Question #45665

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↳Redirected from "An electron in a hydrogen atom drops from energy level n=5 to n=3. What is the energy transition using the Rydberg equation?"

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Answer 1 · 2017-03-15T19:39:29

#E approx 4513 \ eV#

The energy of a photon is:

#E = hf = (hc)/lambda#, as speed #c = f lambda#

....and where Plank's constant is #h approx 6.63 times 10^(-34) m^2 kg s^(-1)#

So we have:

#E = (6.63 times 10^(-34) cdot 3 times 10^8)/(0.275 times 10^(-9)#

#implies E approx 7.23 times 10^(-16) \ J#

And as: #1 \ J approx 6.24 times 10^(18) eV#

#E approx 4513 \ eV#