FLORIDA INTERNATIONAL UNIVERSITY



CHM 3411 – Problem Set 4

Due date: Wednesday, February 23rd

Do all of the following problems. Show your work.

1) The normalized radial wavefunction for the 2s state of a hydrogen atom is given in Table 9.1 of Atkins.

a) Write down the radial wavefunction for this state, substituting the correct values for Z and (. Note that Atkins uses the symbol “a” for the Bohr radius (equn 9.11) when me is replaced by (, the reduced mass.

b) Find rmp, the most probable value for r. (Hint: You can do this by taking the first derivative of the radial probability, P(r) = r2 |R2s|2 , and setting it equal to zero. You will find more than one extreme point. Be sure to identify the value of r corresponding to the absolute maximum in probability, which corresponds to rmp.)

c) Find , the average value for r.

d) Find , the average value for 1/r.

e) Are and rmp equal? Does = 1/?

2) Just as we can construct p orbitals out of the ( = 1 set of spherical harmonics, we can also construct d orbitals out of the ( = 2 set of spherical harmonics.

One of the d orbitals for hydrogen has the following form

d((,() = N sin( cos( sin( (2.1)

a) For what value of N will the above d orbital be normalized? Hint: Recall that normalization for an angular function is the requirement that

1 = (0( sin( d( (02( d( d*((,() d((,() (2.2)

b) For what value(s) of ( and ( does the above d orbital have its maximum magnitude? Based on your answer, identify the d orbital. (Hint: P((,() d( d( = sin ( [d((,()]2 d( d(. Pictures of the d orbitals are given in fig 9.16 of Atkins.)

Also do the following problems from Atkins:

Exercise 9.2b Compute the wavelength, frequency, and energy (in cm-1) of the n = 5 ( n = 4 transition in Li2+.

Exercise 9.5b The wavefunction for the 2s orbital of a hydrogen atom is

(2s = N [ 2 – (r/a0) ]exp(-r/2a0)

Determine the normalization constant N. NOTE: I want you to find N using the normalization condition that applies. Recall normalization for the entire wavefunction, ( = ((r,(,(), is

1 = (0( r2 dr (0( sin( d( (02( d( (2s* (2s

Exercise 9.6b By differentiation of the 3s radial wavefunction, show that it has three extrema in its amplitude, and locate them. NOTE: Also do the following. Identify the extreme points as local maxima in probability or absolute maxima in probability. Also, find the number and location(s) of all of the radial nodes.

Solutions.

1) a) For a 2s state of hydrogen (Table 9.1)

R(r) = (1/8)1/2 (Z/a0)3/2 (2 - () exp(-(/2)

But Z = 1, n = 2, ( = 2Zr/na0 = r/a0, and so

R(r) = (1/8a03)1/2 (2 - r/a0) exp(-r/2a0)

b) Contrary to the hint, it is best to use P(r) dr = r2 |R(r)|2 dr to find the extreme points.

So P(r) = r2 [(1/8a03)1/2 (2 - r/a0) exp(-r/2a0) ]2 = (1/8a03) [ 4r2 – (4r3/a0) + (r4/a02) ] exp(-r/a0)

Extreme points occur when dP(r)/dr = 0 .

So d/dr (1/8a03) [ 4r2 – (4r3/a0) + (r4/a02) ] exp(-r/a0) = 0

= (1/8a03) { [ 4r2 – (4r3/a0) + (r4/a02) ] (-1/a0) exp(-r/a0) + [ 8r – (12r2/a0) + (4r3/a02) ] exp(-r/a0) }

If we multiply both sides of this equation by 8a06r exp(r/a0) we get

0 = - 4a02r + 4a0r2 – r3 + 8a03 – 12a02r + 4a0r2

If we multiply by (-1) and combine terms, we get

0 = r3 – 8a0r2 + 16a02r – 8a03

Hmmm...a cubic equation. Well, we know that one of the extreme points will correspond to a minimum in the probability, which occurs at the node of the radial wavefunction. We may find this node by setting R(r) = 0.

0 = R(r) = (1/8a03)1/2 (2 - r/a0) exp(-r/2a0)

will occur when r = 2a0. If we factor out (r – 2a0) from the cubic equation we obtained for the extreme points, we get

0 = r3 – 8a0r2 + 16a02r – 8a03 = (r – 2a0) (r2 – 6a0r + 4a02)

We may find the other two extreme points, which correspond to maxima, by finding the roots of the quadratic term

0 = (r2 – 6a0r + 4a02) r = (1/2) { 6a0 ( (20a02)1/2 } = [ 3 ( (5)1/2) ]a0

The maximum further with the largest value of r is the absolute maximum, so rmp = [ 3 + (5)1/2 ] a0 ( 5.24 a0

(NOTE: This is in agreement with the answer given in self-test 9.6, p 336 of Atkins)

c) = (0( r2 dr R2,1 r R2,1

= (0( r3 [(1/8a03)1/2 (2 - r/a0) exp(-r/2a0)]2 dr

= (1/8a03) (0( [ 4r3 – (4r4/a0) + (r5/a02) ] exp(-r/a0) dr

The general integral we need (# 661) is

(0( xn exp(-ax) dx = n!/an+1

and so = (1/8a03) [ 4 . 3! a04 – 4 . 4! a04 + 5! a04 ] = 6 a0

d) = (0( r2 dr R2,1 (1/r) R2,1

= (0( r [(1/8a03)1/2 (2 - r/a0) exp(-r/2a0)]2 dr

= (1/8a03) (0( [ 4r – (4r2/a0) + (r3/a02) ] exp(-r/a0) dr

We may use the same general integral as in part c, to get

= (1/8a03) [ 4 . 1! a02 – 4 . 2! a03 + 3! a04 ] = 1/(4a0)

e) rmp = [ 3 + (5)1/2 ]a0 ; = 6a0, so not equal.

