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Lyman Series

Last updated date: 23rd Apr 2024
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Lyman Series: An Introduction

When the electron transitions to its lowest energy state of n = 1 from n = 2 (where n is the main quantum number), the Lyman series in hydrogen spectra changes and ultraviolet emission lines are produced that's why the Lyman series lies in the UV region.

The hydrogen atom experiences a sequence of transitions and emission lines when an electron moves from n = 2 to n = 1 and is known as the Lyman series (where n is the principal quantum number referring to the energy level of the electron). Greek letters are used to identify the transitions in order: Lyman-alpha is the transition from n = 2 to n = 1, followed by Lyman-beta at n = 3, Lyman-gamma at n = 4, etc. Theodore Lyman, the series' discoverer, is honoured with a name for it.

Lyman Series

The Lyman series lies in the UV region (ultraviolet) because of all the wavelengths. The table below shows the wavelength of Lyman series all ultraviolet:

Energy Level(n)










The series was started between 1906 and 1914 by Theodore Lyman. It bears his name as a result. The Lyman series, in accordance with Bohr's theory, emerges as electrons go from higher energy levels (nh = 2,3,4,5,6,...) to nl = 1 energy state.

Lyman Series Formula

The Lyman Series formula is as follows:

\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{{(1)}^2}}} - \dfrac{1}{{{{({n_2})}^2}}}} \right)\]


Z – Atomic Number (hydrogen =1)

\[{n_1}\]- lower energy level (\[{n_1}\]- 1)

\[{n_2}\]- higher energy level

𝜆 – Wavelength

R -Rydberg Constant (1.09737 \[ \times \]107m-1)

Rydberg Formula

The integer Rydberg n numbers indicate electron orbitals at various integral distances from the atom in Bohr's model of the atom. Therefore, the photon energy released or absorbed when an electron jumps from orbital 1 to orbital 2 is represented by a frequency (or spectral energy) released in a transition from n1 to n2.

\[\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{{\left( {{n_1}} \right)}^2}}} - \dfrac{1}{{{{\left( {{n_2}} \right)}^2}}}} \right)\]


Z – Atomic Number

\[{n_1}\]- - lower energy level

\[{n_2}\]- - higher energy level

𝜆 – Wavelength

R -Rydberg Constant (1.09737 \[ \times \] 107m-1)


In 1913, Niels Bohr developed his Bohr model hypothesis, which provided an explanation for why hydrogen spectral lines fit Rydberg's formula. According to Bohr, the hydrogen atom's electron must have quantised energy levels that are represented by the following formula:

\[{E_n} = - \dfrac{{m{e^4}}}{{2{{(4\pi {\varepsilon _o}\hbar )}^2}}}\dfrac{1}{{{n^2}}} = - \dfrac{{13.6}}{{{n^2}}}\left[ {eV} \right]\]

When an electron moves from an initial energy level (Ei) to a final energy level (Ef), according to Bohr's third assumption, the atom must emit radiation with a wavelength of:

\[\lambda = \dfrac{{hc}}{{{E_i} - {E_f}}}\]

When dealing with energy in electron volts and wavelengths in angstroms, there is also a more convenient notation:

\[\lambda = \dfrac{{12430}}{{{E_i} - {E_f}}}\]

By substituting the expression for the energy in the hydrogen atom, where the initial energy corresponds to energy level n and the final energy belongs to energy level m, for the energy in the aforementioned formula, the following result is obtained:

\[\dfrac{1}{\lambda } = \dfrac{{{E_i} - {E_f}}}{{12430}} = \dfrac{{{{(12430)}^{ - 1}}}}{{13.6}}\left( {\dfrac{1}{{{m^2}}} - \dfrac{1}{{{n^2}}}} \right) = R\left( {\dfrac{1}{{{m^2}}} - \dfrac{1}{{{n^2}}}} \right)\]

Where, R is the identical Rydberg constant from Rydberg's well-known equation.

