Match each photon to the result of its absorption

Question

n=1 , 2, 3, 4, infinityE= -63, -27, -12, -9, 0A gaseous sample of this element is bombarded by photons of various energies (in these same units). Match each photon to the result of its absorption (or lack thereof) by an n=1 electron. Photon EnergyA 63 B 51C 47D 36The results can be:n=1 to n=2 n=1 to n=3n=1 to n=4electron ejectednot absorbed

Saul Waters · Accepted Answer

WARNING I am only 15 but do have a good enough knowledge about particle physics and especially quantum mechanics. Use my answer as a guide line as it may not be exactly correct. This however does not mean that it is definitely wrong, it is actually most likely to be correct as I do have knowledge a 15 year old shouldn’t know.ok,so following Paul Dirac’s theory of negative energy and using your data provided we can say that for an electron to escape it’s orbitals (photo-electric effect) and energy of 63 must be exposed or grater.A: electron escapes with almost 0 kinetic energy.B: a photon with 51 energy is put in, so -63 + 51 = -12 which is n=1 to n=3.C: a photon with 47 energy is put in, so -63 + 47 = -16 which is n=1 to n=2.D: a photon with 36 energy is put is, so -63 + 36 = -27 which is n=1 to n=2I’m not really sure about C, as I think that the energy put in was enough to excite the electron from the ground state to its 2nd energy state but not enough to excite it to 3rd. So since electron that were excited with x energy will drop to its lowest state by emitting the x energy photon, also it can only drop in states. So I think that it will just as soon as excited will drop to its 2nd state as it’s the highest it can reach witn given enegy.

Saul Waters · Answer

Maybe, but I’m not 100% certain

Ms. Mara Collier

Match each photon to the result of its absorption

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