An electron is in the hydrogen atom with n=5. (a) Find the possible values of L and Lz for this

An electron is in the hydrogen atom with n=5. (a) Find the...

Key Concepts

Principal Quantum Number

The principal quantum number (n) determines the energy level and size of the electron's orbital in an atom. In the hydrogen atom, n takes on positive integer values, and for a given n the possible orbital shapes and angular momenta are determined. It is a fundamental quantum number in the quantum mechanical model of the atom.

Orbital Angular Momentum Quantum Number (l)

For a given principal quantum number n, the orbital angular momentum quantum number l can take integer values from 0 up to n-1. This quantum number determines the magnitude of the orbital angular momentum of the electron by the relation L = ?(l(l+1))?, and influences the shape of the atomic orbital.

Magnetic Quantum Number (m or Lz Quantum Number)

The magnetic quantum number, often denoted as m, specifies the projection of the orbital angular momentum along a chosen axis (commonly the z-axis). Its allowed values range from -l to +l in integer steps, meaning that for each orbital angular momentum state there are multiple orientations of the angular momentum vector relative to the z-axis.

Quantization of Angular Momentum

In quantum mechanics, angular momentum is quantized, meaning both its magnitude and its projection along a given axis can only take discrete values. The magnitude is given by L = ?(l(l+1))? and the z-component by Lz = m?. This quantization leads to the concept of space quantization of angular momentum which explains why only specific orientations of the angular momentum vector are allowed.

Angular Momentum Vector Orientation

The orientation of the orbital angular momentum vector relative to the z-axis is determined by the ratio between its z-component and its magnitude. The angle ? between the angular momentum vector and the z-axis can be calculated using cos ? = m/?(l(l+1)). This relation shows that for each value of l, the possible angles are dictated by the different possible values of m, illustrating the concept of quantized orientations in space.

Transcript

00:01 Okay, so for this problem, we know an electron has a principal quantum number of n equal to three.

00:09 And our goal is to get the possible angular momentum values of l and lz and unis of h bar.

00:17 And then we want to find all the angles between the l and the z axis.

00:20 So if n equals three, then we can use the equation.

00:24 Let me pull it up, 29 .6 to show or apply that to conclude.

00:31 That l can be anywhere between 0 and n minus 1, so 0 and 2.

00:37 So l can be either 0, 1, or 2.

00:56 And an ml, or lz, can take on, let's see, one second.

01:07 I'm trying to find the part in the book i want to refer to.

01:10 Okay, so based on that same equation, you can get that ml is equal to, so if l equals zero, then ml can be just zero because it has to be less than or equal to l.

01:26 If l equals 1, then ml can be equal to zero plus or minus 1.

01:34 And then if l is equal to two, then ml can be equal to zero plus or minus one and plus or minus two.

01:45 And now we want to find lz and l.

01:49 So we can use equation 29 .3 and 29 .4 to calculate l and lz.

01:59 So in units of h bar l big l is going to be the square root of um so in general l is going to be the square root of l times l plus one and then that's in units of h bar so if l is equal to zero then um big if little l is equal to zero then big l is equal also to zero and then if l is equal to one then big l is equal to the square root of 1 times 2, so that's just the square root of 2.

02:36 And then if l is equal to 2, then big l is the square root of 2 times 3, which is the square root of 6.

02:44 And then we can use also that lz is equal to basically ml in units of h -bar.

03:02 So for this one, lz would have to be 0.

03:07 For this one, you could have lz equal 0 plus or minus 1.

03:12 And for this one, you could have lz equal to 0 plus or minus 1 plus or minus 2.

03:21 And now we want to find the angle between l and the z axis.

03:28 So let's go ahead and do that on another page.

03:33 So for l equals 0, then lz is not even along the z axis.

03:41 So i'm not going to address that case.

03:43 But for l is equal to 1, we know that just to transpose what i wrote before, l is equal to the square root of 2.

03:53 And then lz is equal to 0 plus or minus 1.

04:00 And so therefore you can make an angle between lz and l.

04:09 Let me just double -check the wording.

04:15 So we want to find all the angles between l and the z axis.

04:18 So we know the angular momentum can have projections of zero and plus or minus one.

04:24 So we can say that lz is that would be plus one.

04:30 And then this would be zero and then this would be minus one.

04:36 So these are all my lzs plus one.

04:46 And then let me just pause the video.

04:50 Okay, and so we need like l to have a length of square of two, and then the projections to be zero plus or minus one.

05:00 So therefore l, so for the projection of plus one, and l has to kind of flare out and have a magnitude of square root of two.

05:18 And so if one side is square root of two, let me rewind.

05:24 So the length of this vector is the square root of two.

05:27 The spin is represented by the length of a vector.

05:29 The length of the vector is the square root of two...

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