Replace m by 1 to get the relationship between Bohr, Rydberg, and Lyman:

\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{n^2}}}} \right)\]

Spectral Series

Every time an electron in a hydrogen atom transitions from a higher to a lower energy level, the energy difference between the two levels is released as radiation with a certain wavelength. It's referred to as a spectral line. Different spectral lines are formed because the wavelength of the spectral line relies on the two orbits (energy levels) between which the electron transfer occurs. The various wavelengths combine to form spectral series, which are unique to the atoms generating them.

Interesting Facts

  • Hospitals utilise UV rays to sterilise medical equipment because they are so potent.

  • All live cells are killed by UV light at high dosages.

  • To shield themselves from the sun, elephants cover themselves in mud.

Numerical Questions

1. In the Lyman series the wave number of a hydrogen atom is 82200cm-1. Calculate the electron’s higher energy level.



\[\dfrac{1}{\lambda } = \]82200cm-1

R -Rydberg Constant (1.09737 \[ \times \]107m-1) or 109678 cm-1

Lyman series formula is:

\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{{(1)}^2}}} - \dfrac{1}{{{{({n_2})}^2}}}} \right)\]

Put the values in the above equation, we get:

\[\dfrac{{82200}}{{109678}} = \left( {\dfrac{1}{{{{(1)}^2}}} - \dfrac{1}{{{{({n_2})}^2}}}} \right)\]

\[\dfrac{3}{4} = \left( {\dfrac{1}{{{{(1)}^2}}} - \dfrac{1}{{{{({n_2})}^2}}}} \right)\]

\[{n_2} = 2\]

Hence, the electron's higher energy level is 2.

2. Calculate the wavelength when the electron jumps from the 3rd energy state in the Balmer series of the hydrogen spectrum.

Sol. The formula for the Balmer series:

\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{({n_2})}^2}}}} \right)\]

R -Rydberg Constant (1.09737 \[ \times \]107m-1)


\[{n_2}\] = 3

\[\dfrac{1}{\lambda } = 1.09737 \times {10^7}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(3)}^2}}}} \right)\]

\[\dfrac{1}{\lambda } = 1.09737 \times {10^7}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(3)}^2}}}} \right)\]

\[\lambda = 6561 \times {10^{ - 10}}\]m

Hence, the value of wavelength is \[6561 \times {10^{ - 10}}\] m.


The Rydberg formula has been used to determine the wavelengths of several spectral series that have been created from the emission spectrum of atomic hydrogen. These spectral lines that have been seen are the result of electron transitions between two energy levels within an atom. The development of quantum mechanics was greatly aided by the Rydberg formula's categorisation of the series.

Competitive Exams after 12th Science

FAQs on Lyman Series

1. What does the Lyman series show?

In the ultraviolet spectrum, excited atoms emit electromagnetic energy that is described by the Lyman series of wavelengths. Wavelength is the separation between two adjacent waves' corresponding points. Two points or particles that are in the same phase—i.e., points that have completed equal fractions of their periodic motion—are referred to as "corresponding points." Wavelength is typically measured from crest to crest or from trough to trough in transverse waves from compression to compression or from rarefaction to rarefaction in longitudinal waves.

2. Why can't we observe the Lyman series?

The electromagnetic spectrum's ultraviolet region is where light is emitted when an electron in a hydrogen atom transitions back to its ground state. It is hence invisible. The part of the electromagnetic spectrum known as Uv rays spans from the violet, or short-wavelength, end of the visible light spectrum to the X-ray area. Although ultraviolet (UV) radiation cannot be seen by the human eye, it may cause some materials to fluoresce—that is, to generate electromagnetic radiation with lower energy, like visible light—when it hits them. However, a lot of insects can sense UV light.

3. What are the various forms of the Lyman series?

Greek letters are used to identify the transitions in order: Lyman-alpha, which occurs from n = 2 to n = 1, Lyman-beta, which occurs from n = 3, Lyman-gamma, which occurs from n = 4, and so on. Theodore Lyman was the scientist who discovered it. 

An atomic electron transition occurs when an electron moves (or jumps) from one energy level to another inside of a natural or manufactured atom. As the electron "jumps" from one quantized energy level to another, usually in a few nanoseconds or less, it seems discontinuous. It is sometimes referred to as an atomic transition, an electronic (de-)excitation, or a quantum leap.