1/ = 1/(6a0) ; = 1/(4a0), so not equal.

2) a) Normalization is

1 = (0( sin( d( (02( d( [N sin( cos( sin( ]2

= N2 (0( (sin()3 (cos()2 d( (02( (sin()2 d(

We have no general form for the ( integral. However, since (sin()2 + (cos()2 = 1, we may rewrite the integral as

(0( (sin()3 (cos()2 d( = (0( (sin() (sin()2 (cos()2 d(

= (0( (sin() [ 1 - (cos()2 ] (cos()2 d(

= (0( (sin() (cos()2 d( - (0( (sin() (cos()4 d(

The general form of this integral (# 321) is

( (sin ax) (cos ax)m dx = - (cos ax)m+1

(m+1) a

And so (0( (sin() (cos()2 d( - (0( (sin() (cos()4 d( = [ - (1/3) (cos()3 + (1/5)(cos()5 ]0(

= (2/3) – (2/5) = (4/15)

For the ( integral, the general form (# 296) is

( (sin ax)2 dx = (x/2) – (1/4a) (sin 2ax)

So (02( (sin()2 d( = [ ((/2) – (1/4) sin 2( ]02( = (

Collecting the above results, we get

1 = N2 (4/15) ( , and so N = (15/4()1/2

b) P(() ~ (sin ()3 (cos ()2 (the extra factor of sin ( is from the intgration factor.

We want extreme points, so for the extreme points in (

dP(()/d( = 0 = d/d( [ (sin ()3 (cos ()2 ] = d/d( { (sin ()3 [ 1 - (sin ()2 ] }

= d/d( { (sin ()3 - (sin ()5 ] } = 3 (sin ()2 (cos () – 5 (sin ()4 (cos ()

If we multiply through by 1/[(sin ()2 (cos () ], we get

0 = 3 – 5 (sin ()2 , or sin ( = ( 3/5 or ( ( 36.9 (, 143.1 (

Since P(() ~ (sin ()2, then for the extreme points in ( are where sin ( has its maximum magnitude.

So ( = 90 (, 270 ( (that is, at (/2 and at (3(/2) ). This will be along the y-axis.

Based on the pictures of the d orbitals, the above is a dyz orbital.

Exercise 9.2b.

Ignoring the small difference between me and ( (about 1 part in 14000) we may say that for lithium

En = - (Z2/n2) R( , where Z = 3 for Li, and R( = 109737. cm-1

For n = 4 E4 = - (9/16) R(

For n = 5 E5 = - (9/25) R(

(E = E5 – E4 = [ (9/16) – (9/25) ] R( = (0.2025) R( = (0.2025) (109737. cm-1) = 22222. cm-1

So ( = 1/((E) = 1/(22222. cm-1) = 4.50 x 10-5 cm = 450. nm.

Since (( = c , ( = c/( = (2.998 x 108 m/s)/(450. x 10-9 m) = 6.66 x 1014 s-1

Exercise 9.5b

Normalization is

1 = (0( r2 dr (0( sin( d( (02( d( [N [ 2 – (r/a0) ]exp(-r/2a0) ]2

= N2 (0( sin( d( (02( d( (0( [4r2 – (4r3/a0) + (r4/a02) ]exp(-r/a0) dr

The angular integral is equal to 4(. For the radial integral we use integral #661 (see problem 1c), to get

= 4(N2 { 4 . 2! a03 – 4 . 3! a03 + 4! a03 } = 4(N2 (8a03)

So N = (1/32(a03)1/2 . Note that this is just the product of the normalization factors for R2,0 (Table 9.1) and Y0,0 (Table 8.2).

Exercise 9.6b (NOTE: The textbook is wrong if it is looking for extreme points in the wavefunction, as there are clearly only two such points – see Fig 9.4c, p 329 of Atkins).

The 3s radial wavefunction is (note ( = 2r/3a0)

R(r) = (1/243a03)1/2 [ 6 – (4r/a0) + (4r2/9a02) ] exp(-r/3a0)

We are asked for extreme points in the wavefunction (not the probability) and so

0 = (d/dr) R = d/dr (1/243a03)1/2 [ 6 – (4r/a0) + (4r2/9a02) ] exp(-r/3a0)

= (1/243a03)1/2 [ - (4/a0) + (8r/9a02) – (2/a0) + (4r/3a02) – (4r2/27a03) ] exp(-r/3a0)

= (1/243a03)1/2 [ - (4r2/27a03) + (20r/9a02) – (6/a0) ] exp(-r/3a0)

If we multiply both sides by (-1) (243a03)1/2 (27a03) exp(r/3a0) , we get

0 = 4r2 – 60a0r + 162a02

Using the quadratic equation, r = (1/4) [ 30 ( 3 (28)1/2 ]a0 .

Using Fig 9.4c of Atkins as a guide, the lower value, r = (1/4) [ 30 - 3 (28)1/2 ]a0 , is an absolute maximum, and the higher value, r = (1/4) [ 30 + 3 (28)1/2 ]a0 , is a local maximum.

There are two radial nodes, which will occur when

0 = [ 6 – (4r/a0) + (4r2/9a02) ]

If we multiply both sides by 9a02, we get

0 = 4r2 – 36a0r + 54a02

Using the quadratic formula, the nodes are at r = (a0/2) [ 9 ( 3 (3)1/2 ]

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