When an electron jumps from higher orbits to lower orbits, radiations are emitted at particular frequencies. When an electron jumps from second, third orbits to the first orbit, the spectral lines are in the ultraviolet region. This is identified as Lyman series. When an electron jumps from the outer to the second orbit the series is called Balmer series and lies in the visible region of the spectrum. When the transition is from the outer to the third orbit then the series is termed Paschen and lies in the near-infra-red region. The transitions from the outer to the fourth and fifth orbits are termed as the Brackett and Pfund series respectively. The spectral lines for these two series lie in the very-far infra-red region of the hydrogen atom.

Fig 1.5 shows the energy level diagram of hydrogen.

Fig 1.5 Energy Level Diagram

Though it is theoretically possible to calculate the various energy states of the atoms of simpler elements, these levels must be determined indirectly from spectroscopic and other data for the more complicated atoms. The experimentally determined energy level diagram is shown in fig. 1.5. The number immediately to the right of a line gives the value of integer n while the numbers to the left of each line gives the energy to this level in electron volts. The lowest energy level E, is called normal or the ground state of the atom and the higher energy levels E2, E3, E4, ....... are called exited states.

It is customary to express the energy value of the stationary states in electron volts E rather than in joules W. Also it is more common to specify the emitted radiation by its wavelength l in angstroms rather than by its frequency f in hertz. Equation 1.2 can be rewritten as

l = (12400) / E₂ - E₁ (1.13)

where E₁ and E₂ are energy levels in Electron volts

Since only differences of energy enter into this expression the zero state may by chosen at will. It is convenient and customary to choose the lowest energy level as the zero level.

Find the wavelength of the photon entitled when a hydrogen atom goes from n = 10 state to ground state.

The wavelength in Angstrom is given by

l = 12400 / E₂ - E₁

Since Hydrogen atom goes from n = 10 state to ground state,

l = 12400 / E₁₀ - E₁

The energy of the 10^th state E₁₀

= (-13.6) / 10²

= -0.136 V

The energy in the ground state E₁

= (-13.6) / 1²

= -13.6eV

Thus the wavelength of the emitted photon

= (12400) / (-0.136 - (-13.6))

= 920.97 A.U.

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