00:01
Okay, so for this problem here, we are given a formula for just the spectral lines.
00:10
And we need to find the first three terms in the series.
00:15
Okay, so, and we also need to find the shortest wavelength of all the lines in the series.
00:25
Okay, so let's get started.
00:27
So first let's find lambda 1.
00:30
So 1 over lambda 1 is equal to 1 .097 times 10 to the minus 2, multiplied by 1 over 2 squared minus 1 over 3 squared.
00:44
And 1 over lambda 1 is equal to, after putting this in our calculator, we get that lambda 1 is 0 .0 .0 .0.
00:57
And then taking the reciprocal of this, we see that lambda 1 is equal to 666 .66 nanometers.
01:11
Okay.
01:12
So this is our first one.
01:19
All right.
01:20
So now let's find lambda 2.
01:22
1 over lambda 2 is equal to 1 .097 times 10 to the minus 2.
01:30
1 over 2 squared minus 1 over 4 squared.
01:37
And that will just calculate 2.
01:41
That will be 0 .00 to 0 .20...