Question

We can observe the differences in energy levels by observing the light that is emitted when atoms are heated. When atoms are heated, the electrons can jump up to a higher energy level. When one electron falls back down to a lower energy level, a single photon of light is emitted with an energy equal to the difference in energy of the two levels. We call this atomic emission spectroscopy. If an electron in an atom jumps from the n = 3 energy level which has an energy of -17.57 EV up to the n = 6 energy level which has an energy of -3.5 EV and then falls back down to the n =3 energy level, what is the energy of the photon emitted? Enter your answer in EV with at least 2 decimal places

Answer 1

Ans. The energy released when an electron falls from higher (initial) energy level to a lower (final) energy level is emitted in form of a photon. The energy of emitted photon (= energy released during fall of electron from higher to lower energy level) is given by-

Energy of emitted photon =

Energy of final energy level (n = 3) – Energy of Initial energy level (n = 6)

Or, Energy of emitted photon = -17.57 EV – (-3.5 EV)

Hence, Energy of emitted photon = -14.07 EV