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Energy associated with nth orbit of H-at...

Energy associated with nth orbit of H-atom is given by the expression, `E_(n)=-(13.6)/(n^(2))eV`. Show that `E_((n+l))-E_(n)=(13.6xx2)/(n^(3))eV~, when 'n' in 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))`
If the value of n is very large, then (2n+1)=2n and `(n+1)=n" "therefore E_((n+1))-E_(n)=(13.6xx2n)/(n^(2)xxn^(2))=(13.6xx2)/(n^(3))eV`
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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.

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Knowledge Check

  • The energy of the electron in the hydrogen atom is given by the expression E_N=-1312/(n^2) KJ where n is an integer . The smallest amount of energy that a hydrogen atom in the ground state can absorb is

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    B
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  • For which of the following species, the expression for the energy of electron in nth orbit, E_(n)=(13.6Z^(2))/(n^(2))eV*"atom"^(-1) has the validity-

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    B
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    C
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    D
    tritium

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