Asked in ETEA MDCAT 2010 2010Moderate

The energy of an electron in the excited state n=4 in hydrogen atom is:

Correct answer: C. -0.85eV

  • A. -13.6eV
  • B. -3.4eV
  • C. -0.85eV
  • D. -1.5eV

Explanation

The energy of an electron in the excited state with n=4 in a hydrogen atom can be calculated by using the formula for the energy levels of hydrogen:E = -13.6eV / n²Substituting n=4 into the formula, we get:E = -13.6 / (4)²E= -13.6 / 16E = -0.85 eV

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