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Nov 5, 2018

Rydberg Equation

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,
⋎ = 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(- 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 = 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.

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 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 .
Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom
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 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/nı²) - (1/nıı²)}
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 = 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.

Balmer Series:

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/λ = 109677 {(1/nı²) - (1/nıı²)}
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 {(1/2²) - (1/nıı²)}
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:

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 {(1/3²) - (1/nıı²)}

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 {(1/4²) - (1/nıı²)}

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 wavelength of this spectral lines can be calculated by the following equation,
 = 1/λ = 109677 {(1/5²) - (1/nıı²)}
Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom
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
E = - (1/2) Ze²/r = - (2π²mZ²e⁴)/(n²h²)
Related Post:
Rutherford's Atomic Model

Structure of Atom