Answer
$2.22\times10^{-33}J, 1.39\times10^{-14}eV$.
Work Step by Step
Calculate the angular frequency.
$$\omega=\sqrt{\frac{kâ}{m}}=\sqrt{\frac{110N/m}{0.250kg}}=21.0\frac{rad}{s}$$
The energy levels are given by $E_n=(n+\frac{1}{2})\hbar\omega$.
The ground state energy is when n=0.
$E_0=\frac{1}{2}\hbar\omega=1.11\times10^{-33}J=6.93\times10^{-15}eV $
The spacing between adjacent levels is now calculated.
$\Delta E=\hbar\omega=2.22\times10^{-33}J= 1.39\times10^{-14}eV $
The spacing is so tiny that quantum effects are unimportant.
