Atomic emission spectrum of hydrogen

Atomic absorption and emission spectrum

The most important factor for the determination of the structure of the atom is the emission and absorption spectrum of the element. But the emission energy of hydrogen always produces line spectrum or atomic spectra.

Thus the measurements of the atomic emission spectrum are feasible. Because each hydrogen atom has a definite energy level in which the electron can stay. This is the ground state of an atom.

So on the addition of thermal or electrical energy, the electron moved to the higher energy level of an atom.

Thus when this excited electron returns to the ground state it emits a definite frequency of radiation.

Frequency of atomic spectra

In 1901 plank proposed a hypothesis in which he connected photon energy and frequency of the emitted light.

$\Delta&space;E=&space;h\nu$

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

where ν = frequency of emitted light
h = plank constant

The emission spectrum of hydrogen

The energy corresponding to a particular line in the emission and absorption spectra or spectrum of hydrogen is the energy difference between the ground level and the exited level.

Bohr’s theory provides the energy of an electron at a particular energy level. Thus the energy of an electron in the hydrogen

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

$But\,&space;\Delta&space;E=&space;E_{2}-E_{1}$

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

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

But the frequency of emitted light from the electromagnetic spectrum related to energy by plank equation

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

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

where R = Rydberg constant

Rydberg constant value for hydrogen

Rydberg equation can measure the values of rydberg constant for the hydrogen.

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

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

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

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

$where\,&space;wavenumber,\bar{\nu&space;}=&space;\frac{\nu&space;}{c}$

and c = velocity of light

Thus the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second.

But wavenumber 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 showing a remarkable agreement between the experiment and values from the atomic theory.

Emission and absorption spectra of hydrogen

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.

Thus the transitions energies calculated from the Rydberg equation exhibited several series of lines.

How many spectral lines are in hydrogen?

Hydrogen is given several spectral lines 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 of an atom that is n = 1 (K – Shell).

When heat or electrical energy is supplied to hydrogen, it absorbed different amounts of energy to give absorption spectra or spectrum.

Some of the atoms absorbed such energy to shift their electron to third energy level, while some others may absorb a large amount of energy to shift their electron to the fourth, fifth, sixth and seventh energy levels.

But the electrons of hydrogen in the excited state are relatively unstable and hence drop back to the ground state by the emission of energy in the form of the spectrum. Thus for different drop back given different spectral lines for hydrogen.

Lyman series of spectral lines

Transition to the ground state where n = 1, from the exited state where n = 2, 3, 4, etc exited states constitute the Lyman series of spectral lines. These emission atomic spectra lie in the ultraviolet region, frequency less than 4000 Å.

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

Thus when n2 = 2, it gives the wavelength of the lower line of the spectrum.

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

$or,&space;\lambda&space;=&space;\frac{4}{109677\times&space;3}$

= 1215 Å

But when n2 = ∞ gives the frequency and wavelength of the higher line of the spectrum.

ν = 109677 cm-1
and λ = 912 Å

Balmer series emission spectrum

Transition to n = 2, from the exited state where n > 2 states constitute the Balmer series of spectral lines. These spectral lines lie in the visible region, frequency 4000 Å to 7000 Å.

Again when the transition occurs from n =3 to n =2 constitute Hα – the line of the spectrum. Thus the wavelength and frequency of the Hα -line

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

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

= 6.564 × 10-5 cm = 6564 Å.

But when the transition occurs from n =4 to n =2 constitute Hβ – the line of the spectrum. Thus the wavelength and frequency of the 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 region

When n1 = 3 and n2 >3, it constitutes the Paschen series lies in the near-infrared region. Thus the wavelength of these spectral lines calculated by the Rydberg equation.

Brackett series of spectral lines

When n1 = 4 and n2 >5, it constitutes the Brackett series lies in the far-infrared region. So the wavelength of these spectral lines also calculated from the Rydberg equation.

Pfund series lies in which region

Transition to the quantum number n = 5 from n >5, which constitutes the Pfund series lies situated in the far-infrared spectrum region.

This is the theoretical basis for the formation of emission line spectra or the atomic spectrum of hydrogen. The above discussion also tells us that as we go to the higher to still higher energy level the energy gap between two-level is continued decreases.