The particle is a standing wave along its circular path. Given \(x_c\), each \(x_d\) makes a unique angle \(\theta\). The spread of \(\theta\) depended on the range of \(x_d\) which in turn depends on the value of \(v^2_{min}\) from \(c^2\). (Post "A Pump!" dated 25 Jul 2015.) At \(v^2_{min}\) and \(x=x_{min}\) is the orbit of the particle at its lowest \(KE\), but at the furthest point on its oscillation path.
We have a wave of lowest \(KE\) around a circle of radius \(x_{min}\),
\(2\pi x_{min}=n\lambda_{min}\)
This wavelength, \(\lambda_{min}\) is likely to be the wavelength we associate color with. Valid values of \(x\) around \(x_{min}\) spread the color spectrum on both sides of \(x_{min}\). For each value of \(x\) there is a unique value for \(\theta\).
The particle is in a helical path of radius \(x_c\). The specific motion along \(\small{2\pi x_c}\) is changed, but \(x_d\) that indicates oscillation remains unchanged and the wavelength we associate color with is still \(\lambda_{min}\).
In both cases, they are still de Broglie standing wave of radius \(x_{min}\).
This explanation allows for two different material of different energies at the quantum level to have the same color. And it allows for the finite spectrum width of the emission from a laser, otherwise the emission spectrum from a laser associated with one energy level change is a very sharp line.
Note: \(2\pi x_c=n\lambda\) serve to illustrate de Broglie standing wave. The actual standing wave that we associate color with is at \(v^2_{min}\) with \(x=x_{min}\).
It does not matter whether the particle has a wave or a helical path around \(x_c\), for the discussion so far. The helical path gives the particle a spin.
