Types of Spectra

When white light passes through a prism, it splits into a continuous band of colours (VIBGYOR). This is called a spectrum. There are two broad types:

1. Continuous Spectrum

A spectrum that contains radiation of all wavelengths without any gaps. White light from the Sun or an incandescent bulb gives a continuous spectrum — a smooth, unbroken rainbow of colours.

2. Line Spectrum (Discrete Spectrum)

A spectrum that consists of only certain specific wavelengths (bright lines separated by dark gaps). Each element produces its own unique set of spectral lines — like a fingerprint.

Key Point: The line spectrum of an element is unique. This is the basis of spectral analysis — identifying elements by their spectral lines.

Emission vs Absorption Spectra

Emission Spectrum

When atoms are excited (by heating or electrical discharge), they emit light. The emitted light, when passed through a prism, gives bright coloured lines on a dark background. This is the emission spectrum.

Absorption Spectrum

When white light passes through a sample of gas, certain wavelengths are absorbed by the atoms. The spectrum shows dark lines at exactly those wavelengths where bright lines appear in the emission spectrum. This is the absorption spectrum.

[JEE Tip] Emission and absorption spectra of the same element are complementary — dark lines in absorption correspond to bright lines in emission. This is called Kirchhoff's law of spectroscopy.

Line Spectrum of Hydrogen

The hydrogen atom produces the simplest and most studied atomic spectrum. When hydrogen gas at low pressure is subjected to an electrical discharge, it emits light that, when passed through a prism, gives a line spectrum — only specific discrete wavelengths are present.

Why is this important?

The observation that atoms emit only specific wavelengths means that atoms can have only certain specific energy values — energy in atoms is quantised. This was a major clue that led to the development of quantum theory.

Spectral Series of Hydrogen

The hydrogen spectrum consists of several series of lines, each named after their discoverer. The lines in each series converge as they approach a series limit (the shortest wavelength in that series).

Series Transition to n1n_1 Region Discovery
Lyman n1=1n_1 = 1 Ultraviolet 1906
Balmer n1=2n_1 = 2 Visible 1885
Paschen n1=3n_1 = 3 Infrared 1908
Brackett n1=4n_1 = 4 Infrared 1922
Pfund n1=5n_1 = 5 Far infrared 1924

In each series, the electron transitions end at the level n1n_1 and start from any higher level n2>n1n_2 > n_1.

[NEET Tip] Only the Balmer series falls in the visible region. The mnemonic for visible lines (from longest to shortest wavelength): Hα_\alpha, Hβ_\beta, Hγ_\gamma, Hδ_\delta corresponding to transitions from n=3,4,5,6n = 3, 4, 5, 6 to n=2n = 2.

The Rydberg Formula

All spectral lines of hydrogen can be described by a single formula, proposed by Johannes Rydberg:

νˉ=1λ=RH(1n12−1n22)\boxed{\bar{\nu} = \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)}

where:

  • νˉ\bar{\nu} = wave number (m−1^{-1} or cm−1^{-1})
  • λ\lambda = wavelength
  • RHR_H = Rydberg constant = 1.097×1071.097 \times 10^7 m−1^{-1} = 109,677109,677 cm−1^{-1}
  • n1n_1 = lower energy level (1, 2, 3, …)
  • n2n_2 = upper energy level (n2>n1n_2 > n_1)

Applying the Formula to Each Series

Lyman series (n1=1n_1 = 1): νˉ=RH(112−1n22)n2=2,3,4,…\bar{\nu} = R_H\left(\frac{1}{1^2} - \frac{1}{n_2^2}\right) \quad n_2 = 2, 3, 4, \ldots

Balmer series (n1=2n_1 = 2): νˉ=RH(122−1n22)n2=3,4,5,…\bar{\nu} = R_H\left(\frac{1}{2^2} - \frac{1}{n_2^2}\right) \quad n_2 = 3, 4, 5, \ldots

Paschen series (n1=3n_1 = 3): νˉ=RH(132−1n22)n2=4,5,6,…\bar{\nu} = R_H\left(\frac{1}{3^2} - \frac{1}{n_2^2}\right) \quad n_2 = 4, 5, 6, \ldots

Important Points

  • The first line (longest wavelength, least energy) in each series corresponds to n2=n1+1n_2 = n_1 + 1
  • The series limit (shortest wavelength, maximum energy) corresponds to n2=∞n_2 = \infty
  • For the series limit: νˉlimit=RH/n12\bar{\nu}_{\text{limit}} = R_H / n_1^2

[JEE Tip] The total number of spectral lines when an electron falls from level nn to ground state is n(n−1)2\frac{n(n-1)}{2}.

