# Emission hydrogen spectrum

### Emission line in the 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.

${\color{Blue}&space;\Delta&space;H=h\nu&space;}$

${\color{Blue}&space;or,&space;\nu&space;=\frac{\Delta&space;E}{h}}$

#### Wavelengths of the hydrogen spectrum

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

${\color{Blue}E&space;_{n}=\frac{-2\pi&space;^{2}me^{4}}{n^{2}h^{2}}}$

${\color{Blue}&space;\therefore&space;\Delta&space;E=E_{2}-E_{1}}$

${\color{Blue}=\frac{-2\pi&space;^{2}me^{4}}{n_{2}^{2}h^{2}}-\left&space;(&space;\frac{-2\pi&space;^{2}me^{4}}{n_{1}^{2}h^{2}}&space;\right&space;)}$

${\color{Blue}&space;=\frac{2\pi&space;^{2}me^{4}}{h^{2}}\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

Frequency of emitted light from the spectrum related to energy by plank equation

${\color{Blue}&space;\nu&space;=\frac{\Delta&space;E}{h}}$${\color{Blue}&space;=\frac{2\pi&space;^{2}me^{4}}{h^{2}}\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

${\color{Blue}&space;=\frac{2\pi&space;^{2}me^{4}}{h^{3}}\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

${\color{Blue}&space;\therefore&space;\nu&space;=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

where R is the Rydberg constant.

#### Rydberg constant for the hydrogen atom

This is the Rydberg equation for the measurement of the emission spectral region of the hydrogen atom.

$R=\frac{2\pi&space;^{2}me^{4}}{h^{3}}$

$=\frac{2\times&space;\left&space;(&space;3.1416&space;\right&space;)^{2}\times&space;9.108\times&space;10^{-28}gm\times&space;\left&space;(&space;4.8\times&space;10^{-10}esu&space;\right&space;)^{4}}{\left&space;(&space;6.627\times&space;10^{-27}erg&space;sec&space;\right&space;)^{^{3}}}$

${\color{Red}&space;=3.2898\times&space;10^{15}cycles&space;sec^{-1}}$

${\color{Blue}&space;\lambda&space;\nu&space;=c,\nu&space;=\frac{c}{\lambda&space;}=c\bar{\nu&space;}}$

${\color{Blue}&space;Wavenumber,\bar{\nu&space;}=\frac{\nu&space;}{c}}$

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).

${\color{Blue}&space;\therefore&space;R=\frac{3.2898\times&space;10^{15}sec^{-1}}{2.9979\times&space;10^{10}cmsec^{-1}}}$

= 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.

#### Spectra of 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 calculated from the Rydberg equation. The experimental hydrogen spectrum exhibited several series of lines.

#### How many spectral lines are in hydrogen?

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.

#### Lyman series emission 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 calculated by the following equation.

${\color{Blue}&space;\therefore&space;\bar{\nu&space;}&space;=\frac{1}{\lambda&space;}=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

For the Lyman series, we have n1 = 1 and n2 = 2, 3, 4…

When n2 = 2,

${\color{Blue}&space;\therefore\frac{1}{\lambda&space;}=109677\left&space;(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right&space;)cm^{-1}}$

${\color{Blue}&space;or,\frac{1}{\lambda}&space;=109677\times&space;\frac{3}{4}cm^{-1}}$

${\color{Blue}&space;\therefore&space;\lambda&space;=&space;\frac{4}{3109677\times3}&space;cm}$

= 1215 × 10⁻⁸ cm

= 1215 Å.

When n2 = ∞

${\color{Blue}&space;\therefore\frac{1}{\lambda&space;}=109677\left&space;(\frac{1}{1^{2}}-\frac{1}{\infty&space;^{2}}\right&space;)cm^{-1}}$

${\color{Blue}&space;or,\frac{1}{\lambda&space;}=109677cm^{-1}}$

${\color{Blue}&space;\therefore&space;\lambda&space;=\frac{1}{109677}cm^{-1}}$

= 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).

