### Principles of the emission hydrogen spectrum

Measurements in the emission hydrogen spectrum are feasible because each hydrogen atom has definite quantum energy in which electron resides. Ordinarily, they are in the ground state with the lowest energy level of an atom.On the addition of energy by thermal or electrical one electron is removed to the higher energy level of the hydrogen atom.

However, this exited electron returns to the ground state and in this process extra energy emitted in the form of a photon.

The source of excitation influence the form and intensity of emission in such measurement. Through we get a line spectra of the hydrogen atom for spectroscopic study.

Photon energy and frequency of the emitted light from the hydrogen spectrum are connected by Plank relation.

Î”E = hÎ½

or, Î½ = Î”E/h

or, Î½ = Î”E/h

#### The wavelength of the spectrum of a hydrogen atom

The energy corresponding to a particular line in the**emission spectrum**of a

**hydrogen atom**is the energy difference between the ground level and the exited level.

Bohr's model provides the energy of an electron at a particular energy level. The energy of an electron in the hydrogen atom

En = - 2Ï€²me⁴/n²h².

∴ Î”E = E₂ - E₁

= - (2Ï€²me⁴/n

= (2Ï€²me⁴/h²)[1/n

∴ Î”E = E₂ - E₁

= - (2Ï€²me⁴/n

_{₁}²h²) - [- (2Ï€²me⁴/n_{₂}²h²)]= (2Ï€²me⁴/h²)[1/n

_{₁}² - 1/n_{₂}²]Frequency of emitted light from the spectrum related to energy by plank equation

Î½ = Î”E/h

= (2Ï€²me⁴/h³)[(1/n

∴ Î½ = R[(1/n

where R is the Rydberg constant.

= (2Ï€²me⁴/h³)[(1/n

_{₁}²) - (1/n_{₂}²)].∴ Î½ = R[(1/n

_{₁}²) - (1/n_{₂}²)]where R is the Rydberg constant.

This is the Rydberg equation for the measurement of the emission

**spectral region**of the hydrogen atom.

Wavelength of the hydrogen spectrum |

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⁻¹.

Î»Î½ = c and Î½ = c/Î» = c⊽

Wavenumber, ⊽ = Î½/c

where c is the velocity of light.

= {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.

Î»Î½ = c and Î½ = c/Î» = c⊽

Wavenumber, ⊽ = 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. Wavenumber (⊽) 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).

∴ R = (3.2898 × 1015 sec⁻¹)/(2.9979 × 1010 cm sec⁻¹).

= 109737 cm⁻¹

= 10973700 m⁻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 model.

### The spectral line in the hydrogen atom

Putting n = 1, n = 2, n = 3, etc in the 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 exhibited several series of lines.

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 normal conditions, the electron of each hydrogen atom remains in the ground state near the nucleus that is n = 1 (K - Shell).

When thermal or electrical 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.

Chemistry articles for school-college courses

- What is Chemical equilibrium?
- Law of mass action
- What is chemical kinetics?
- Study properties of gases online
- What is the heat capacity of gases?

### The spectral region of hydrogen

Emission hydrogen spectrum region |

#### The spectral region in 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 these spectral lines can be calculated by the following equation.⊽ = 1/Î» = R[(1/n

For Lyman series, we have n

When n

1/Î» = 109677{1 - (1/4)}

or, 1/Î» = {(109677 × 3)/4} cm⁻¹

∴ Î» = {4/(109677 × 3)}

= 1215 × 10⁻⁸ cm

= 1215 â„«.

When n

1/Î» = 109677{1 - (1/∞2)}

or, 1/Î» = 109677 cm⁻¹

∴ Î» = (1/(109677)

= 912 × 10⁻⁸ cm

= 912 â„«.

_{₁}²) - (1/n_{₂}²)]For Lyman series, we have n

_{₁}= 1 and n_{₂}= 2, 3, 4...When n

_{₂}= 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„«.

When n

_{₂}= ∞,1/Î» = 109677{1 - (1/∞2)}

or, 1/Î» = 109677 cm⁻¹

∴ Î» = (1/(109677)

= 912 × 10⁻⁸ cm

= 912 â„«.