Balmer Series — The Visible Lines

The Balmer series is special because it falls in the visible region. The individual lines are:

Line Transition Colour Wavelength
Hα_\alpha 3→23 \to 2 Red 656.3 nm
Hβ_\beta 4→24 \to 2 Blue-green 486.1 nm
Hγ_\gamma 5→25 \to 2 Blue-violet 434.0 nm
Hδ_\delta 6→26 \to 2 Violet 410.2 nm
Series limit ∞→2\infty \to 2 UV boundary 364.6 nm

Notice:

  • The lines get closer together as n2n_2 increases
  • The wavelength decreases (energy increases) as n2n_2 increases
  • Beyond the series limit, the spectrum becomes continuous (ionisation)

Number of Spectral Lines

When hydrogen atoms are excited to energy level nn, the total number of possible spectral lines as electrons cascade down is:

Total lines=n(n−1)2\text{Total lines} = \frac{n(n-1)}{2}

For example, if n=4n = 4: Total lines =4×32=6= \frac{4 \times 3}{2} = 6

These correspond to transitions: 4→34 \to 3, 4→24 \to 2, 4→14 \to 1, 3→23 \to 2, 3→13 \to 1, 2→12 \to 1.

Key Point: The line spectrum of hydrogen was one of the most important experimental observations that led to the development of Bohr's model and ultimately quantum mechanics.

Solved Examples

Example 1: First Line of Lyman Series

Calculate the wavelength of the first line in the Lyman series of hydrogen. (RH=1.097×107R_H = 1.097 \times 10^7 m−1^{-1})

Solution: Lyman series: n1=1n_1 = 1, first line: n2=2n_2 = 2

1λ=RH(112−122)=1.097×107(1−14)=1.097×107×34\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 1.097 \times 10^7 \left(1 - \frac{1}{4}\right) = 1.097 \times 10^7 \times \frac{3}{4}

1λ=8.228×106 m−1\frac{1}{\lambda} = 8.228 \times 10^6 \text{ m}^{-1}

λ=18.228×106=1.215×10−7 m=121.5 nm\lambda = \frac{1}{8.228 \times 10^6} = 1.215 \times 10^{-7} \text{ m} = 121.5 \text{ nm}

Answer: λ=121.5\lambda = 121.5 nm (UV region).

Example 2: First Line of Balmer Series

Calculate the wavelength of Hα_\alpha line (first line of Balmer series).

Solution: Balmer series: n1=2n_1 = 2, first line: n2=3n_2 = 3

1λ=RH(122−132)=1.097×107(14−19)\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{9}\right)

=1.097×107×9−436=1.097×107×536= 1.097 \times 10^7 \times \frac{9 - 4}{36} = 1.097 \times 10^7 \times \frac{5}{36}

1λ=1.524×106 m−1\frac{1}{\lambda} = 1.524 \times 10^6 \text{ m}^{-1}

λ=656.3 nm\lambda = 656.3 \text{ nm}

Answer: λ=656.3\lambda = 656.3 nm (red light — this is the famous Hα_\alpha line).

Example 3: Series Limit of Balmer Series

Calculate the wavelength of the series limit of the Balmer series.

Solution: Series limit: n1=2n_1 = 2, n2=∞n_2 = \infty

1λ=RH(122−1∞2)=RH×14=1.097×1074\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = R_H \times \frac{1}{4} = \frac{1.097 \times 10^7}{4}

1λ=2.7425×106 m−1\frac{1}{\lambda} = 2.7425 \times 10^6 \text{ m}^{-1}

λ=364.6 nm\lambda = 364.6 \text{ nm}

Answer: λ=364.6\lambda = 364.6 nm (at the UV boundary of visible region).

Example 4: Identifying the Series

A hydrogen atom emits a photon of wavelength 1094 nm. To which spectral series does this line belong and what are the nn values?

Solution: Given, λ=1094 nm=1094×10−9 m\lambda = 1094 \text{ nm} = 1094 \times 10^{-9} \text{ m}

So, νˉ=1λ=11094×10−9=9.14×105 m−1\bar{\nu} = \frac{1}{\lambda} = \frac{1}{1094 \times 10^{-9}} = 9.14 \times 10^5 \text{ m}^{-1}

Now test the Paschen series (n1=3n_1 = 3):

νˉ=RH(132−1n22)\bar{\nu} = R_H\left(\frac{1}{3^2} - \frac{1}{n_2^2}\right)

Trying n2=6n_2 = 6, νˉ=1.097×107(19−136)=1.097×107×336=9.14×105 m−1\bar{\nu} = 1.097 \times 10^7\left(\frac{1}{9} - \frac{1}{36}\right) = 1.097 \times 10^7 \times \frac{3}{36} = 9.14 \times 10^5 \text{ m}^{-1}

This matches the given value.

