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Question
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.
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Solution
According to Bohr’s postulates, in a hydrogen atom, a single alectron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit os a given radius, the centripetal force is provided by Columb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small.
So,
`(mv^2)/r = (ke^2)/r^2`
or `mv^2 = (ke^2)/r .............. (1)`
where m = mass of electron
r = radius of electronic orbit
v = velocity of electron.
Again,
`mvr = (nh)/(2π)`
or `v = (nh)/(2πmr)`
From eq.(1), we get,
`m((nh)/(2πmr)^2) = (ke^2)/r`
`=> r = (n^2h^2)/(4π^2kme^2).....................(2)`
(i) Kinetic energy of electron,
`E_k = 1/2 mv^2 = (ke^2)/(2r)`
Using eq (2), we get
`Ek =ke^2/2 (4π^2kme^2)/(n^2h^2)`
=`(4π^2kme^2)/(n^2h^2)`
`(2π^2k^2me^4)/(n^2h^2)`
(ii) Potential energy
`E_p = -(k(e) xx (e))/r = - (ke^2) / r `
Using eq (2), we get
`E^p =-ke^2 xx (4π^2kme^2)/(n^2h^2)`
= `-(4π^2k^2me^4)/(n^2h^2)`
Hence, total energy of the electron in the nth orbit
`E =E_p+E_k =(4π^2k^2me^4)/(n^2h^2)+(2π^2k^2me^4)/(n^2h^2) =- (2π^2k^2me^4)/(n^2h^2) =- (13.6)/n^2 eV `
When the electron in a hydrogen atom jumps from higher energy level to the lower energy level, the difference of energies of the two energy levels is emitted as a radiation of particular wavelength. It is called a spectral line.
In H-atom, when an electron jumps from the orbit ni to orbit nf, the wavelength of the emitted radiation is given by,
`1/λ = R (1/n_f^2 -1/n_i^2)`
Where,
R → Rydberg’s constant = 1.09678 ×107 m−1
For Balmer series, nf = 2 and ni = 3, 4, 5, …
`1/λ = R (1/2^2 -1/n_i^2)`
Where, ni = 3, 4, 5, …
These spectral lines lie in the visible region.
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