# Rydberg Equation

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/n12 - 1/n22]
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.

### Energy of an Electron in Bohr Orbits:

The Energy associated with the permitted orbits is given by ,
En = - 1/2(e2/r), where r = n2h2/4Ï€2me2
Thus, En = - (2Ï€2me4/n2h2)
Again E1 = - (2Ï€2me4/h2)
 ∴ En = E1/n2
Thus the values of E2, E3, E4, E5 etc. in terms of E1(- 13.6 eV) are given as,
 E2 = -13.6/22 = - 3.4 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.

### 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 = 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 initial state 1 and the final state 2.
So that, E= E2 - E1
= - (2Ï€2me4/n12h2) - [- (2Ï€2me4/n22h2)]
= (2Ï€2me4/h2)[1/n12 - 1/n22]
Then Î½ corresponding to the energy E is given by,
Î½ = E/h = (2Ï€2me4/h3)[1/n12 - 1/n22]
 ∴ Î½ = R[1/n12 - 1/n22]
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.
• 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€2me4/h3)
= {2 × (3.1416)2 × (9.108 × 10-28 gm) × (4.8 × 10-10 esu)4}/(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,
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-1) or per metre (m-1).
Thus, 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 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 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 Lyman Series in the ultraviolet region.
The wavelength of this spectral lines can be calculated by the following equation,
 ⊽ = 1/Î» = R[1/n12 - 1/n22]
For Lyman Series, we have n1 = 1 and n2 = 2, 3, 4 ....
 Thus, (i) When n2 = 2, 1/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 n2 = ∞, 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 Balmer Series in the visible region.
The wavelength of this spectral lines can be calculated by the following equation,
 ⊽ = 1/Î» = 109677[(1/22) - (1/n22)]
• Pschen Series:
Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Pschen Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
 ⊽ = 1/Î» = 109677[(1/32) - (1/n22)]
• Brackett Series:
Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Brackett Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
 ⊽ = 1/Î» = 109677[(1/42) - (1/n22)]
• Pfund Series:
Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
 ⊽ = 1/Î» = 109677[(1/52) - (1/n22)]
 Series of Lines n1 n2 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-1. Show that the transition occur to the Ground state (n =1) from n = 2.
(R = 109600 cm-1)
We know that, Wave number,
= R[1/n12 - 1/n22]
For Lyman series n = 1 and
Wave Number () = 82200 cm-1
Thus, 82200 = 109600(1/12 - 1/n22)
or, 1/n22 = 1 - (822/1096)
= 274/1096
=1/4
∴ n2 = 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 × 10-8 cm = 4 × 10-5 cm and 7000 â„« = 7 × 10-5 cm
Therefore wave number,
= 1/Î» = (1/4) × 105 to (1/7) × 105 cm⁻¹
Transition energy of the Balmer lines,
 ⊽ = 1/Î» = 109677[(1/22) - (1/n22)]
Taking () = (1/4) × 105 cm-1
We have,
(1/4) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/4 ⋍ 1.1{(1/4) - (1/n22)}
or, n2 ⋍ 44
So n2 = 7 ( nearest whole number).
Taking () = (1/7) × 105 cm-1
We have,
(1/7) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/7 ⋍ 1.1 {(1/4) - (1/n22)}
or, n2 ⋍ 3
So n2 = 3 ( nearest whole number).
So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.
Calculate the wave length of HÉ‘ and HÎ² of the Balmer Series.
We know that, for the Balmer Series,
 ⊽ = 1/Î» = 109677[(1/22) - (1/n22)]
Again for HÉ‘ line n2 = 3 and for HÎ² line n2 = 4
Thus the wave length for HÉ‘ - line,
1/Î» = 109677 {(1/22) - (1/32)}
= 109677 × (5/36)
Î» = 36/(5 × 109677)
= 6.564 × 10-5 cm
= 6564 â„«
Again the wave length for HÎ² - line,
1/Î» = 109677 {(1/22) - (1/42)}
= 109677 × (3/16)
Î» = 16/(3 × 109677)
= 4.863 × 10-5 cm
= 4863 â„«

Rydberg Equation: Energy of an Electron in Bohr Orbits, Explanation of the Rydberg Equation, Value of the Rydberg Constant and Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom(Lyman Series, Balmer Series, Pschen Series, Brackett Series and Pfund Series)

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