Start learning →

In hydrogen atom, energy of n-th orbit:

A-13.6 × n eV
BAlways 0
C-13.6/n² eV
D-13.6 × n² eV
Answer & Solution
Correct answer: C. -13.6/n² eV
E_n = -13.6/n² eV. Ground state n=1: -13.6 eV (most bound). Excited states approach 0 as n → ∞. Ionization energy = 13.6 eV (energy to free electron from ground state).
Practise this question with Quill — JEE Advanced →

Keep practising Atoms and Nuclei

Browse all Atoms and Nuclei questions for JEE Advanced, or explore more Physics topics.

Start a Atoms and Nuclei lesson →