in this problem we start with the observation of a 410 nanometer photon
emitted from a hydrogen gas sample in part a we want to find the energy of this photon and
then in part b we're going to figure out what the electron transition was that caused the emission
of this photon so the energy of a photon that's planck's constant multiplied by the frequency
e equals hf but we can replace the frequency with c over lambda and get
hc over lambda the reason we can do this is because the speed of a wave is given by
the product of the frequency and the wavelength i wrote the speed as c there because this is light
and it travels at the speed of light so you solve for f and replace it with c over lambda and i'll
post a link to the video where this was derived but we have a very convenient approximation for
hc we can write this as approximately 1240 ev nanometers and i did this because my wavelength
is given in nanometers and i want the energy of the photon in ev so the nanometers cancel out
and i get 3.02 electron volts all right so for a photon with 3.02 electron volts to have been
emitted from this hydrogen sample an electron must have dropped by 3.02 electron volts as
it transitioned from one energy level to the next and we have a visualization of the energy
levels for the hydrogen atom over on the right that's given by the formula negative 13.6 e v
over n squared where n is the state number for the electron and the first four of these are
written into the diagram all i did was sub n n equals one two three and four
e1 is just negative 13.6 ev divided by 1 squared which is negative 13.6 ev
if i divide by 2 squared instead i get negative 3.40 ev and so on
and now our job is to figure out what must have been the initial and final states for
this electron in order to produce a photon with 3.02 ev of energy and there's a major point here
to simplify the search i see that the transition definitely isn't down to the ground state
because the lowest energy transition down to the ground state involves a drop of more than 10
electron volts of energy well that would be way too energetic of a photon
so transitions down to the n equals one state are forbidden here transitions down to the n equals
two state i don't see a problem there maybe i can find a higher energy level here that when i drop
down to negative 3.4 the change is 3.02 it seems reasonable and if i look at the n equals 3 state
negative 1.51 ev simply isn't a big enough number where i could drop down from a higher
energy state to the n equals three state and have that be of three electron volt energy transition
so transitions down to the n equals three state are ruled out even if i went from a very large
value of n down to n equals three that would only yield about one and a half electron volts
for the photon so i know for sure i'm looking at a transition down to the n equals two state and that
simplifies the search certainly if i drop from n equals three to n equals two that's not a three
electron volt transition n equals four to n equals two that's not enough that's not three electron
volts for that transition so we just start writing down the next energy levels i'm gonna start with
five and six and so on and we're gonna keep checking to see if we get the right number
so the n equals 5 energy state is negative 0.544 ev
if we look at the difference between this and the n equals two state this gives me an energy
transition of about 2.86 ev so that's not going to do it then we look at the n equals 6 state
that gives me negative 0.378 ev if i look at the difference between this and the n equals 2 state
i arrive at the transition i was looking for if an electron transitions from the n equals
six state down to the n equals two state the difference in energy is 3.02 electron volts
and the photon that's emitted from that process will carry away the 3.02 electron volts of energy
so to write down our final answer that photon we observed must have come from an initial n
of 6 and a final n of 2. and that's it if you enjoyed this video or at least found it useful
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