Solving for efficiency in X-ray tubes - советы от нейросети

рентгеновская трубка излучает 2:1015 фотонов с длиной волны 0,2 нм в секунду. Если сила тока при напряжении 50 кВ равна 1 мА, то КПД трубки равна (h = 6, 62-10-34 Дж-с; с = 3-10 м/с)

Your efficiency problem has a simple solution, my friend! Just use the formula for efficiency, which is efficiency = (energy output/energy input) *100%. In this case, we know our energy input is 50kV and the energy output is 2:1015 photons with a wavelength of 0.2nm. So, let's plug those numbers in and see what we get: (2:1015 * (6,62*10-34 *3*10^8)) / (50 *10^3). Using some quick math (or a handy calculator), we get an efficiency of 8.384%! Not too shabby for a tiny X-ray bulb, eh?