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[BLANK_AUDIO].

Â Hi there.

Â So, this is just a quick run through on the treatment of the data

Â you've obtained for the hydrogen emission simulator.

Â So, if you, on page three of the, the,

Â the descriptor you'll find the, the title data treatment.

Â Now the first thing we're supposed to use is, well first use its, the

Â simulator, and you obtain the wavelengths in

Â nanometers of four lines in the Balmer series.

Â So these were obtained on the simulator by you, you, you, taught it how to do that.

Â And, and, and the simulator program.

Â And, you position the cross hairs as best you can.

Â 0:53

And then, the first thing you have to do is convert these to electron volts.

Â So, a good way to treat the data, again as we mentioned

Â lower down on the descriptor, is to put the data into a spreadsheet.

Â So, here are the four lines.

Â Using the same list that I measured.

Â 1:12

So you, four, four nanometer values.

Â And then the first thing I asked you to do was convert these to electron volts.

Â And this is just simply a, a simple conversion factor.

Â There's loads of these on the Iinternet.

Â The one I use is given here.

Â So this, all you have to do is simply type in the values in nanometers that you get

Â from the simulator, and it will give you the electron volt's values.

Â 1:41

So if we go back to the, descriptor again, so

Â the next step then, was you had to identify the transitions.

Â And then you had to use a little bit of what we found out in the lecture.

Â You know, the lower energy has a principal quantum number, n equals 2.

Â because the Balmer series corresponds to n equals 2.

Â So therefore, you know the lines

Â must correspond to transitions from higher levels.

Â So, basically, you've observed tje first, four lines.

Â So the transitions are from three to two, four to two, five to two, and six to two.

Â And then you were given how to assign them so

Â which transition in the Balmer series is lowest energy change.

Â 2:20

Which of the observed lines corresponds to the lowest energy.

Â And which is the next lowest transition and

Â which line corresponds to the next lowest energy.

Â So again, if you look back on the lectures, you know that the,

Â the lowest energy, the lower energy gap corresponds at three to two level.

Â Then you go.

Â 2:51

I've done this in the spreadsheet here.

Â So here we have the highest wavelength band.

Â The highest wavelength corresponds to the lowest energy.

Â So, that's from the three-to-two transition so

Â N2 in this case is going to be three.

Â Next one, N2's going to be four.

Â Next one's going to be five.

Â And the next one is going to be, is going to be, to be six.

Â [NOISE] So going back then to the description again.

Â Next thing you're asked is to plot a graph

Â of the energy of each transition in electron volts.

Â But I've already calculated the transition energy of electron volts.

Â And then you have to work out 1 over N2 squared.

Â So using Excel or a spreadsheet.

Â And your oh, asked to obtain the gradient.

Â And the, and the intercept.

Â 3:39

So first let's show that on Excel.

Â So here we see we, in addition to the, the

Â energy is in nanometers, the energy is in electron volts here.

Â We have the N2 values.

Â Then we did the N2 squared values.

Â And now we have one over N2 squared because that's what we were asked to plot.

Â So we were asked to plot one over N2 squared against electron volts.

Â And using the plotting routine in, Excel, you

Â obtain the graph given here where you have the

Â energy electron volts on the Y axis and one over N2 squared on the, on the X axis.

Â And here you have your straight line.

Â And then you have the equation for the straight line.

Â Y is equal to this equation for a straight line.

Â It's y is equal to m x plus c.

Â And the next step then was to use this plot to obtain the, the Rydberg constant.

Â 4:33

So going back then to our, our descriptor again.

Â So we've just finished here plotted a

Â energy electron volts versus one over n squared.

Â And we have the gradient and we have the intercept.

Â Now it says to adapt equation six and express in the form of y equals m x +

Â c and use it to evaluate the Rydberg constant,

Â r, separately from both your calculated gradient and your intercept.

Â So let's first go up to equation six here.

Â 5:23

So here you have delta-e.

Â So we can say that's equal to, we're just going to multiply out by R.

Â So it's R over n1-squared

Â minus R over n2 squared.

Â So what I'm going to do now is slightly rearrange that to make it clearer for you.

Â I'm just going to set this equal to minus R into

Â 1 over n2 squared plus R over n1 squared.

Â Now n1 from the Balmer series is going to be two, so

Â n1 squared, two squared, is going to be, it's going to be four.

Â And this is your classic y is equal to m x plus c plot.

Â So this is delta-e is to y.

Â And here you have m x.

Â M is going to be minus r and x is, so

Â what we're plotting on the x axis is one over n2 squared.

Â Plus c, our intercept is going to be r over four.

Â 6:30

So if we plotted that, what you'd get, say here is a rough sketch.

Â So here you have your y, which is going to be a delta-e in electron volts.

Â Here you are going to have one over n2 squared.

Â And you have four points, so your four points are that.

Â You draw a straight line through them.

Â Your slope then, is going to be, minus R, with a negative after Rydberg, constant.

Â And your intercept here, is going to be the Rydberg constant divided by 4.

Â So now if we move back here to our plot.

Â [SOUND] We can see we've plotted this already.

Â So now we know that the the negative of the

Â slope is the Rydberg constant, and it's in electron volts.

Â So it's -13.588 or -13.6 electron volts, which is the value, of course.

Â You know everybody from, from the lectures, and also the slope is, 3, 3.4,

Â which is equal to of course the Rydberg constant, R, divided by, divided by 4.

Â [BLANK_AUDIO].

Â And so the last thing then, so is this just two points down here.

Â We've done this.

Â We've shown how to adapt the equation six.

Â And you've estimated R from the gradient and the intercept.

Â And then, you used the value of R to estimate the ionization energy.

Â Of course you know that the ionization energy

Â is the, is the energy required to remove the

Â electron, completely from the ground state, so that's

Â going to be just simply, the value of R.

Â So the ionization energy for say, for the hydrogen atom is 70.6 electron voles.

Â And then what would those do is convert that into other, another values.

Â Energy in kilojoules per mole, or centimeters minus one.

Â And here you're given the conversion factors.

Â So for the estimate the ionization of hydrogen,

Â you'd just multiply 13.6 by this value here.

Â 96.487 kilojoules per mole.

Â You want it in cm minus 1 [SOUND].

Â You can multiply it by 8065.6 cm minus one, and then you should be

Â able to compare your result to values that you find in the, in the lecture.

Â [BLANK_AUDIO]

Â