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

Î½ = R[^{1}/n_{1}^{2} - ^{1}/n_{2}^{2}] |

Where

**n**and_{1}**n**are integers and_{2}**R**is a constant, called**after the name of the discoverer. The existence of discrete spectral emission could hardly be explained by any model at Rydberg's time.**__Rydberg constant__###
__Energy of an Electron in Bohr Orbits:__

__Energy of an Electron in Bohr Orbits:__

The Energy associated with the permitted orbits is given by ,

**E**, where r = n

_{n}= - 1/2(e^{2}/r)^{2}h

^{2}/4Ï€

^{2}me

^{2}

Thus,

**E**

_{n}= - (2Ï€^{2}me^{4}/n^{2}h^{2})Again

**E**= - (2Ï€

_{1}^{2}me

^{4}/h

^{2})

∴ E_{n} = E_{1}/n^{2} |

Thus the values of

**E**,_{2}**E**,_{3}**E**,_{4}**E**etc. in terms of_{5}**E**(_{1}**- 13.6 eV**) are given as,E_{2} = -13.6/2^{2} = - 3.4 eV |

E_{3} = -13.6/3^{2} = - 1.51 eV |

E_{4} = -13.6/4^{2} = - 0.85 eV |

E_{5} = -13.6/5^{2} = - 0.54 eV |

As

**n**increases the energy becomes less negative and hence the system becomes less stable.###
__Explanation of the Rydberg Equation:__

__Explanation of the Rydberg Equation:__

The explanation of

When

Provided the right amount of energy is supplied the electron may jump from

Since energy and frequency of the emitted light are connected by

**is now simple. The most stable orbit must be the one with the lowest energy (**__Rydberg equation__**).**__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 = hÎ½ or, Î½ = E/h |

The energy corresponding to a particular line in the emission spectrum of hydrogen atom is the energy difference between the

So that,

= - (2Ï€

= (2Ï€

Then Î½ corresponding to the energy E is given by,

**initial state 1**and the**final state 2**.So that,

**E= E**_{2}- E_{1}= - (2Ï€

^{2}me^{4}/n_{1}^{2}h^{2}) - [- (2Ï€^{2}me^{4}/n_{2}^{2}h^{2})]= (2Ï€

^{2}me^{4}/h^{2})[1/n_{1}^{2}- 1/n_{2}^{2}]Then Î½ corresponding to the energy E is given by,

**Î½**= E/h = (2Ï€^{2}me^{4}/h^{3})[1/n_{1}^{2}- 1/n_{2}^{2}]∴ Î½ = R[1/n_{1}^{2} - 1/n_{2}^{2}] |

Where

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

**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.

__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 = (2Ï€**

^{2}me^{4}/h^{3})= {2 × (3.1416)

^{2}× (9.108 × 10

^{-28}gm) × (4.8 × 10

^{-10}esu)

^{4}}/(6.627 × 10

^{-27}erg sec)

^{3}

= 3.2898 × 10^{15} cycles/sec |

Evaluate

**R**in terms of wave number (**⊽**) instead of frequency. Wave number and frequency are connected with,**Î»Î½ = c**and

**Î½ = c/Î» = c⊽**

Thus,

**⊽ = Î½/c**

where

**c**is the velocity of light.

Recall the

**frequency (Î½)**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**) or per metre (^{-1}**m**).^{-1}
Thus,

**R**= 3.2898 × 10^{15}sec^{-1}/2.9979 × 10^{10}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:__

__Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom:__

Putting

**n = 1**,**n = 2**,**n = 3**, etc in**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 |

**Hydrogen atom**contains only

**one electron**but its spectrum gives a

**large number**of lines - Explain.

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

When energy is supplied to this sample of

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.

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 ground state (

**n = 1**) from**n = 2**,**n = 3**,**n = 4**, etc exited states constitute the**in the ultraviolet region.**__Lyman Series__
The wavelength of this spectral lines can be calculated by the following equation,

⊽ = 1/Î» = R[1/n_{1}^{2} - 1/n_{2}^{2}] |

For

**Lyman Series**, we have n_{1}= 1 and n_{2}= 2, 3, 4 ....Thus, (i) When n,_{2} = 21/Î» = 109677{1 - (1/4)}or, 1/Î» = {(109677 × 3)/4} cm^{-1}∴ Î» = {4/(109677 × 3)}= 1215 × 10 ^{-8} cm= 1215 â„« |

Thus, (i) When n,_{2} = ∞1/Î» = 109677{1 - (1/∞^{2})}or, 1/Î» = 109677 cm^{-1}∴ Î» = (1/(109677)= 912 × 10 ^{-8} cm= 912 â„« |

__Balmer Series:__

Thus the transition occurs to the ground state (

**n = 1**) from**n =2**.
Thus the transition occurs to the ground state (

**n = 1**) from**n =2**.
Transition to the

**n = 2**level from**n = 3**,**n = 4**,**n = 5**, etc exited states constitute the**in the visible region.**__Balmer Series__
The wavelength of this spectral lines can be calculated by the following equation,

⊽ = 1/Î» = 109677[(1/2^{2}) - (1/n_{2}^{2})] |

__Pschen Series:__

Transition to the

**n = 3**level from**n = 4**,**n = 5**,**n = 6**, etc exited states constitute the**in the Near Infrared region.**__Pschen Series__
The wavelength of this spectral lines can be calculated by the following equation,

