For some one-electron system with $\mathrm{l}=0$ and $\mathrm{m}=0$, the functions $\mathrm{N}_{0} \mathrm{e}^{-\sigma}$ and $\mathrm{N}_{1}(2-\sigma) \mathrm{e}^{-\sigma}$ refer respectively to the ground $\left(\mathrm{E}_{0}\right)$ and fist excited $\left(\mathrm{E}_{1}\right)$ energy levels. If a variational wave function $\mathrm{N}_{2}(3-\sigma) \mathrm{e}^{-\sigma}$ yields an average energy $\overline{\mathrm{E}}$, it will satisfy
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
(3 − σ)e^{−σ} is a linear combination of the ground function e^{−σ} and the first excited function (2 − σ)e^{−σ}, so its energy is a weighted average of E₀ and E₁: E₀ ≤ E ≤ E₁.