The energy for n = 4 and ℓ = 2 state is greater than the energy for n = 5 and ℓ = 0 state.

Question

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The energy of an electron in a hydrogen atom is given by the equation E = -13.6/n^2 eV, where n is the principal quantum number. The principal quantum number determines the energy level of the electron, with higher n values corresponding to higher energy levels. The angular momentum quantum number, ℓ, determines the shape of the electron's orbital. Orbitals with higher ℓ values have more angular momentum and are more spread out.In the given question, the energy for n = 4 and ℓ = 2 state is greater than the energy for n = 5 and ℓ = 0 state. This is because the n value is more important than the ℓ value in determining the energy of an electron. The n = 4 state is a higher energy level than the n = 5 state, so the electron in the n = 4 state has more energy than the electron in the n = 5 state, even though the ℓ = 2 state has more angular momentum than the ℓ = 0 state.

The Energy for N = 4 and ℓ = 2