Probe forming angle and resolution of a Scanning Electron Microscope

A SEM operates at $30keV$ and uses an objective lens with spherical aberration constant, $C_s=2mm$.

1. Calculate the optimal probe forming angle for this system.
2. Calculate the resolution of this SEM.

This is a tutorial question for my Master's Imaging course. In the solutions, first the de Broglie wavelength is computed (which is fine), then:

1. An equation is introduced for the optimal probe forming angle which I can't find anywhere (in my notes or online) and so don't understand its origin, this is: $$\alpha_{opt}=1.1\left(\frac{\lambda}{C_s}\right)^{\frac{1}{4}}$$
2. An equation (or actually 2) is introduced for the resolution which, again, I can't seem to find anywhere and so, again, fail to understand, these are: $$resolution=\frac{d_{min}}{2}=\frac{1.3\times\lambda^{\frac{3}{4}}\times C_s^{\frac{1}{4}}}{2}$$

If anyone could shed some light as to where these are coming from it would be very much appreciated.

Cheers

Turns out the total probe size is given by combining the effects of diffraction and spherical aberration, i.e. ${d_{tot}}^2={d_{C_s}}^2+{d_{diff}}^2$ and. $$d_{diff}=\frac{1.22\lambda}{\alpha},$$$$d_{C_s}=\frac{\alpha^3C_s}{2}.$$
Differentiating ${d_{tot}}^2$ w.r.t. $\alpha$ and setting this equal to zero gives a minimum. This can be re-arranged for $\alpha$ to give the optimisation angle, $\alpha_{opt}$, in terms of $\lambda$ and $C_s$ (which is the 1st expression I asked about in the question).
Then setting $\alpha =\alpha_{opt}$ in the original expression, $d_{min}$ can be found in terms of $\lambda$ and $C_s$ (which was the numerator of the 2nd expression I asked about in the question).
Resolution is then just defined to be $\delta_{min}=\frac{d_{min}}{2}$ which concludes all my original problems.