## University Physics with Modern Physics (14th Edition)

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.