**was determined experimentally and it was possible to relate the frequencies of different spectral lines by a simple equation known as**

*Hydrogen spectrum***.**

*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,

Î”E = hÎ½

or, Î½ = Î”E/h

or, Î½ = Î”E/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 = - 2Ï€²me⁴/n²h²

So that, Î”E= E₂ - E₁

= - (2Ï€²me⁴/n

= (2Ï€²me⁴/h²)[1/n

= - (2Ï€²me⁴/n

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

_{₁}² - 1/n_{₂}²]Then Î½ corresponding to the energy E is given by,

Î½ = Î”E/h

= (2Ï€²me⁴/h³)[(1/n

= R[(1/n

Where R is the Rydberg Constant.

Î½ = Î”E/h

= (2Ï€²me⁴/h³)[(1/n

_{₁}²) - (1/n_{₂}²)]= R[(1/n

_{₁}²) - (1/n_{₂}²)]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 = 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

= {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,

Î»Î½ = c and Î½ = c/Î» = c⊽

Thus, ⊽ = Î½/c

where c is the velocity of light.

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 meter (m⁻¹).

Thus, 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 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

**of the atom also exhibited several series of lines.**

*hydrogen spectrum**Hydrogen spectrum series*

*Hydrogen spectrum series*

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/Î» = R[(1/n

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

_{₁}= 1 and n_{₂}= 2, 3, 4...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„«

_{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

1/Î» = 109677{1 - (1/∞2)}

or, 1/Î» = 109677 cm⁻¹

∴ Î» = (1/(109677)

= 912 × 10⁻⁸ cm

= 912 â„«

_{₂}= ∞,1/Î» = 109677{1 - (1/∞2)}

or, 1/Î» = 109677 cm⁻¹

∴ Î» = (1/(109677)

= 912 × 10⁻⁸ cm

= 912 â„«

*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/Î» = R[(1/n

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

_{₁}= 2 and n_{₂}= 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/Î» = R[(1/n

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

_{₁}= 3 and n_{₂}= 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/Î» = R[(1/n

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

_{₁}= 4 and n_{₂}= 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/Î» = R[(1/n

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

_{₁}= 5 and n_{₂}= 6, 7, 8, ...Hydrogen spectrum |

*Hydrogen Spectrum questions answers*

*Hydrogen Spectrum questions answers*

*Question*

- Why a large number of lines appear in the

**although it contains only one electron?**

*hydrogen spectrum**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 wavenumber of the experimental

**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)**

*hydrogen spectrum**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*Question*

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

*Answer*

- The Balmer Series of the

**comprises the transition from n 〉2 levels to the n = 2 level.**

*hydrogen spectrum*- 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/n

_{₂}²)]

Taking (⊽) = (1/4) × 10⁵ cm⁻¹

We have,(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

So n

We have,(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 ⋍ 44So n

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

We have,(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

So n

We have,(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 ⋍ 9So n

_{₂}= 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 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„«

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„«

1/Î» = 109677 [(1/2²) - (1/4²)]

= 109677 × (3/16)

∴ Î» = 16/(3 × 109677)

= 4.863 × 10⁻⁵ cm

= 4863 â„«

- More questions answers on

*hydrogen spectrum*