Cepstrum, peak picking
Cepstrum argument is called quefreency, which is in fact a time domain. So for example if you are looking for the fundamental frequency then you are searching for a peak in a specific range. In your case that's $[0.002; 0.2] \;sec$, which corresponds to $[50; 500] \; Hz$., knowing that $f=\dfrac{1}{t}$. 

So when you are searching for fundamental frequency, you can find peak in quefrency domain and then convert it to corresponding frequency. You can also use a different approach. We know that our cepstrum is symmetrical around center (similarly to Fourier Transform), and also we have our sampling frequency $f_s$. Let's say that you have $N$ cepstrum bins, therefore distance between each of them is: $\dfrac{f_s}{N}$. That's why you can find the corresponding fundamental frequency $f_0$ by dividing sampling frequency by the index of a peak.

For example our peak is at index number $10$, and sampling frequency is $1000 Hz$. Corresponding $f_0$ is $100 Hz$. If peak is at bin $50$ (more to the right, meaning lower frequency - remember quefrency is like time), then $f_0 = 20$ Hz. The subraction of 1 in denominator is cause by fact that MATLAB is indexing from 1 not from 0, like most of the programming languages.