JEE Main 2019 Chemistry question
The ground state energy of hydrogen atom is −13.6 eV. The energy of second excited state of He ion in eV is:
Answer: (A)
Solution
n = 3: −13.6 × 4/9 = −6.04 eV.
The ground state energy of hydrogen atom is −13.6 eV. The energy of second excited state of He ion in eV is:
n = 3: −13.6 × 4/9 = −6.04 eV.