Aren’t you fascinated when you see a rainbow? The colours we see in it never fail to captivate us! Did you know that even though we identify the distinct colours of a rainbow, it is actually a continuous range of colours? A similar range of colours appears when white light passes through a prism. This range of colours is a spectrum. Let’s learn about the different forms of spectra in more detail.

### Suggested Videos

## Spectrum

Ordinary white light consists of waves of all wavelengths in the visible range. This is why, when white light passes through a prism, a series of coloured bands are seen called **spectrum**. This spectrum of white light ranges from violet at 7.5 x 10^{14} Hz to red at 4 x 10^{14} Hz.

Since the colours merge into each other i.e. violet merges into blue, blue into green and so on, we call it a **continuous spectrum**. When this light passes through an object or medium, the wave with the shortest wavelength (violet) deviates the most than the one with the longest wavelength (red).

The interaction of electromagnetic radiation with matter causes the atoms and molecules to absorb energy and go to a higher energy state. Since this state is unstable, they need to emit radiations to return to their normal states. This gives rise to emission and absorption spectra. Let’s learn about them in detail.

### Emission and Absorption Spectra

The** emission spectrum **is the spectrum of radiation emitted by a substance that has absorbed energy. Atoms, molecules, and ions that have absorbed radiation are called ‘**excited**‘. The** absorption spectrum **is the opposite of the emission spectrum.

It is the spectrum formed by electromagnetic radiation that has passed through a medium, in which radiation of some frequencies is absorbed. **Spectroscopy **is the study** **of emission and absorption spectra.

### Line Spectrum

Unlike visible light which shows a continuous spectrum of all wavelengths, the emission spectra of atoms in the gas phase emit light only at specific wavelengths with dark spaces between them. This is called **line spectra **or **atomic spectra** since the emitted radiation is identified by bright lines in the spectra.

Learn about Atomic Spectra here in detail.

Each element has its own unique **line emission spectrum**. Did you know that just the way fingerprints are used to identify people, the characteristic lines in an atomic spectrum are used to identify unknown atoms!

Robert Bunsen, a German chemist was the first to identify elements using line spectra. Using spectroscopic methods, we discovered elements such as helium, rubidium, scandium, gallium, thallium, etc.

Learn more about Group 18 Elements here in detail.

## Line Spectrum of Hydrogen

Hydrogen molecules dissociate when we pass electric discharge through gaseous hydrogen. Subsequently, the energetically excited H_{2} atoms emit electromagnetic radiation of discrete frequencies giving rise to a spectrum. The hydrogen spectrum has many series of lines. These lines are named after their discoverers.

In 1885, the scientist Balmer showed that if spectral lines are expressed as wavenumber, then the visible lines of the hydrogen spectrum obey the following formula –

Wavenumber = 109,677 (1/2^{2 }– 1/n^{2}) cm^{-1}

where n ≥ 3.

We call this series of lines, **Balmer series**. These lines are the only lines in the hydrogen spectrum that appear in the visible region of electromagnetic radiation. Johannes Rydberg, a Swedish spectroscopist, showed that all series of lines in the hydrogen spectrum can be described by the formula –

Wavenumber = 109,677 (1/n_{1}^{2} -1/n_{2}^{2}) cm^{-1}

where n_{1} = 1, 2…. and n_{2} = n_{1} + 1, n_{1} + 2……

The value 109,677 is the **Rydberg constant** for hydrogen. The lines that correspond to n_{1} = 1, 2, 3, 4, 5 are called Lyman, Balmer, Paschen, Brackett and Pfund series, respectively.

Series |
n_{1} |
n_{2} |
Spectral Region |

Lyman | 1 | 2,3… | Ultraviolet |

Balmer | 2 | 3,4… | Visible |

Paschen | 3 | 4,5… | Infrared |

Brackett | 4 | 5,6… | Infrared |

Pfund | 5 | 6,7… | Infrared |

The hydrogen atom has the simplest line spectrum of all elements. For heavier atoms, the line spectrum becomes more and more complex. However, there are certain features that are common to all line spectra –

- Line spectrum of every element is unique.
- There is regularity in the line spectrum of each element.

Now, that we understand the line spectrum of hydrogen, let’s understand the features of the hydrogen atom, its structure, and its spectrum.

## Bohr’s Model For Hydrogen Atom

In 1913, Neils Bohr was the first to explain the features of the hydrogen atom and its spectrum. His model is based on the following postulates –

- Electrons in a hydrogen atom can move around the nucleus in a circular path that has a fixed radius and energy. These paths, arranged concentrically around the nucleus, are
**orbits**. - The energy of an electron in orbit does not change with time. However, when the electron absorbs a required amount of energy, it moves from a lower stationary state to a higher stationary state. On the other hand, when the electron emits energy, it moves from a higher stationary state to a lower stationary state. This energy change is not continuous.
- The following equation gives the frequency of radiation absorbed or emitted when an electron transitions between two stationary states that differ in energy by ∆E –

ν = ∆E/h = (E_{2} – E_{1})/h

where, E_{1} and E_{2} are the energies of the lower and higher energy states, respectively. This equation is commonly called **Bohr’s frequency rule.**

- In a given stationary state, the angular momentum of an electron is given as –

m_{e} v r = n.(h/2π) where n = 1,2,3……

Thus, an electron can only move in those orbits whose angular momentum is an integral multiple of h/2π. This is why there are only certain orbits.

