# Units used in X-ray energy spectrums

I am trying to understand the X-ray emission spectrum based on medical X-ray tubes and there are several variants of the graph function given in several books as in this image:

At a specific energy (in KeV) the plot shows the number of photons emitted by an element in the units of $Hash/ [KeV*cm^2 *mAs]$ where $Hash$ is the numeric value of the number of photons.

My question is regarding the units of the dependent variable of the graph function. How is there a $KeV$ unit added to it?. Shouldn't the number of photons have the units of $Hash/ [cm^2 *mAs]$ where, the photons cross an area over a particular current-time product and therefore $cm^2*mAs$ in the denominator is understood but why a KeV term(unit) is also present?

• what you really see there is a spectral distribution function - the total intensity is $I = \int I(E) dE$. So the integral takes care of the energy dimension. – Sanya Oct 25 '17 at 21:06
• Yes, I would get the total number of photons if I do $∫I(E)dE$ but if I want to know the number of photons at a specific energy, the units makes no sense to me. – dykes Oct 25 '17 at 21:46
• there is no "number of photons at a specific energy". For that, you can either remember your calculus (a single number has a Lebesgue-measure of zero) or (with more handwaiving) the Heisenberg uncertainty principle (there is never a absolutely discrete energy). See also robs great answer. – Sanya Oct 25 '17 at 21:54
• Very similar to (duplicate?): physics.stackexchange.com/questions/334970/… – dmckee --- ex-moderator kitten Oct 25 '17 at 22:16

What you've plotted here is a differential spectrum: the number of x-ray events observed in each energy window from $E_i$ to $E_i + \Delta E$. The most common way to construct such a spectrum is to build a histogram. But in a histogram, the number of events in each bin depends on the width of the bins, $\Delta E$. For example, if you start with a histogram that has 100 energy bins, decide it's too noisy, and combine the odd bins with their even neighbors to produce a new histogram with 50 bins, the number of events in each bin will approximately double. But you haven't changed your data, in that case, just your representation of that data.
You can avoid that blurring between data and representation by displaying, rather than raw counts per bin, the counts divided by the bin width. In our $100\to50$ rebinning example, the number of counts in the larger bins would approximately double, but the width doubles as well: the differential spectrum doesn't change.
Essentially, each point in your spectrum here is $$\frac{dN}{dE\cdot dA \cdot dt\cdot dI}$$ where the integral of this spectrum, $N$, is the total number of events you'd expect from the experiment. You're not bothered by the fact that if you ran the experiment twice as long, or used a detector with twice the area, or drove twice as much current through the source, that you'd double the number of photons you find. For energy the relationship is nonlinear: the total number of events is $N = \int \frac{dN}{dE} dE$, and what's plotted here is the differential $dN/dE$.
I learned this as a graduate student when I tried, unsuccessfully, to turn a plot of neutron flux versus wavelength, $dN/d\lambda$, into a plot of neutron flux versus energy, $dN/dE$, using the de Broglie relation $p = h/\lambda$ and the kinetic energy $E = p^2/2m$. You can't just re-map the points on the spectrum to the new values; you have to take into account that the denominator is different as well in order to have the two spectra integrate the to same $N$.