Suppose we somehow knock out an electron from an atom. The energy level of the e
ID: 2163431 • Letter: S
Question
Suppose we somehow knock out an electron from an atom. The energy level of the electron is 720 keV. Another electron in that atom makes a transition from its level at ?1.8 keV to that empty level and emits a photon. (a) Find the energy (keV) of the emitted photon. (b) Find the wavelength (nm) of the photon. This is how x rays are produced in commercial instruments. A beam of high-energy electrons are directed at a target where they knock electrons out of atoms. Other electrons in the atoms make transitions to these empty levels, emitting photons.Explanation / Answer
first you have two energy levels. -1.8 kev this has an electron after the collision 720 kev this electron is knocked out of the atom part a) the -1.8 kev electron moves down to the -720 kev level that means it has to loose energy -1.8kev - (720 kev) = -721.8 kev difference this is the energy of the photon that is generated part b) we know that a photon of energy is E = hf E is energy in joules f being frequency h being 6.626*10^-34 c being speed of light 3.0*10^8 but we want wavelength so we use f*L = c L = lamba or the wavelength f = c/L so the equation E = hf = h(c/L) so rearanging we get L = hc/E since E is in joules but our energy levels were in eV you have to convert the 11 kev to joules by Ej = eV* 1.602*10^-19 solving for L = 6.626*10^-34*3.0*10^8/(721.8 *10^3*1.6*10-19) 1.72 * 10 ^-12meters wavelength or .0172 angstroms seems a pretty reasonable answer but we want this in nm that is 10^-9 so that is = 0.00172 nm this seems reasonable because light is in the 300 to 1000 nm range