We know that wavenumber,

${\color{Blue}&space;\therefore&space;\bar{\nu&space;}&space;=\frac{1}{\lambda&space;}=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

For Lyman series n = 1 and wavenumber

${\color{Blue}&space;\bar{\nu&space;}=82200&space;cm^{-1}}$

${\color{Blue}&space;\therefore&space;82200=109600\left&space;(&space;\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

${\color{Blue}&space;or,\frac{1}{n_{2}^{2}}=1-\frac{822}{1096}=\frac{274}{1094}=\frac{1}{4}}$

∴ n2 = 2

#### Balmer series emission 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 calculated by the following equation.

${\color{Blue}&space;\therefore&space;\bar{\nu&space;}&space;=\frac{1}{\lambda&space;}=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

n1 = 2 and n₂ = 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).

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.

$\therefore&space;\bar{\nu&space;}=\frac{1}{\lambda&space;}=\frac{1}{4}\times&space;10^{5}\rightarrow\frac{1}{7}\times&space;10^{5}cm^{-1}$

$\bar{\nu&space;}=\frac{1}{\lambda&space;}=109677\left&space;(&space;\frac{1}{2^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

When, $\bar{\nu&space;}=\frac{1}{4}\times&space;10^{5}cm^{-1}$

$\therefore&space;\frac{1}{4}\times&space;10^{5}=109737\left&space;(&space;\frac{1}{4}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

$=1.1\times&space;10^{5}\left&space;(&space;\frac{1}{4}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

$\therefore&space;\frac{1}{4}=1.1\left&space;(&space;\frac{1}{4}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

$or,n_{2}^{2}\simeq&space;44$

∴ n₂ = 7 ( nearest whole number).

When,

$\bar{\nu&space;}=\frac{1}{7}\times&space;10^{5}cm^{-1}$

$\therefore&space;\frac{1}{7}\times&space;10^{5}=109737\left&space;(&space;\frac{1}{4}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

$=1.1\times&space;10^{5}\left&space;(&space;\frac{1}{4}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

$\therefore&space;\frac{1}{7}=1.1\left&space;(&space;\frac{1}{4}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

$or,n_{2}^{2}\simeq&space;9$

n2 = 3 ( nearest whole number).

The transition from n =7 to n = 3 while falling to n = 2 will generate the Balmer series.

#### The minimum wavelength of Balmer series

Transition energy of the Balmer lines,

$\bar{\nu&space;}=\frac{1}{\lambda&space;}=109677\left&space;(&space;\frac{1}{2^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)$

Again for Hα line n2 = 3 and for Hβ line n2 = 4

Thus the wavelength for Hɑ – line,

$\bar{\nu&space;}=\frac{1}{\lambda&space;}=109677\left&space;(&space;\frac{1}{2^{2}}-\frac{1}{3^{2}}&space;\right&space;)$

$\therefore&space;\frac{1}{\lambda&space;}=109677\times&space;\frac{5}{36}$

∴ λ = 6.564 × 10-5 cm = 6564 Å.

Again the wavelength for Hβ – line,

$\bar{\nu&space;}=\frac{1}{\lambda&space;}=109677\left&space;(&space;\frac{1}{2^{2}}-\frac{1}{4^{2}}&space;\right&space;)$

$\therefore&space;\frac{1}{\lambda&space;}=109677\times&space;\frac{3}{16}$

∴ λ = 4.863 × 10-5 cm = 4863 Å.

#### Paschen series lies in which region?

Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Paschen series lies in the near-infrared region. The wavelength of these spectral lines calculated by the Rydberg equation.

${\color{Blue}&space;\therefore&space;\bar{\nu&space;}&space;=\frac{1}{\lambda&space;}=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

n1 = 3 and n2 = 4, 5, 6, …

#### Brackett series of hydrogen spectral lines

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 calculated by the Rydberg equation.

${\color{Blue}&space;\therefore&space;\bar{\nu&space;}&space;=\frac{1}{\lambda&space;}=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

n1 = 4 and n2 = 5, 6, 7, …

#### Pfund series lies in which region

The wavelength of these spectral lines calculated by the Rydberg equation.

${\color{Blue}&space;\therefore&space;\bar{\nu&space;}&space;=\frac{1}{\lambda&space;}=R\left&space;(&space;\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&space;\right&space;)}$

n1 = 5 and n2 = 6, 7, 8, …

Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund series lies in far-infrared spectrum region.

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