Emission hydrogen spectrum

A+ A-

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.

ΔE = hν
or, ν = ΔE/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 = - 2π²me⁴/n²h².

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

Frequency of emitted light from the spectrum related to energy by plank equation
ν = ΔE/h
= (2π²me⁴/h³)[(1/n²) - (1/n²)].

∴ ν = R[(1/n²) - (1/n²)]
where R is the Rydberg constant.

This is the Rydberg equation for the measurement of the emission spectral region of the hydrogen atom.
frequency and wavelength of the hydrogen spectrum
Wavelength of the hydrogen spectrum
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.

λν = c and ν = c/λ = c⊽
Wavenumber, ⊽ = 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. 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.

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

The spectral region of hydrogen

Emission hydrogen spectrum region
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/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.

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/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.

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/λ = R[(1/n²) - (1/n²)]
n = 2 and n = 3, 4, 5, ...

Question
Which of the Balmer lines get spectrum fall in the visible region with wavelength 4000 to 7000 Å?(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 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.

⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
(⊽) = (1/4) × 10⁵ cm⁻¹.

∴ (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.
∴ n = 7 ( nearest whole number).

(⊽) = (1/7) × 10⁵ cm⁻¹
∴ (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.
n = 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 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.

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/λ = R[(1/n²) - (1/n²)]

n = 3 and n = 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/λ = R[(1/n²) - (1/n²)]

n = 4 and n = 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/λ = R[(1/n²) - (1/n²)]
n = 5 and n = 6, 7, 8, ...

Principles of the emission hydrogen spectrum, the wavelength of the spectrum of a hydrogen atom, the study of the spectral region of hydrogen

Study chemistry online

Contact us

Name

Email *

Message *

Powered by Blogger.