__near the nucleus that is n = 1 (K - Shell). When energy is supplied to this sample of hydrogen gas by passing an electron discharge through the gas in the discharge tube,__individual atom absorbed different amounts of energy__.__Some of atoms absorbed such energy to shift their electron to third energy level (n = 3)__while some others may__absorbed large amount of energy to shift their electron__to the__Fourth__(n = 4),__Fifth__(n = 5),__Sixth__(n = 6) and__Seventh__(n = 7) energy level. The electrons in the__higher energy level are relatively unstable__and__hence drop back to the lower energy level with the emission of energy__in the__form of line spectrum containing various lines in of particular frequency and wave length__.####
__Lyman Series:__

Transition to the __Lyman Series:__

__ground state__(n = 1)__from n = 2, n = 3, n = 4, etc exited states constitute the Lyman Series in the ultraviolet region.__The

__wavelength__of this spectral lines can be calculated by the following equation,*=*

**⊽****1/λ**= 109677 {(1/

**n**ı²) - (1/

**n**ıı²)}

For Lyman Series, we have nı = 1 and nıı = 2, 3, 4 ....

Thus, (i) When nıı = 2, 1/λ = 109677{1 - (1/4)} = {(109677 × 3)/4} cm⁻¹

∴ λ = {4/(109677 × 3)} = 1215 × 10⁻⁸ cm =

**1215 Å**
(ii) When nıı = ∞, 1/λ = 109677{1 - (1/∞²)} = 109677 cm⁻¹

∴ λ = (1/(109677) = 912 × 10⁻⁸ cm =

####

Transition to the n = 2 level

**912 Å**

__Problem:__
The wave number of the experimental spectra of hydrogen atom in Lyman Series is 82200 cm⁻¹. Show that the transition occur to the Ground state (n =1) from n = 2. (R = 109600 cm⁻¹)

__Answer:__
We know that, Wave number (

**⊽**) = R(1/nı² - 1/nıı²),
For Lyman series n = 1 and Wave Number (⊽) = 82200 cm⁻¹

Thus, 82200 = 109600(1/1² - 1/nıı²)

or, 1/nıı² = 1 - 822/1096 = 274/1096 =1/4

∴ nıı = 2

__Thus the transition occurs to the ground state (n = 1) from n =2__.

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__Balmer Series:__

Transition to the n = 2 level __Balmer Series:__

__from n = 3, n = 4, n = 5, etc exited states constitute the Balmer Series in the visible region.__

The

__wavelength__of this spectral lines can be calculated by the following equation,*=*

**⊽****1/λ**= 109677 {(1/

**n**ı²) - (1/

**n**ıı²)}

__Problem:__
Which of the Balmer lines fall in the visible region of the spectrum wave length 4000 to 7000 Å ? (Given R = 109737 cm⁻¹

__Answer:__
The Balmer Series of the hydrogen spectrum comprise the transition from n 〉 2 levels to the n = 2 level.

4000 Å = 4000 × 10⁻⁸ cm = 4 × 10⁻⁵ cm and 7000 Å = 7 × 10⁻⁵ cm

Therefore wave number (

*⊽*) = 1/λ = (1/4) × 10⁵ to (1/7) × 10⁵ cm⁻¹
Transition energy of the Balmer lines,

*=*

**⊽****1/λ**= 109937 {(1/

**2**²) - (1/

**n**ıı²)}

Taking (⊽) = (1/4) × 10⁵ cm⁻¹ we have:

(1/4) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/nıı²)} = 1.1 × 10 ⁵{(1/4) - (1/nıı²)} cm⁻¹

∴ 1/4 ⋍ 1.1 {(1/4) - (1/nıı²)}

or, nıı ⋍ 44

So

**nıı = 7**( nearest whole number).
Taking (⊽) = (1/7) × 10⁵ cm⁻¹ we have:

(1/7) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/nıı²)} = 1.1 × 10 ⁵{(1/4) - (1/nıı²)} cm⁻¹

∴ 1/7 ⋍ 1.1 {(1/4) - (1/nıı²)}

or, nıı ⋍ 7.2

So

**nıı = 3**( nearest whole number).__So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.__

__Problem:__
Calculate the wave length of H⍺ and Hᵦ of the Balmer Series.

__Answer:__
We know that, for the Balmer Series 1/λ = R {(1/2²) - (1/nıı²)}

Again for H⍺ line nıı = 3 and for Hᵦ line nıı = 4

Thus the wave length for H⍺ - line,

1/λ = 109677 {(1/2²) - (1/3²)} = 109677 × (5/36)

∴ λ = 36/(5 × 109677) =

**6.564 × 10⁻⁵ cm**=**6564 Å**
Again the wave length for Hᵦ - line,

1/λ = 109677 {(1/2²) - (1/4²)} = 109677 × (3/16)

∴ λ = 16/(3 × 109677) =

####

**4.863 × 10⁻⁵ cm**=**4863 Å**####
__Pschen Series:__

__Pschen Series:__

Transition to the n = 3 level

__from n = 4, n = 5, n = 6, etc exited states constitute the Balmer Series in the Near Infrared region__.
The

####

__wavelength__of this spectral lines can be calculated by the following equation,*=*

**⊽****1/λ**= 109677 {(1/

**3**²) - (1/

**n**ıı²)}

####
__Brackett Series:__

__Brackett Series:__

Transition to the n = 4 level

__from n = 5, n = 6, n = 7, etc exited states constitute the Balmer Series in the Far Infrared region__.
The

####

__wavelength__of this spectral lines can be calculated by the following equation,*=*

**⊽****1/λ**= 109677 {(1/

**4**²) - (1/

**n**ıı²)}

####
__Pfund Series:__

__Pfund Series:__

Transition to the n = 5 level

__from n = 6, n = 7, n = 8, etc exited states constitute the Balmer Series in the Infrared region__.
The

For this case, Centrifugal force = Electronic Force of Attraction

__wavelength__of this spectral lines can be calculated by the following equation,*=*

**⊽****1/λ**= 109677 {(1/

**5**²) - (1/

**n**ıı²)}

Electronic Transitions and values of n₁ and n₂ for various spectral lines of Hydrogen Atom |

__A General of an Electron Moving about a Nucleus of Charge +Ze:__For this case, Centrifugal force = Electronic Force of Attraction

∴ mv²/r = Z e²/r²

Thus the

__Radius of the Bohr's Orbit (r)__= n²h²/4π²mZe²__And the Energy of an Electron__,

E = - (1/2) Ze²/r = - (2π²mZ²e⁴)/(n²h²)

__Related Post:__

__Rutherford's Atomic Model____Structure of Atom__