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  • When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula

When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula

\bar{v}=109677\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}

What points of Bohr's model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.

Answers (1)

When through the hydrogen gas, an electric discharge is passed, the molecules produce the excited hydrogen atoms by disassociating. Electromagnetic radiation is emitted by these excited atoms of discrete frequencies. It can be represented by the following formula :

\bar{v}=109677\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}

2 basic postulates of Bohr's model of an atom can be used to reach this interpretation of the spectrum of atomic hydrogen:

1. The following principle is followed by the criterion for selecting the electron's stationary orbits, "The angular momentum of the electron should be an integral multiple ofh/2\pi (h= Planck's \; constant)."

If we assume that at an orbit of radius r, the mass and velocity of the electron is m and v, then its angular momentum can be given by the following formula: -

mvr=n.h/2\pi , where n is a non-zero positive integer.

2. In accordance with the quantum theory of radiation, when an electron changes its orbit from one orbit to another, there will be a difference of energy between the two energy levels which will be either absorbed or emitted.

Therefore, when from an orbit with energy E2, an electron jumps to an orbit of energy E1 (E2>E1), the difference in energy is given out in the form of quantized radiation. If we assume the frequency of the radiation to be ϑ, then the energy will be given by :

E_{2}-E_{1}=h\vartheta

With these two points, Rydberg's constant can be reached –

In a Bohr type system orbit, consider 2 electron orbits that have quantum numbers n1 and n2, in such a way that n2>n1. The corresponding energies are En1 and En2 and En1<En2.

Therefore, during the transfer of electrons from n2 to n1 orbit, the energy emitted will be: -

E_{2}-E_{1}=h\vartheta, wherein, ϑ is the frequency of radiation.

As per the Bohr atomic model for the hydrogen atom (Z=1):

E_{1}=\frac{-me^{4}}{8\epsilon _{0}^{2}h^{2}n_{1}^{2}},E_{2}=\frac{-me^{4}}{8\epsilon _{0}^{2}h^{2}n_{2}^{2}}

Thus, E_{2}-E_{1}=h\vartheta=\frac{me^{4}}{8\epsilon _{0}^{2}h^{2}}\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]

\vartheta =\frac{me^{4}}{8\epsilon _{0}^{2}h^{3}}\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ] and

\bar{v} =\frac{\vartheta }{c}=\frac{me^{4}}{8\epsilon _{0}^{2}h^{3}}\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]

Now, on substituting the values form, e, \epsilon _{0} and h

\bar{v}=109677\frac{1}{n_{2}^{2}} cm^{-1}

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