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In this problem, we're looking at the line spectrum of a hydrogen atom and
we're going to analyze one of the lines, and it's at 485.1 nanometers.
And we're going to determine where that electron is coming from in terms of
what energy level is it coming from.
We know that in the Balmer series and we see that that's listed here.
That is when an electron starts an a higher energy level and
it goes to the n equals 2 state.
So if we got a picture in our mind of what was going on here,
we would have an electron starting at some higher level that I don't know, and
I'm going to call it n sub i, because that's where it's starting.
And it's going to travel down to the n equals 2 level.
So that electron is going to travel down to the n equals 2 level.
And when it does, it will emit a photon of light.
And it says here that that photon that we're going to examine has
485.1 nanometers as its wavelength.
Now, what we know about the concept that's going on here is that the change
in energy of that electron as it travels from a high state to the low state.
Whatever that energy gap is and as it travels down, it is a negative value.
If I were to take the absolute value of that value.
[LAUGH] You get what I mean.
You would get the energy of the photon that's emitted.
So what we're going to do is we're going to first calculate the energy of
the photon.
From that, we'll know the energy change of the electron.
And then we'll have another equation that will relate to the change in energy to
the n values of those energy levels.
So let's begin by determining the energy of the photon.
Now, we know the energy of the photon is equal to h times nu, but
I don't know nu, which is the frequency.
I do know the wavelength, so that is equal to, if I were to
substitute the frequency for c over lambda, I'll get the wavelength in there.
So h is 6.626 times 10 to the minus 34 joules times seconds.
Now, I didn't give it to you as that many significant figures in our notes, but
I'm going to carry four, because my wavelength was with four.
The speed of light, which I usually use 3 times 10 to the minus 8.
I'm going to use 2.998 times 10 to the minus 8 meters per second,
because that'll have four significant figures.
And then my wavelength is 485.1 nanometers, but I need it in meters.
So I'll go ahead and put in the work to get that to meters.
So when I multiply and divide these values out,
I will have a energy of the photon equal to 4.095 times
10 to the minus 19, and then I'll have units of joule.
So that's the energy of one photon.
That energy of one photon would equal to the gap between these two.
Since the electron is traveling down, the change in energy of the electron
is actually equal to a negative 4.095 times 10 the minus 19 joules.
So the photon always has to have a positive energy, but
a change can be negative and
since the electron is traveling down, its change in energy is a negative value.
Now, the equation I want to use is that the change in energy of
an electron is equal to minus R sub H, which is the Rydberg constant,
times 1 over nf squared minus 1 over ni squared.
So ni is what I'm trying to determine.
nf is where it finishes up.
So it's finishing up at a n equals 2 state.
So let's plug in the numbers that we know here.
We have for the energy of the electron, negative 4.095
times 10 to the minus 19, and that is units of joules.
We have for the Rydberg constant a negative 2.179 times 10 to the minus 18.
And again, I'm giving you an extra significant figure that I
didn't give you in your notes to the Rydberg consent that has units of joules.
1 over final, that's 2, squared minus 1 over n initial squared.
So if I divide the Rydberg constant into
the change in energy, I will have a value of 0.1879.
And that's going to be equal to, and the joules will cancel, so it's unitless.
1 over 4 minus 1 over n sub i squared.
Now I want to subtract one fourth from both sides.
One fourth of this is the same as 0.25.
So when I subtract it from this side, I'm going to get a negative 0.0621.
And that will be equal to a negative 1 over n sub i squared.
because that's what's left, the minus sign hasn't gone anywhere.
So I'll change the sign, okay?
And I'll do the reciprocal.
So this is going to give me 1 over 0.0621 is equal to n sub i squared.
And then if I take the square root of both sides, I'd be left with n sub i.
And to the nearest whole number, which these n values have to be whole,
I have a n equal 4.
So the electron is coming from the number four state energy level down to
the number two energy level, and
when it does that, it will emit a photon with a wavelength of 485.1 nanometers.
Dr. Allison SoultLecturer
Dr. Kim WoodrumSr. Lecturer
