Answer
See explanation.
Work Step by Step
For a period of 0.500 s, the frequency is 2.00 Hz.
The energy levels are given by $E_n=(n+\frac{1}{2})hf$.
The ground level energy, n = 0, is $\frac{1}{2}hf$.
$$ E_o = \frac{1}{2} (6.626\times10^{-34}J \cdot s)(2.00 Hz) =6.63\times10^{-34}J=4.14\times10^{-15}eV $$
The spacing between energy levels is $hf$.
$$ \Delta E = (6.626\times10^{-34}J \cdot s)(2.00 Hz) =1.33\times10^{-33}J=8.28\times10^{-15}eV $$
The spacing is undetectable; it is so tiny that quantum effects are unimportant for this everyday pendulum.