### Hydrogen Atom described by Bohr’s Model

For the hydrogen atom, Bohr’s theory describes that –

- The stationary states for the electron are numbered n = 1,2,3…These are
**Principal Quantum numbers**. - The radii of the stationary states are given as r
_{n}= n^{2}a_{0}, where a_{0}= 52,9pm. Therefore, the radius of the first stationary state i.e. the**Bohr orbit**is 52.9pm. The electron in the hydrogen atom is in this orbit i.e. n=1. As*n*increases,*r*will increase and the electron will be found away from the nucleus. - The energy of the stationary state is given by the equation –

E_{n} = – R_{H} (1/n^{2}) where n=1,2,3……

R_{H} is the **Rydberg constant **and the value is 2.18 x 10^{-18} J. Using this equation, you can calculate the energy of any state. For example, the energy of the stationary state for n=2 will be:

E_{2} = –2.18×10^{–18}J ( 1/2^{2})= –0.545×10^{–18} J

The energy of an electron is taken as zero when it is not under the influence of the nucleus. In this situation, n=∞ and the atom are called an ionized hydrogen atom.

- Bohr’s theory is also applicable to ions with only one electron, for example, the hydrogen atom. For these kinds of ions, the energies of stationary states are –

E_{n} = -2.18 x 10^{-18} (Z^{2}/n^{2})J

The radii are given by r_{n} = 52.9(n^{2})/Z pm, where Z is the atomic mass number.

- With Bohr’s theory, it is also possible to calculate the velocities of electrons moving in orbits. The magnitude of the velocity of the electron increases with an increase in positive charge on the nucleus and decreases with an increase in principal quantum number.

Learn more about Bohr’s Model of Hydrogen Atom here in detail.

### Explanation of Line Spectrum of Hydrogen

Bohr’s model can explain the line spectrum of the hydrogen atom. According to assumption 2, radiation is absorbed when an electron goes from orbit of lower energy to higher energy; whereas radiation is emitted when it moves from higher to lower orbit. The energy gap between the two orbits is –

∆E = E_{f} – E_{i} where f is the final orbit, i is the initial orbit

Since, E_{n} = – R_{H} (1/n^{2}), we can now say ∆E = (-R_{H} /n^{2}_{f}) – (-R_{H} /n^{2}_{i}) = R_{H }(1/n^{2}_{i }– 1/n^{2}_{f})

= 2.18 x 10^{-18} J (1/n^{2}_{i }– 1/n^{2}_{f})

The frequency and wavenumber associated with the absorption and emission of the photon can also be calculated –

ν = ∆E/h = R_{H}(1/n^{2}_{i} – 1/n^{2}_{f}) = 2.18 x 10^{-18} J/6.626 x 10^{-34} J s (1/n^{2}_{i} – 1/n^{2}_{f})

= 3.29 x 10^{15} (1/n^{2}_{i} – 1/n^{2}_{f}) Hz

Wavenumber = ν/c = R_{H}/hc (1/n^{2}_{i} – 1/n^{2}_{f}) = 3.29 x 10^{15} s^{-1}/ 3 x 10^{8} ms^{-s} (1/n^{2}_{i} – 1/n^{2}_{f})

= 1.09677 x 10^{7} (1/n^{2}_{i} – 1/n^{2}_{f}) m^{-1}

- When n
_{f}> n_{i}, the term in brackets is positive and energy is absorbed. - When n
_{i}> n_{f}, ∆E is negative and energy is released.

### Limitations of Bohr’s Model

Bohr’s model, although an improvement over Rutherford’s model, failed to account for the following points –

- It fails to explain the finer details of the hydrogen atom spectrum observed by spectroscopic techniques. Bohr’s model is unable to explain the spectrum of atoms other than hydrogen. It also fails to explain the concepts of Zeeman and the Stark effect.
- It does not explain the ability of atoms to form molecules with chemical bonds.

## Solved Example For You

Question: An electron in a hydrogen atom transitions from energy level with n=4 to n=2. What is the wavelength of the emitted light?

Solution: Wavenumber = R (1/n_{1}^{2} -1/n_{2}^{2}) cm^{-1}

Here, R = 109,677, n_{1} = 4, n_{2} = 2

Therefore, wavenumber = 109,677 (1/4^{2} – 1/2^{2}) cm^{-1 }= 20564.4 cm^{-1}

Now, λ = 1/wavenumber = 1/20564.4 = 486 x 10^{-7}cm = 486 nm.

## Leave a Reply