Therefore, the transition is: n2=6→n1=3n_2 = 6 \to n_1 = 3

Answer: The line belongs to the Paschen series, corresponding to the transition 6→36 \to 3.

Example 5: Energy of a Spectral Line

Calculate the energy of the photon emitted when an electron transitions from n=4n = 4 to n=2n = 2 in hydrogen.

Solution: 1λ=RH(14−116)=1.097×107×316=2.057×106 m−1\frac{1}{\lambda} = R_H\left(\frac{1}{4} - \frac{1}{16}\right) = 1.097 \times 10^7 \times \frac{3}{16} = 2.057 \times 10^6 \text{ m}^{-1}

E=hcνˉ=6.626×10−34×3×108×2.057×106=4.09×10−19 JE = hc\bar{\nu} = 6.626 \times 10^{-34} \times 3 \times 10^8 \times 2.057 \times 10^6 = 4.09 \times 10^{-19} \text{ J}

In eV: E=4.09×10−191.6×10−19=2.56 eVE = \frac{4.09 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.56 \text{ eV}

Answer: E=4.09×10−19E = 4.09 \times 10^{-19} J = 2.56 eV. (This is the Hβ_\beta line at 486 nm — blue-green).

Example 6: Total Number of Spectral Lines

If hydrogen atoms are excited to n=5n = 5, how many spectral lines can be observed?

Solution: Total lines=n(n−1)2=5×42=10\text{Total lines} = \frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10

These are: 5→45 \to 4, 5→35 \to 3, 5→25 \to 2, 5→15 \to 1, 4→34 \to 3, 4→24 \to 2, 4→14 \to 1, 3→23 \to 2, 3→13 \to 1, 2→12 \to 1.

Answer: 10 spectral lines.

Example 7: Longest Wavelength in Lyman Series

Calculate the longest wavelength in the Lyman series.

Solution: The longest wavelength (least energy) corresponds to the smallest transition: n2=2→n1=1n_2 = 2 \to n_1 = 1.

1λ=RH(1−14)=1.097×107×34=8.228×106\frac{1}{\lambda} = R_H\left(1 - \frac{1}{4}\right) = 1.097 \times 10^7 \times \frac{3}{4} = 8.228 \times 10^6

λ=121.5 nm\lambda = 121.5 \text{ nm}

Answer: λ=121.5\lambda = 121.5 nm.

Example 8: Shortest Wavelength in Paschen Series

Find the shortest wavelength line in the Paschen series.

Solution: Shortest wavelength (series limit): n1=3n_1 = 3, n2=∞n_2 = \infty

1λ=RH×19=1.097×1079=1.219×106\frac{1}{\lambda} = R_H \times \frac{1}{9} = \frac{1.097 \times 10^7}{9} = 1.219 \times 10^6

λ=8.20×10−7 m=820 nm\lambda = 8.20 \times 10^{-7} \text{ m} = 820 \text{ nm}

Answer: λ=820\lambda = 820 nm (infrared region).

Example 9: Ratio of Wavelengths

Find the ratio of the longest wavelength of the Balmer series to the shortest wavelength of the Lyman series.

Solution: Longest Balmer (n2=3→n1=2n_2 = 3 \to n_1 = 2): 1λB=RH(14−19)=RH×536\frac{1}{\lambda_B} = R_H\left(\frac{1}{4} - \frac{1}{9}\right) = R_H \times \frac{5}{36}

Shortest Lyman (series limit, n2=∞→n1=1n_2 = \infty \to n_1 = 1): 1λL=RH(1−0)=RH\frac{1}{\lambda_L} = R_H(1 - 0) = R_H

So, λB=365RH,λL=1RH\lambda_B = \frac{36}{5R_H}, \quad \lambda_L = \frac{1}{R_H}

Therefore, λBλL=365=7.2\frac{\lambda_B}{\lambda_L} = \frac{36}{5} = 7.2

Answer: λB/λL=36/5=7.2\lambda_B / \lambda_L = 36/5 = 7.2.

Example 10: Lines in Visible Region

When hydrogen atoms are excited to n=6n = 6, how many lines appear in the visible region?

Solution: Visible lines belong to the Balmer series (n1=2n_1 = 2). From n=6n = 6, the transitions to n=2n = 2 are: 6→2,  5→2,  4→2,  3→26 \to 2, \; 5 \to 2, \; 4 \to 2, \; 3 \to 2

That gives 4 lines in the visible region (Hα_\alpha, Hβ_\beta, Hγ_\gamma, Hδ_\delta).

Answer: 4 lines in the visible region.