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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.
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उत्तर
In the hydrogen atom,
Radius of electron orbit, r =`(n^2h^2)/(4pi^2kme^2)`................. (1)
Kinetic energy of electron, `E_k = 1/2 mv^2 = (ke^2)/(2x)`
Using eq (1), we get
`E_k = (ke^2)/2 (4pi^2kme^2)/(n^2h^2)`
`=(2pi^2k^2me^4)/(n^2h^2)`
Potential energy, `E_p = (k(e)xx (e))/r = -(ke^2)/r`
Using eq (1),we get
`E_p = -ke^2 xx (4pi^2kme^2)/(n^2h^2)`
`= - (4pi^2k^2 me^4)/(n^2h^2)`
Total energy of electron, `E = (2pi^2k^2me^4)/(n^2h^2) - (4pi^2k^2me^4)/(n^2h^2)`
`= (2pi^2k^2me^4)/(n^2h^2)`
`=(2pi^2k^2me^4)/(h^2) xx (1/n^2)`
Now, according to Bohr’s frequency condition when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni) to the lower state (nf) is,
`hv = e_(n_i) - E_(n_i`
`or hv = - (2pi^2k^2me^4)/h^2 xx 1/n_f^2 - ((2pi^2k^2me^4)/h^2 xx 1/n_f^2)`
`or hv= (2pi^2k^2me^4)/h^2 xx (1/n_f^2 -1/n_f^2)`
`or v= (2pi^2k^2me^4)/h^3 xx (1/n_f^2 -1/n_f^2)`
`or v= (c2pi^2k^2me^4)/(ch^3) xx (1/n_f^2 -1/n_f^2)`
`(2pi^2 k^2me^4)/(ch^3) =`R = Rydherg constant = `1.097 xx 10^7m^-1`
Thus,
`v = Rc xx (1/n_f^2 -1/n_f^2)`
Now, higher state ni = 4, lower state, nf = 3, 2, 1
For the transition,
ni = 4 to nf = 3:→ Paschen series
ni = 4 to nf = 2:→ Balmer series
ni = 4 to nf = 1:→ Lyman series
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