K.
Remember that for X-ray spectra there are 2 different features. The part that you may have been looking at is the continuous part.
The continuous portion as mentioned in the beginning of the reading is produced by bremsstrahlung radiation.
The events that are mentioned in the question are different from the continuous nature of bremsstrahlung radiation because it is an actual collision between an X-ray photon and an electron.
So, lets look at each answer choice:
B) P1 b/c peak 1 has lower intensity
although peak 1 having a lower intensity may be true, P1 is not the most probable.
C) P2, b/c the peak has the longer wavelength
again, it is true that P2 has a longer wavelength, but that is not the reason why it is more probable.
Note: The graph is designed specifically for the question. The wavelength is the dependent variable and is located on the x-axis. So if you really have NO idea at all, the graph may hint at intensity being the indicator.
D) The reason is given in the solution.
But going back to why the intensity (rather than wavelength) indicates probability, I guess this requires a bit more thinking.
The intensity is proportional to the # of electrons emitted at a given wavelength. We know that the x-ray will emit at a multitude of wavelengths. So the x-ray photons are at many different wavelengths and have many different energies. Your task is to determine which event has a higher probability. The event the refer to is the collision of the x-ray and the electron. So if more electrons are ejected for P2, then it goes to show that the event occurs more.
