Energy Level Diagram
A diagram which represents the total energies of electron in different stationary orbits of an atom is called the energy level diagram of that atom. In this diagram, total energies of electron in various stationary orbits are represented by parallel horizontal lines drawn according to some suitable scale.
The vertical line connecting any two states represents the transition of the electron from one to the other of these states. Difference of energies of two states gives the amount of energy emitted/absorbed according as the electron goes from higher to lower energy state or from lower to higher energy level, in an atom.
Total energy of electron in nth orbit of hydrogen atom is
E = -2∏2 mK2 e4/n2h2
On substituting the standard values, we get
E = (13.6/n2) eV (1)
Putting n = 1, 2, 3……….. We get the energies of electrons in various orbits as:
E1 = -13.6/12 = 13.6 eV
E2 = - 13.6/22 = -3.4 eV
E3 = - 13.6/32 = -1.51 eV
E4 = - 13.6/42 = 0.85 eV
E5 = - 13.6/52 = - 0.54 eV
E6 = - 13.6/62 = - 0.37 eV
E7 = - 13.6/72 = - 0.28 eV
Clearly, as n increases, En becomes less negative until at n = ∞, En = 0.
The energy levels of hydrogen atom are represented in energy level of the principle quantum number (n)labels the stationary states in ascending order of energy. The highest energy state corresponds to n = ∞and has energy E = 13.6/∞2 = 0 eV. This is the energy of the atom, when the electron is completely removed (r = ∞) from the nucleus and the electron is at rest. As n increases, energies of the excited states come closer and closer together.
Note that an electron can have any total energy above E = 0 eV. In such a situation, the electron is free. And there is a continuum of energy states above E = 0 eV.
The various spectra series of hydrogen atom have also been from this diagram we can calculate the energy, frequency, wave number etc. of any line of spectral series of hydrogen atom.
Example 1: calculate the radius of the third Bohr orbit of hydrogen atom and the energy of electron in that orbit.
Solution: here, r =? n = 3, E =?
As r = n2h2/4∏2 mke2
r = 9 × (6.6 × 10-34)2 × 9 × 109/ [4 × (22/7) × (22/7) × 9.1 × 10-31 (1.6 × 10-19)2
= 4.775 × 10-10 m
= 4.775 Å
E = - 2∏2 mk2 e4/n2h2
On putting the standard values, we get
E = - 2.43 × 10-19 J
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