Answer
See explanation.
Work Step by Step
The energy levels are given by $E_n=(n+\frac{1}{2})hf$.
a. Find the quantum number n.
$$\frac{1}{2}mv^2=(n+\frac{1}{2})hf $$
$$\frac{1}{2}(0.020kg)(0.480m/s)^2=(n+\frac{1}{2})(6.626\times10^{-34}J \cdot s)(1.50 Hz) $$
$$n=\frac{0.5 (0.020kg)(0.480m/s)^2}{ (6.626\times10^{-34}J \cdot s)(1.50 Hz)}-\frac{1}{2} $$
$$n=2.3\times10^{30}$$
b. The energy difference is hf.
$$\Delta E = (6.626\times10^{-34}J \cdot s)(1.50 Hz) = 9.94\times10^{-34}J $$
The spacing is undetectable; it is so tiny that quantum effects are unimportant.