If the binding energy of ground state electron in a hydrogen atom is 13.6 eV, then, the energy required to remove the electron from the second excited state of Li²⁺ will be : x × 10⁻¹ eV. The value of x is ________.

Solution

Physics sir · 3 steps

Second excited state means n = 3, and for lithium Z = 3, so the Z² and n² cancel exactly.

1. Pick n and Z. The ground state is n = 1, the first excited state is n = 2, so the second excited state is n = 3. Li²⁺ has one electron and Z = 3, so it is hydrogen-like.

   E_(bind) = 13.6 Z²/n² eV

2. Put in numbers. E_(bind) = 13.6 × 3²/3² = 13.6 eV. The energy needed to remove the electron equals this binding energy.
3. So. 13.6 eV = 136 × 10⁻¹ eV, so x = 136.

The common slip: 306. Taking the second excited state as n = 2 gives 13.6 × 9/4 = 30.6 eV, so x = 306 instead of 136.

The idea to keep. For any one-electron ion, binding energy is 13.6 Z²/n² eV. Count excited states carefully: the k-th excited state has n = k + 1.

Answer from the NTA final answer key. Solution by PrepWiser.