X-ray spectra and the presenence of L, M,.. etc series? Consider the diagram below:

(Public Domain)
On diagrams like why are the $L$, $M$, and so on series not present? And is their any way to make them present e.g. reduce the tube voltage etc.
 A: The figure seems to be a cartoon, not a real spectrum. And/or it is not of a single element. The pair of lines at higher energy look like $L_\alpha$ and $L_\beta$ but then it is an element with impossibly high $Z$. The other pair of lines has an intensity ratio typical of $K_\alpha$ and $K_\beta$ but then the weaker one ($K_\beta$) should be at higher photon energy.
Tungsten is a common material for x-ray anodes in this range. The W $K_{\alpha 1}$ energy is at 59.31 keV and $K_{\beta 1}$ is at 67.23 keV. But maybe it also shows the $K_{\alpha 2}$ line at 58.0 keV? In that case the alpha1-alpha2 splitting is not to scale.
The point of the cartoon seems to be that one needs electrons with at least the $1s$ binding energy to excite the $K_\alpha$ lines. So 60 keV is not enough to get the K emission lines of tungsten.
A: Since the answer to this question is mixed up in the answers and comments, and since I have found some other points concerning this question I thought I would 'summarize' them in this answer.
The intensity of a line depends on 4 main things:
1. The Affect of filtering 
(Thanks to Jon Cluster and Pieter for explaining) Filtering can be put in place deliberately or as a natural consequence of the apparatus used e.g. the glass window, air etc. Although this filtering can be determined by the filter used, as a rule higher energies x-rays (and therefore K-series) tend to be filtered less. 
2. The Affect of Ionization Cross-section
(See reference {1}). The ionization cross-section, $\sigma(E)$, determines the probability that if you send in a beam of electrons (energy $E$) you get an ionization. The ionization cross section follows the following rough pattern.


*

*Increases rapidly above $E_{ion}$ (the ionization energy)

*Reaches a maximum at $E\sim 3E_{ion}$

*Decreases slowly and monotonically above there.


3. Probability the vacant site will be filled by a given outer state. 
(see reference {2}) Let us say we have a vacant site in the $K$ then the intensity of the lines (ignoring other factors) formed by transitions to this line is proportional to the statistical wight $2J+1$ of the starting level. This is known as the 'sum rule'.
4. Fluorescence Yield
(Thanks to Pieter, see also reference {2}) The probability of emitting characteristic radiation is balanced by that of emitting an augar electron instead. In all cases expect that of the K-series the probability of emitting an augar electron is much greater then that of emitting an x-ray reducing their intensities. This can be quantified in something called the 'fluorescence yield' defined as the ratio of the radiative probability, $\Gamma_r$, to the total transmission probability, $\Gamma_{tot}$:
$$\omega=\frac{\Gamma_r}{\Gamma_{tot}}$$
Conclusion
The above factors (and others such as detector efficiency etc.) determine the intensity of the lines and thus their presence in the spectrum.
References
{1} Bell, D.C. and Garratt-Reed, A.J., 2003. Energy dispersive X-ray analysis in the electron microscope (Vol. 49). Garland Science. ($\S$2.2) 
{2} Verma, H.R., 2007. Atomic and Nuclear Analytical Methods:  XRF, Mössbauer, XPS, NAA and Ion-Beam Spectroscopic Techniques. Springer ($\S$1.3.2)
A: The temperature may be high enough to strip most L electrons, while K electrons remain (assuming high-Z element). I'd say it somehow matches the value of 100 kV.
Now to obtain K lines that strong, you need other effects in your setup, such as non-thermal electrons. They could generate those high-energy lines, while not interfering much with the thermal emission.
