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JEE Physics
Spectral Series

Frequently Asked Questions

Only specific wavelengths are observed because electrons in hydrogen can only occupy certain quantized energy levels. Transitions between these levels release photons with discrete energies, corresponding to particular wavelengths.

The Balmer series lies in the visible region because the energy differences between levels where electrons fall to n = 2 correspond to photon wavelengths between about 400 nm and 700 nm, which are within the range detectable by the human eye.

Hydrogen emits sharp, discrete lines because its electrons transition between quantized energy levels, releasing photons with precise energies. There are no intermediate states between levels to produce a continuous spectrum.

As energy levels increase (i.e., higher n), the spacing between energy levels decreases, so the spectral lines get closer together (converge), especially near the series limit.

No, hydrogen can only absorb photons whose energy exactly matches the energy gap between its quantized energy levels. This means it can only absorb specific frequencies (wavelengths).

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Spectral Series

The Spectral Series refers to the set of wavelengths of light emitted or absorbed by electrons in atoms as they transition between energy levels. Each element has its own unique spectral lines, forming distinct series such as the Lyman, Balmer, and Paschen series in hydrogen. These lines appear as bright or dark lines in a spectrum and are key to understanding atomic structure, identifying elements in stars, and exploring quantum mechanics. The study of spectral series has played a crucial role in the development of modern physics and continues to be essential in fields like astronomy, chemistry, and spectroscopy.

1.0Definition of Spectral Series

  • A spectral series of the hydrogen atom is a group of spectral lines corresponding to electronic transitions where electrons fall from higher energy levels to a fixed lower energy level, emitting photons of specific wavelengths characteristic of hydrogen.
  • It has been shown that the energy of the outer orbit is greater than the energy of the inner ones. When the Hydrogen atom is subjected to external energy, the electron jumps from lower energy State i.e. the hydrogen atom is excited. The excited state is not stable hence the electron returns to its ground state in about 10^{-8} seconds. The excess of energy is now radiated in the form of radiations of different wavelength.The different wavelengths constitute spectral series. Which are characteristic of atom emitting, then the wavelength of different members of series can be found from the following relations,

νˉ=λ1​=R[n12​1​−n22​1​]

This relation explains the complete spectrum of hydrogen. A detailed account of the important radiations are listed below.

Spectrum of Hydrogen

2.0Classification of Spectral Series

1.Lyman Series

The series consists of wavelengths which are emitted when an electron jumps from an outer orbit to the first orbit i. e. the electron jumps to K orbit giving rise to the Lyman series.

Here n1​=1 and n2​=2,3,4,…,∞.

The wavelengths of different members of Lyman series are:

  1. First Member

In this case n1​=1 and n2​=2 hence

λ1​=R[121​−221​]=43R​

Or

λ=3R4​

λ=3×10.97×1064​=1216×10−10 m=1216 A˚

  1. Second Member

Here n1​=1 and ​​n2​=3 hence

λ1​=R[121​−321​]=98R​ or λ=8R9​

λ=8×10.97×1069​=1026×10−10 m=1026 A˚

  1. Limiting Member

Here n1​=1 and n2​=∞ hence

λ1​=R[121​−∞21​]=R or λ=R1​

λ=10.97×1061​=912×10−10 m=912 A˚

This series lies in ultraviolet region

2.Balmer Series

This series consists of all wavelengths which are emitted when an electron jumps from an outer orbit to the second orbit i. e. the electron jumps to L orbit give rise to the Balmer series. n1​=2 and n2​=3,4,5…∞

The wavelength of different members of the Balmer series.

  1. First Member

In this case n1​=2 and n2​=3 hence

λ1​=R[221​−321​]=365R​ or λ=5R36​

λ=5×10.97×10636​=6563×10−10 m=6563 A˚

  1. Second Member

Here n1​=2 and n2​=4 hence

λ1​=R[221​−421​]=163R​ or λ=3R16​

λ=3×10.97×10616​=4861×10−10 m=4861 A˚

  1. Limiting Member

Here n1​=2 and n2​=∞ hence

λ1​=R[221​−∞21​]=4R​ or λ=R4​=3646 A˚

This series lies in the visible and near ultraviolet region.

3.Paschen Series

This series consists of all wavelengths emitted when an electron jumps from an outer orbit to the third orbit i. e. the electron jumps to M orbit give rise to paschen series.Here

n1​=3 and n2​=4,5,6…∞

The different wavelengths of this series can be obtained from the formula

λ1​=R[321​−n22​1​] where n2​=4,5,6…∞

For the first member, the wavelength is 18750Å. This series lies in the infra-red region.

4.Brackett Series

This series consists of all wavelengths which are emitted when an electron jumps from an outer orbit to the fourth orbit i. e. the electron jumps to N orbit giving rise to the Brackett series.Here n1​=4 and n2​=5,6,7…∞

The different wavelengths of this series can be obtained from the formula

λ1​=R[421​−n22​1​] where n2​=5,6,7…∞

This series lies in the infra-red region of the spectrum.

5.Pfund series

The series consists of all wavelengths which are emitted when an electron jumps from an outer orbit to the fifth orbit i. e. the electron jumps to O orbit give right to Pfund series.Here n1​=5 and n2​=6,7,8…∞

The different wavelengths of this series can be obtained from the formula

λ1​=R[521​−n22​1​] where n2​=6,7,8…∞

This series lies in the infra-red region of the spectrum.

S. No.

Series Observed

Value of n1​

Value of n2​

Position in the Spectrum

1.

Lyman Series

1

n2​=2,3,4…∞

Ultra Violet

2.

Balmer Series

2

n2​=3,4,5…∞

Visible

3.

Paschen Series

3

n2​=4,5,6…∞

Infra-red

4.

Brackett Series

4

n2​=5,6,7…∞

Infra-red

5.

Pfund Series

5

n2​=6,7,8…∞

Infra-red

Spectrum

Illustration-1.Find the maximum wavelength of the Brackett series of hydrogen atoms.

Solution: n1​=4 and n2​=5

∴λmax​1​=R[421​−521​]

λmax​=9×10.97×10625×16×1010​=40400 A˚

Q-2.For the given transitions of electrons, obtain the relation between λ1​,λ2​ and λ3​.

Transitions of Electrons

Solution:For given condition

E3​−E1​=(E3​−E2​)+(E2​−E1​)

⇒λ3​hc​=λ2​hc​+λ1​hc​⇒λ3​1​=λ2​1​+λ1​1​

λ3​=λ2​+λ1​λ2​λ1​​

Table of Contents


  • 1.0Definition of Spectral Series
  • 2.0Classification of Spectral Series