00:01
In this exercise, we are given the figure, i've given an abbreviated figure from the book here, that represents the emission spectrum of a one electron ion in a gas phase.
00:13
All the lines result from electronic transitions from excited states to the n equals three state.
00:21
So we're being asked to figure out what electric transition corresponds to lines a and b, and then also going to figure out if the wavelength of line b is 142 .5 nanometers, figure out the wavelength of line a.
00:38
Okay, so first thing we need to do is figure out our equation.
00:43
So we're going to be working with the equation e or energy equals negative 2 .178 times 10 to the negative 18th.
00:57
Times z squared over n f squared minus z squared over n i squared.
01:08
Now we don't know what z is or what e is.
01:12
However, we do know that the wavelength of b is 142 .5 nanometers.
01:28
So we can actually figure out our energy using e equals planks constant times the speed of light divided by our wavelength.
01:40
So in this case, we're going to have e equals 6 .626 times 10 to the negative 34th times 3 .0 times 10 .000 times 3 .0 times 10 to 0 .4th .5 .0.
01:59
The 8th, all divided by 142 .5 times 10 to the negative 9th meters.
02:09
We do have to put it into meters so that it goes with the speed of light.
02:13
And that gives us an energy of 1 .39 times 10 to the negative 18th joules...