⊽ = 1/Î» = 109677[(1/3^{2}) - (1/n_{2}^{2})] |

__Brackett Series:__

Transition to the

The wavelength of this spectral lines can be calculated by the following equation,

**n = 4**level from**n = 5**,**n = 6**,**n = 7**, etc exited states constitute the**in the Near Infrared region.**__Brackett Series__The wavelength of this spectral lines can be calculated by the following equation,

⊽ = 1/Î» = 109677[(1/4^{2}) - (1/n_{2}^{2})] |

__Pfund Series:__

Transition to the

**n = 5**level from**n = 6**,**n = 7**,**n = 8**, etc exited states constitute the**in the Near Infrared region.**__Pfund Series__
The wavelength of this spectral lines can be calculated by the following equation,

⊽ = 1/Î» = 109677[(1/5^{2}) - (1/n_{2}^{2})] |

Series of Lines |
n_{1} |
n_{2} |
Spectral Region |
Wavelength |

Lyman Series | 1 | 2,3,4..etc | Ultraviolet | ã„‘4000 â„« |

Balmer Series | 2 | 3,4,5..etc | Visible | 4000 â„« to 7000 â„« |

Paschen Series | 3 | 4,5,6..etc | Near Infrared | 〉7000 â„« |

Brackett Series | 4 | 5,6,7..etc | Far Infrared | 〉7000 â„« |

Pfund Series | 5 | 6,7,8..etc | Far Infrared | 〉7000 â„« |

The

**wave number**of the experimental spectra of hydrogen atom in**Lyman Series**is**82200 cm**. Show that the transition occur to the^{-1}**Ground state**(**n =1**) from**n = 2**.**(R = 109600 cm**^{-1})
We know that, Wave number,

For Lyman series

Wave Number (

Thus, 82200 = 109600(1/1

or, 1/n

= 274/1096

=1/4

**⊽ = R[1/n**_{1}^{2}- 1/n_{2}^{2}]For Lyman series

**n = 1**andWave Number (

**⊽**) = 82200 cm^{-1}Thus, 82200 = 109600(1/1

^{2}- 1/n_{2}^{2})or, 1/n

_{2}^{2}= 1 - (822/1096)= 274/1096

=1/4

**∴ n**_{2}= 2
Which of the Balmer lines fall in the visible region of the spectrum wave length

**4000**to**7000 â„«**? (Given**R = 109737 cm⁻¹**)
The Balmer Series of the hydrogen spectrum comprise the transition from

**n 〉2**levels to the**n = 2**level.
4000 â„« = 4000 × 10

Therefore wave number,

Transition energy of the Balmer lines,

^{-8}cm = 4 × 10^{-5}cm and 7000 â„« = 7 × 10^{-5}cmTherefore wave number,

**⊽**= 1/Î» = (1/4) × 10^{5}to (1/7) × 10^{5}cm⁻¹Transition energy of the Balmer lines,

⊽ = 1/Î» = 109677[(1/2^{2}) - (1/n_{2}^{2})] |

Taking (

We have,

(1/4) × 10

= 1.1 × 10

∴ 1/4 ⋍ 1.1{(1/4) - (1/n

or, n

So

Taking (

We have,

(1/7) × 10

= 1.1 × 10

∴ 1/7 ⋍ 1.1 {(1/4) - (1/n

or, n

So

So all transition from

**⊽**) = (1/4) × 10^{5}cm^{-1}We have,

(1/4) × 10

^{5}cm^{-1}= 109737{(1/4) - (1/n_{2}^{2})}= 1.1 × 10

^{5}{(1/4) - (1/n_{2}^{2})} cm^{-1}∴ 1/4 ⋍ 1.1{(1/4) - (1/n

_{2}^{2})}or, n

_{2}⋍ 44So

**n**( nearest whole number)._{2}= 7Taking (

**⊽**) = (1/7) × 10^{5}cm^{-1}We have,

(1/7) × 10

^{5}cm^{-1}= 109737{(1/4) - (1/n_{2}^{2})}= 1.1 × 10

^{5}{(1/4) - (1/n_{2}^{2})} cm^{-1}∴ 1/7 ⋍ 1.1 {(1/4) - (1/n

_{2}^{2})}or, n

_{2}⋍ 3So

**n**( nearest whole number)._{2}= 3So all transition from

**n =7**to**n = 3**while falling to the**n = 2**will generate Balmer series.
Calculate the wave length of

**H**and_{É‘}**H**of the Balmer Series._{Î²}
We know that, for the Balmer Series,

⊽ = 1/Î» = 109677[(1/2^{2}) - (1/n_{2}^{2})] |

Again for

Thus the wave length for

= 109677 × (5/36)

∴

= 6.564 × 10

Again the wave length for

= 109677 × (3/16)

∴

= 4.863 × 10

**H**line_{É‘}**n**and for_{2}= 3**H**line_{Î²}**n**_{2}= 4Thus the wave length for

**H**- line,_{É‘}**1/Î»**= 109677 {(1/2^{2}) - (1/3^{2})}= 109677 × (5/36)

∴

**Î»**= 36/(5 × 109677)= 6.564 × 10

^{-5}cm**= 6564 â„«**Again the wave length for

**H**- line,_{Î²}**1/Î»**= 109677 {(1/2^{2}) - (1/4^{2})}= 109677 × (3/16)

∴

**Î»**= 16/(3 × 109677)= 4.863 × 10

^{-5}cm**= 4863 â„«**