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प्रश्न
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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उत्तर
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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संबंधित प्रश्न
Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni) to the lower state, (nf).
When electron in hydrogen atom jumps from energy state ni = 4 to nf = 3, 2, 1, identify the spectral series to which the emission lines belong.
The light emitted in the transition n = 3 to n = 2 in hydrogen is called Hα light. Find the maximum work function a metal can have so that Hα light can emit photoelectrons from it.
Draw energy level diagram for a hydrogen atom, showing the first four energy levels corresponding to n=1, 2, 3 and 4. Show transitions responsible for:
(i) Absorption spectrum of Lyman series.
(ii) The emission spectrum of the Balmer series.
Mention demerits of Bohr’s Atomic model.
Which of these statements correctly describe the atomic model according to classical electromagnetic theory?
The wavelength of the first time line of Ballmer series is 6563 A°. The Rydberg constant for hydrogen is about:-
The Bohr model for the spectra of a H-atom ______.
- will not be applicable to hydrogen in the molecular from.
- will not be applicable as it is for a He-atom.
- is valid only at room temperature.
- predicts continuous as well as discrete spectral lines.
Find the ratio of energies of photons produced due to transition of an election of hydrogen atom from its (i) second permitted energy level to the first level. and (ii) the highest permitted energy level to the first permitted level.
The line at 434 nm in the Balmer series of the hydrogen spectrum corresponds to a transition of an electron from the nth to second Bohr orbit. The value of n is ______.
The energy of an electron in the first Bohr orbit of the H-atom is −13.6 eV. The energy value of an electron in the excited state of Li2+ is ______.
