The energy of an electron in the first Bohr orbit of H atom is -13.6 eV. The possible energy value of the excited state for the electron in Bohr orbits of hydrogen is:

- 3.4 e V

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4.2 e V

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- 6.8 e V

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+ 6.8 e V

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Solution

The correct option is A $- 3.4 e V$
Excited state is given as = $\frac{- 13.6 e V}{n^{2}}$

e.g, n = 2, E = $\frac{- 13.6}{4}$ = $- 3.4 e V$

The rest of the given values are not possible.
Hence, the correct option is A.

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The energy of an electron in the first Bohr orbit of H atom is -13.6 eV. The possible energy value of the excited state for the electron in Bohr orbits of hydrogen is:

The correct option is A −3.4 eVExcited state is given as = −13.6eVn2e.g, n = 2, E = −13.64 = −3.4eVThe rest of the given values are not possible.Hence, the correct option is A.

The correct option is A $- 3.4 e V$
Excited state is given as = $\frac{- 13.6 e V}{n^{2}}$

e.g, n = 2, E = $\frac{- 13.6}{4}$ = $- 3.4 e V$

The rest of the given values are not possible.
Hence, the correct option is A.