The energy of the electron in the excited state n=3 in hydrogen atom is:
Correct answer: B. -1.51eV
- A. -13.6eV
- B. -1.51eV
- C. 3.40eV
- D. 0.54eV
Explanation
The energy of an electron in the excited state ( n=3 ) for a hydrogen atom can be calculated using the formula: En =−13.6/n2 eVPlugging in ( n=3 ), we get: E3 =−13.6 /(3)2 eV E3 =−13.6/9 eV E3 =−1.51 eVSo, the energy of the electron in the excited state ( n=3 ) in a hydrogen atom is ( -1.51 eV ).
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