Question

The wavenumber of the get spectrum of hydrogen in the 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/Î» = R[(1/n

For Lyman series n = 1 and wave Number (⊽) = 82200 cm⁻¹

Thus, 82200 = 109600[(1/1²) - (1/n

or, 1/n22 = 1 - (822/1096)

= 274/1096

=1/4

∴ n

_{₁}²) - (1/n_{₂}²)]For Lyman series n = 1 and wave Number (⊽) = 82200 cm⁻¹

Thus, 82200 = 109600[(1/1²) - (1/n

_{₂}²)]or, 1/n22 = 1 - (822/1096)

= 274/1096

=1/4

∴ n

_{₂}= 2.#### The spectral region in 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 these spectral lines can be calculated by the following equation.⊽ = 1/Î» = R[(1/n

n

_{₁}²) - (1/n_{₂}²)]n

_{₁}= 2 and n_{₂}= 3, 4, 5, ...Question

Which of the Balmer lines get spectrum fall in the visible region with wavelength 4000 to 7000 â„«?(Given R = 109737 cm⁻¹).

Answer

The Balmer series of the hydrogen spectrum get comprises 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.

⊽ = 1/Î» = 109677[(1/2²) - (1/n

_{₂}²)](⊽) = (1/4) × 10⁵ cm⁻¹.

∴ (1/4) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n

= 1.1 × 10⁵{(1/4) - (1/n

∴ 1/4 ⋍ 1.1{(1/4) - (1/n

or, n

∴ n

(⊽) = (1/7) × 10⁵ cm⁻¹

∴ (1/7) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n

= 1.1 × 10⁵{(1/4) - (1/n

∴ 1/7 ⋍ 1.1{(1/4) - (1/n

or, n

n

∴ (1/4) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n

_{₂}²)}= 1.1 × 10⁵{(1/4) - (1/n

_{₂}²)} cm⁻¹∴ 1/4 ⋍ 1.1{(1/4) - (1/n

_{₂}²)}or, n

_{₂}² ⋍ 44.∴ n

_{₂}= 7 ( nearest whole number).(⊽) = (1/7) × 10⁵ cm⁻¹

∴ (1/7) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n

_{₂}²)}= 1.1 × 10⁵{(1/4) - (1/n

_{₂}²)} cm⁻¹∴ 1/7 ⋍ 1.1{(1/4) - (1/n

_{₂}²)}or, n

_{₂}² ⋍ 9.n

_{₂}= 3 ( nearest whole number).The transition from n =7 to n = 3 while falling to n = 2 will generate the Balmer series.

Question

Calculate the wavelength of HÉ‘ and HÎ² of the Balmer series.

Answer

Transition energy of the Balmer lines, ⊽ = 1/Î» = 109677[(1/2²) - (1/n

_{₂}²)]

Again for HÉ‘ line n2 = 3 and for HÎ² line n2 = 4

Thus the wavelength 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 wavelength for HÎ² - line,

1/Î» = 109677 [(1/2²) - (1/4²)]

= 109677 × (3/16)

∴ Î» = 16/(3 × 109677)

= 4.863 × 10⁻⁵ cm

= 4863 â„«.

Thus the wavelength 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 wavelength for HÎ² - line,

1/Î» = 109677 [(1/2²) - (1/4²)]

= 109677 × (3/16)

∴ Î» = 16/(3 × 109677)

= 4.863 × 10⁻⁵ cm

= 4863 â„«.

#### The spectral region in Paschen series

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/Î» = R[(1/n

n

_{₁}²) - (1/n_{₂}²)]n

_{₁}= 3 and n_{₂}= 4, 5, 6, ...#### The spectral region in Bracket series

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/Î» = R[(1/n

n

_{₁}²) - (1/n_{₂}²)]n

_{₁}= 4 and n_{₂}= 5, 6, 7, ...#### The spectral region Pfund series

Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund series in the far-infrared spectrum region. The wavelength of these spectral lines can be calculated by the following equation,⊽ = 1/Î» = R[(1/n

n

_{₁}²) - (1/n_{₂}²)]n

_{₁}= 5 and n_{₂}= 6, 7, 8, ...