the energy of an electron in the first bohr orbit of 4 atoms is -13.6 eV. The possible energy value of the excited state for electron in hydrogen orbits is

Dear Student,

Please find below the solution to the asked query:

The energy of an electron in Bohr orbit of an Hydrogen atom can be calculated using the expression: 

$E=\frac{-13.6 eV}{n^2}$

Since, electron in H-atom at the ground state is present at n = 1, then the possible energy value of the excited state i.e. at n = 2 will be:

$E=\frac{-13.6 eV}{2^2}=\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{ }\mathit{e}\mathit{V}$

Hope this information will clear your doubts about Structure of Atom.
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