The arrangement of all electromagnetic radiation in a definite order (increasing or decreasing wavelength or frequency) is called the electromagnetic spectrum. Every region of the spectrum is characterised by a band of wavelengths or frequencies and by specific applications or physical effects.
When light from a source is passed through a prism, radiations of different wavelengths are deviated through different angles and become separated. The angle of deviation is proportional to the refractive index of the prism material, so radiations of different wavelengths are dispersed.
This process of separating light into its constituent wavelengths is called dispersion. The visible result-the set of separated colours or wavelengths-is called a spectrum.
Splitting of lightSpectra are broadly classified into two types:
Continuous spectrum: When white light from a source is dispersed, a bright spectrum continuously distributed on a dark background is obtained. The colours change continuously from violet to red without sharp boundaries; adjacent colours appear merged. This is called a continuous spectrum.

Discontinuous (line) spectrum: When atoms are excited (for example by heating or electric discharge), electrons are promoted to higher energy levels. When these electrons return to lower levels they emit radiation of certain definite wavelengths corresponding to energy differences. The emitted spectrum consists of discrete lines and is called a line spectrum or atomic spectrum.

The hydrogen atom shows a discontinuous line spectrum with lines observed in ultraviolet, visible and infrared regions. Different groups of lines were found to form series and were named after discoverers.
Rydberg proposed an empirical formula applicable to all series of hydrogen-like species. The Rydberg constant for hydrogen is
RH = 109677 cm-1
In the Rydberg expression, n1 and n2 are integers with n2 > n1.
Bohr proposed the first atom model based on Planck's quantum theory. The model successfully explained the stability of the atom and the line spectrum of hydrogen.
Bohr's Atomic ModelBohr's model is applicable only to one-electron species (for example H, He+, Li2+, Be3+).
The derivation below follows Bohr's assumptions and uses Coulomb's law together with quantised angular momentum.
Consider a nucleus of charge +Ze and an electron of charge -e moving in a circular orbit of radius r.
The Coulomb force provides the centripetal force:
e²/(4πε0 r²) = m v² / r
The quantisation condition for angular momentum is:
m v r = n (h/2π)
Eliminate v using the angular momentum expression and Coulomb-centripetal relation to get r in terms of n and Z.
The radius of the nth orbit is therefore proportional to n²/Z:
rn = 0.529 Å × n² / Z
Thus r ∝ n² for a particular atom (fixed Z).
Radius of the first orbit of hydrogen (n = 1, Z = 1):
r = 0.529 Å.
Example. Calculate the ratio of the radius of the 1st orbit of the H atom to that of Li2+.
Sol.
For hydrogen (Z = 1) and Li2+ (Z = 3) with n = 1 for both,
r ∝ n² / Z
Therefore r(H) : r(Li2+) = (1²/1) : (1²/3) = 1 : 1/3 = 3 : 1.
Using m v r = n (h/2π) and the expression for rn, the speed v of electron in nth orbit is obtained.
m v r = n h / 2π
Putting the value of r gives v in terms of n and Z:
v = (Z e²)/(2 ε0 h) × (2π / n)
Numerically for hydrogen (n = 1, Z = 1):
v = 2.18 × 106 m s-1
Thus v ∝ Z / n.
Total energy T.E. of the electron-nucleus system is the sum of kinetic energy (K.E.) and potential energy (P.E.).
K.E. = 1/2 m v²
Using the Coulomb-centripetal relation we obtain K.E. = + (Z² × 13.6 eV) / n².
Potential energy of the system (electron and nucleus) is:
P.E. = - (Z² × 27.2 eV) / n²
Total energy T.E. = K.E. + P.E. = - (Z² × 13.6 eV) / n² per atom.
For hydrogen (Z = 1) the energy of the nth level is
En = - 13.6 eV / n².
In SI units:
En = - 2.18 × 10-18 J / n² per atom.
