0:00

In this problem, we're looking at the line spectrum of a hydrogen atom and

Â we're going to analyze one of the lines, and it's at 485.1 nanometers.

Â And we're going to determine where that electron is coming from in terms of

Â what energy level is it coming from.

Â We know that in the Balmer series and we see that that's listed here.

Â That is when an electron starts an a higher energy level and

Â it goes to the n equals 2 state.

Â So if we got a picture in our mind of what was going on here,

Â we would have an electron starting at some higher level that I don't know, and

Â I'm going to call it n sub i, because that's where it's starting.

Â And it's going to travel down to the n equals 2 level.

Â So that electron is going to travel down to the n equals 2 level.

Â And when it does, it will emit a photon of light.

Â And it says here that that photon that we're going to examine has

Â 485.1 nanometers as its wavelength.

Â Now, what we know about the concept that's going on here is that the change

Â in energy of that electron as it travels from a high state to the low state.

Â Whatever that energy gap is and as it travels down, it is a negative value.

Â If I were to take the absolute value of that value.

Â [LAUGH] You get what I mean.

Â You would get the energy of the photon that's emitted.

Â So what we're going to do is we're going to first calculate the energy of

Â the photon.

Â From that, we'll know the energy change of the electron.

Â And then we'll have another equation that will relate to the change in energy to

Â the n values of those energy levels.

Â So let's begin by determining the energy of the photon.

Â Now, we know the energy of the photon is equal to h times nu, but

Â I don't know nu, which is the frequency.

Â I do know the wavelength, so that is equal to, if I were to

Â substitute the frequency for c over lambda, I'll get the wavelength in there.

Â So h is 6.626 times 10 to the minus 34 joules times seconds.

Â Now, I didn't give it to you as that many significant figures in our notes, but

Â I'm going to carry four, because my wavelength was with four.

Â The speed of light, which I usually use 3 times 10 to the minus 8.

Â I'm going to use 2.998 times 10 to the minus 8 meters per second,

Â because that'll have four significant figures.

Â And then my wavelength is 485.1 nanometers, but I need it in meters.

Â So I'll go ahead and put in the work to get that to meters.

Â So when I multiply and divide these values out,

Â I will have a energy of the photon equal to 4.095 times

Â 10 to the minus 19, and then I'll have units of joule.

Â So that's the energy of one photon.

Â That energy of one photon would equal to the gap between these two.

Â Since the electron is traveling down, the change in energy of the electron

Â is actually equal to a negative 4.095 times 10 the minus 19 joules.

Â So the photon always has to have a positive energy, but

Â a change can be negative and

Â since the electron is traveling down, its change in energy is a negative value.

Â Now, the equation I want to use is that the change in energy of

Â an electron is equal to minus R sub H, which is the Rydberg constant,

Â times 1 over nf squared minus 1 over ni squared.

Â So ni is what I'm trying to determine.

Â nf is where it finishes up.

Â So it's finishing up at a n equals 2 state.

Â So let's plug in the numbers that we know here.

Â We have for the energy of the electron, negative 4.095

Â times 10 to the minus 19, and that is units of joules.

Â We have for the Rydberg constant a negative 2.179 times 10 to the minus 18.

Â And again, I'm giving you an extra significant figure that I

Â didn't give you in your notes to the Rydberg consent that has units of joules.

Â 1 over final, that's 2, squared minus 1 over n initial squared.

Â So if I divide the Rydberg constant into

Â the change in energy, I will have a value of 0.1879.

Â And that's going to be equal to, and the joules will cancel, so it's unitless.

Â 1 over 4 minus 1 over n sub i squared.

Â Now I want to subtract one fourth from both sides.

Â One fourth of this is the same as 0.25.

Â So when I subtract it from this side, I'm going to get a negative 0.0621.

Â And that will be equal to a negative 1 over n sub i squared.

Â because that's what's left, the minus sign hasn't gone anywhere.

Â So I'll change the sign, okay?

Â And I'll do the reciprocal.

Â So this is going to give me 1 over 0.0621 is equal to n sub i squared.

Â And then if I take the square root of both sides, I'd be left with n sub i.

Â And to the nearest whole number, which these n values have to be whole,

Â I have a n equal 4.

Â So the electron is coming from the number four state energy level down to

Â the number two energy level, and

Â when it does that, it will emit a photon with a wavelength of 485.1 nanometers.

Â