1.

If R_(H) represents Rydberg constant, then the energy of the electron in the ground state of hydrogen atom is

Answer»

`- (hc)/(R_(H))`
`- (1)/(R_(H) ch)`
`-R_(H) ch`
`- (R_(H)C)/(h)`

Solution :ACCORDING to Rydberg formula,
`bar(v) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`....(i)
According to Bohr model, for an electronic TRANSITION,
`Delta E = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
But `Delta E = hv = (hc)/(LAMDA) = hc bar(v)`
`:. hc bar(v) = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
or `bar(v) = - (2pi^(2) me^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`...(ii)
COMPARING eqns. (i) and (ii)
`R_(H) = - (2pi^(2) me^(4))/(ch^(3)) :. E_(1) = - (2pi^(2) me^(4))/(h^(2)) = - R_(H) ch`


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