- #1
kelly0303
- 580
- 33
Hello! This should be an easy calculation, but I want to make sure as the result seems a bit weird. Say I have an individual atom, for which I know the energy with a given uncertainty. In my case say I have ##1000 \pm 2.5## eV. Lets say that I perform collinear spectroscopy on this atom (for example using 2 lasers, the first one to excite the transition of interest, the other one to ionize and detect the ions, but this is not relevant for my question) and the laser used for spectroscopy has a transition frequency of ##f_0 = 6000## cm##^{-1}##. Let's assume the atom has a mass of 200 amu. Using: ##\frac{dE}{E} = 2\frac{dv}{v}## (the energy is low so I can use non-relativistic formulas), I get for this case ##v \approx 31062## m/s and ##dv \approx 39## m/s. Now using the Doppler formula: ##f = (1+\frac{v}{c})f_0##, where ##f_0## is the laser frequency mentioned above and ##f## is the rest frame frequency (might have messed up the +/- sign but it doesn't matter for the question). I also get, ignoring the uncertainty in ##f_0##, ##df = \frac{dv}{c}f_0##. If I plug this in I get ##f = 6000.62124## cm##^{-1}## and ##df = 0.00078##. So I am able to get the rest frame frequency, with only one measurement with an uncertainty of ##\approx 24## MHz. However, ##2.5## eV, correspond to ##29000## K! That is a huge temperature. How can I get such a narrow linewidth with such a huge temperature. What am I doing wrong? Thank you!