As n increases the absolute value of T.E. and P.E. decreases and approaches zero at n → ∞. The negative sign indicates the electron is bound (under attractive force of nucleus).
K.E. = + (Z² × 13.6 eV) / n²
T.E. = - K.E.
(i) n = 1 (ground state)
K.E. = 13.6 eV
P.E. = -27.2 eV
T.E. = -13.6 eV
(ii) n = 2 (first excited state)
K.E. = 13.6/4 eV = 3.4 eV
P.E. = -6.8 eV
T.E. = -3.4 eV
E2 - E1 = (-3.4) - (-13.6) = 10.2 eV
(iii) n = 3 (second excited state)
K.E. = 13.6/9 eV = 1.51 eV
P.E. = -3.02 eV
T.E. = -1.51 eV
Energy difference E3 - E2 = (-1.51) - (-3.4) = 1.89 eV
(iv) n = 4
K.E. = 13.6/16 eV = 0.85 eV
P.E. = -1.70 eV
T.E. = -0.85 eV
E4 - E3 = 0.66 eV
As n increases the energy of a level increases (becomes less negative), but the energy difference between successive levels decreases for large n.
When an electron falls from a higher energy level n2 to a lower level n1, it emits a photon of energy equal to the difference between the two levels:
hν = ΔE = En2 - En1 = -13.6 eV (1/n2² - 1/n1²)
Using En = -13.6 eV / n², the emitted or absorbed photon frequency is given by the Rydberg formula
1/λ = RH Z² (1/n1² - 1/n2²)
The theoretical values obtained from Bohr's model agree closely with experimental observations for one-electron species.
Equivalently in wave number (ṽ):
ṽ = RH Z² (1/n1² - 1/n2²)
Different Series in Hydrogen Spectrum.If transitions occur between levels up to principal quantum number n (i.e., from n to all lower levels), the total number of different spectral lines produced by all possible transitions among n levels is:
N = n(n - 1)/2
Example. The energy levels of an atom for 1st, 2nd and 3rd levels are E, 4E and 2E respectively. If a photon of wavelength λ is emitted for a transition 3 → 1, calculate the wavelength for transition 2 → 1 in terms of λ.
Sol.
Energy difference for 3 → 1 is E3 - E1 = 2E - E = E, corresponding to photon of wavelength λ.
Energy difference for 2 → 1 is E2 - E1 = 4E - E = 3E, so photon energy is 3 times that of the 3 → 1 transition.
Since E ∝ 1/λ, the wavelength for 2 → 1 is λ/3. Thus λ2→1 = λ/3.
Example. Let n1 be the frequency of the series limit of the Lyman series, n2 be the frequency of the first line of the Lyman series, and n3 be the frequency of the series limit of the Balmer series. Find the relation between n1, n2 and n3.
Sol.
Using energy (or frequency) relations for the series limits and first lines, one obtains
n1 = n2 - n3 (or equivalent relations depending on sign conventions).
Example. A hydrogen-like species emits 6 wavelengths originating from all possible transitions between a group of levels. These energy levels lie between -0.85 eV and -0.544 eV (inclusive). Calculate (i) the quantum numbers of the levels involved, and (ii) the atomic number Z of the species.
Sol.
From the count of emitted lines and the energy range one finds the difference in principal quantum numbers and then identifies n1 and n2. Solving the combinatorial relation for the number of lines and matching energy values leads to
n1 = 12, n2 = 15 (as determined by the given numerical solution method) and the corresponding Z is found from the level energies.
Arnold Sommerfeld introduced three important refinements that improved agreement with experiment and accounted for small deviations (fine structure) in spectral lines:
Fig: Sommerfeld model of Atom.| 1. What is the electromagnetic spectrum? | ![]() |
| 2. What is Bohr's model of the atom? | ![]() |
| 3. What is the Sommerfeld model of the atom? | ![]() |
| 4. What is an atomic spectrum? | ![]() |
| 5. How does Bohr's model explain the hydrogen atom spectrum? | ![]() |