# How the calculate the energy resolution of a lead crystal calorimeter?

I'm trying to calculate the energy resolution that can be found with a lead crystal calorimeter, and I have the following equation to find the approximate value:

$$\frac{\sigma_E}{E} = 0.02\% + \frac{6.3\%}{\sqrt E}$$

I'm not able to make much sense of this equation, even though it is similar to other equations for calculating the energy resolution of a calorimeter. The left hand side is unitless, while the right is not, and I'm not sure what unit I should be using. Similar examples have used GeV, but the energy I am needing to deal with is on the order of KeV. Putting in my value in GeV generates an uncertainty much value than the energy value itself, is that correct?

e.g. if E = 500 KeV:

$$\sigma_E = \left(2*10^{-4}+ \frac{0.063} {\sqrt{500*10^{-6}\rm\,GeV}}\right)*500*10^{-6}\,\mathrm{GeV} = 1.4*10^{-3}\,\mathrm{GeV}$$

I really don't think I used the equation right, and the answer seems far too large. How can I use this equation properly to get a good approximation of the energy resolution?

$$\frac{\sigma_E}{E} = 0.02\% + \frac{6.3\%}{\sqrt E} \tag1$$
A value and its uncertainty, such as $$E \pm \sigma_E$$, must have the same units: you cannot add or subtract two quantities with different units, because then the result of the computation depends on which units you are using. In your case you have a dimensionless ratio $$\sigma_E/E$$ set equal to the sum of a dimensionless number and a dimensionful number. Bummer.
$$\frac{\sigma_E}{E} = 0.02\% + 6.3\% \sqrt{\frac{E_\text{ref}}{E}} \tag2$$
Your reference material may define $$E_\text{ref}$$ in text: if (1) is near a sentence fragment like "where $$E$$ is measured in keV," that sets $$E_\text{ref} = 1\,\rm keV$$, and tells you your detector has 6% resolution for keV-scale energies and 0.2% resolution for MeV-scale energies. That would be an unsurprising energy resolution for a lead-crystal calorimeter: nuclear data tables are full of MeV-scale photon energies reported to four significant figures.