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Energy associated with the n-th orbit of...

Energy associated with the n-th orbit of H-atom is given by the expresion, `E_(n)=(-13.6)/(n^2)eV`. Show that `E_((n+1))-E_(n)=(13.6xx2)/(n^(3))eV`, when 'n' is very large.

Text Solution

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`E_((n+1))-E_(n)=[-(13.6)/((n+1)^(2))-(-(13.6)/(n^(2)))]`
`=[(13.6)/(n^(2))-(13.6)/((n+1)^(2))]eV=(13.6(2n+1))/(n^(2)(n+1)^(2))eV`
If the value of n is very large, then `(2n+1)=2n and (n+1)congn`.
`therefore E_((n+1))-E_(n)=(13.6xx2n)/(n^(2)xxn^(2))=(13.6xx2)/(n^(3))eV`
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