Hydrogen spectrum Rydberg equation

Hydrogen spectrum was determined experimentally and it was possible to relate the frequencies of different spectral lines by a simple equation known as Rydberg Equation.
Hydrogen spectrum and Rydberg equation
Rydberg equation

Rydberg equation derivation

    The explanation of the 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 excited 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 excited electron returns to its ground state.
    Since the energy and frequency of the emitted light are connected by Plank Relation,
ΔE = hν
or, ν = ΔE/h
    The energy corresponding to a particular line in the emission spectrum of a hydrogen atom is the energy difference between the initial state 1 and the final state 2.
    From the Bohr's model of hydrogen atom energy of an electron in a particular energy level,
En = - 2π²me⁴/n²h²

So that, ΔE= E₂ - E₁
= - (2π²me⁴/n²h²) - [- (2π²me⁴/n²h²)]
= (2π²me⁴/h²)[1/n² - 1/n²]

Then ν corresponding to the energy E is given by,
ν = ΔE/h
= (2π²me⁴/h³)[(1/n²) - (1/n²)]

= R[(1/n²) - (1/n²)]
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.

Rydberg Constant

    Using the mass of an electron =9.108 × 10⁻²⁸ gm, Bohr Theory Predicts the following value of Rydberg Constant.
R = 2π²me⁴/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⁻¹
    Evaluate R in terms of wavenumber (⊽) instead of frequency. Wavenumber 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 meter (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 are 109677 cm⁻¹ (10967700 m⁻¹) showing a remarkable agreement between the experiment and Bohr's theory.

Hydrogen spectrum

    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 hydrogen spectrum of the atom also exhibited several series of lines.

Hydrogen spectrum series

Hydrogen Spectrum Rydberg equation
Energy Level Diagram for the Hydrogen Spectrum

Lyman series spectrum

    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 these spectral lines can be calculated by the following equation,
⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Lyman Series, we have n = 1 and n = 2, 3, 4...

When n = 2,
1/λ = 109677{1 - (1/4)}
or, 1/λ = {(109677 × 3)/4} cm⁻¹
∴ λ = {4/(109677 × 3)}
= 1215 × 10⁻⁸ cm
= 1215 Å

When n = ∞,
1/λ = 109677{1 - (1/∞2)}
or, 1/λ = 109677 cm⁻¹
∴ λ = (1/(109677)
= 912 × 10⁻⁸ cm
= 912 Å

Balmer series spectrum

    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 these spectral lines can be calculated by the following equation,
⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Balmer Series, we have n = 2 and n = 3, 4, 5, ...

Paschen series spectrum

    Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Paschen Series in the near infrared region. The wavelength of these spectral lines can be calculated by the following equation,
⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Paschen Series, we have n = 3 and n =  4, 5, 6, ...

Bracket series spectrum

    Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Bracket Series in the far infrared region. The wavelength of these spectral lines can be calculated by the following equation,
⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Bracket Series, we have n = 4 and n = 5, 6, 7, ...

Pfund series spectrum

    Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund Series in the far infrared region. The wavelength of these spectral lines can be calculated by the following equation,
⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Pfund Series, we have n = 5 and n = 6, 7, 8,  ...
Hydrogen spectrum series
Hydrogen spectrum

Hydrogen Spectrum questions answers

  • Question
    Why a large number of lines appear in the hydrogen spectrum although it contains only one electron?
  • Answer
    This is because any given sample of hydrogen contains an almost infinite number of atoms.
    Under the normal conditions, the electron of 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 absorb a 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 wavelength.
  • Question
    The wave number of the experimental hydrogen spectrum in Lyman Series is 82200 cm⁻¹. Show that the transition occurs to the Ground state (n =1) from n = 2. (R = 109600 cm⁻¹)
  • Answer
We know that wavenumber,⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Lyman series n = 1 and wave Number (⊽) = 82200 cm⁻¹
Thus, 82200 = 109600[(1/1²) - (1/n²)]
or, 1/n22 = 1 - (822/1096)
= 274/1096
=1/4
∴ n = 2
  • Question
    Which of the Balmer lines fall in the visible region of the spectrum wavelength 4000 to 7000 Å?(Given R = 109737 cm⁻¹)
  • Answer
    The Balmer Series of the hydrogen spectrum comprises 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 = 109677[(1/2²) - (1/n²)]
Taking (⊽) = (1/4) × 10⁵ cm⁻¹
We have,(1/4) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n²)}
= 1.1 × 10⁵{(1/4) - (1/n²)} cm⁻¹
∴ 1/4 ⋍ 1.1{(1/4) - (1/n²)}
or, n² ⋍ 44
So n = 7 ( nearest whole number).

Taking (⊽) = (1/7) × 10⁵ cm⁻¹
We have,(1/7) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n²)}
= 1.1 × 10⁵{(1/4) - (1/n²)} cm⁻¹
∴ 1/7 ⋍ 1.1{(1/4) - (1/n²)}
or, n² ⋍ 9
So n = 3 ( nearest whole number).
    So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.
  • Question
    Calculate the wavelength of Hɑ and Hβ of the Balmer Series.
  • Answer
    Transition energy of the Balmer lines, ⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
Again for Hɑ line n2 = 3 and for Hβ line n2 = 4
Thus the wavelength for Hɑ - line,
1/λ = 109677 [(1/2²) - (1/3²)]
= 109677 × (5/36)
∴ λ = 36/(5 × 109677)
= 6.564 × 10⁻⁵ cm
= 6564 Å

Again the wavelength for Hβ - line,
1/λ = 109677 [(1/2²) - (1/4²)]
= 109677 × (3/16)
∴ λ = 16/(3 × 109677)
= 4.863 × 10⁻⁵ cm
= 4863 Å
    More questions answers on hydrogen spectrum

Rydberg equation, Rydberg constant, hydrogen spectrum series, Lyman, Balmer, Paschen, Bracket, and Pfund series spectrum with questions answers

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