Tad Riley Jessica BelzerJustin CourtrightPHYS 2053Professor ElsterMarch 20, 2014

Homework VI: Hydrogen Atom and Analytic Considerations

1. Look at the n = 1, l =m = 0 case. Hold the mouse down on the radial wave function graph (upper right window) and find the value of the wave function at distances 1, 2, 3, 4 (in units of 4/ao where ao is the Bohr radius). Now move the mouse to the probability density window (lower left) and note the intensity of the picture at the same distances from the center. How does the probability density relate to the curve shown in the radial wave function window?

Distance in Units of 4/ ao Wave Function Probability Density

1 .79 .206

2 .27 .076

3 .1 .028

4 .04 .01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90

0.05

0.1

0.15

0.2

0.25

f(x) = 0.259495123350545 x + 0.00215146299483648R² = 0.999060642589815

Wave Function vs. Probablility Density

Wave Function

Pro

bab

ilit

y D

ensi

ty

As shown, the values and graphical representation of the relationship between the wave function and probability density at Bohr radius values of 1, 2, 3, and 4. Using least-square linear regression, it was found that they have a direct relationship with one another; where a marginal increase in the wave function, leads to an incremental increase of 0.2595 in the probability density. The coefficient of determination then states that 99.906% of the variation in the density is dictated by the variation in the wave function.

From there, the shape of the wave function vs. radius curve looks like an exponential decay, where the greatest values of the wave function are at the center of the probability density cloud.

Look at the n = 2, l =m = 0 case. What does the black band in the probability density window (lower left) represent? How does it relate to the radial wave function (upper right window)? (Hint: Hold the mouse down on the probability density in the center of the dark band, record the radius (in units of Bohr radii) and then find the location where the radial wave function is zero by the same method.) 

2.

At a Bohr radius of ~2.3, is where this band exist, which is where the amplitude of the probability density is 0; therefore, this is where it is not possible for electron to be in existence here. This is basically where there is a gap in between the electron clouds, a sort of boundary. As for the wave function representation, this is where its value is 0, or the location where the wave function crosses from positive to negative values; at this point, a node is occurring as defined where a standing wave has zero amplitude and is halfway between two areas of maximum amplitude, antinodes.

3. Look at the cases n = 2, l = 1, m = 0 and n = 2, l = 1, m = 1.

a. Case 1: n=2, l = 1, m = 0. How does the angular wave function (upper left window) relate to the probability density? At approximately what distance does the probability for finding the electron diminish to zero?

It seems as if the angular wave function is directly related to the probability density’s shape; this is represented in the figure above, which shows that the angular function consist of two circles symmetrical about the z-axis and the probability density nearly being two circles symmetrical about a black band. There is also symmetry about the x-axis for the angular function (obvious by it being a circle) and as the angular function moves in either direction from that axis, the amplitude of the probability decreases proportionally. The probability for finding an election is almost zero at +/- 12 Bohr radii.

Additionally, another differentiation from the two problems, there is a zero probability band between the positive and negative clouds; this is where the nucleus is located, which is logical due to it meaning there is nearly a zero probability for there to be an electron located in the nucleus.

b. For How does the angular wave function (upper left window) relate to the probability density? At approximately what distance does the probability for finding the electron diminish to zero?

This is graphically identical to part a), but instead of the angular function’s circles being symmetrical about the z-axis, it is about the x-axis.

As the angular function moves in either direction from the z-axis, the probability density goes towards zero symmetrically, for either positive or negative initial amplitudes. Also, there is a zero probability band vertical this time, creating vertical symmetry, and this correlates directly to the x-axis on the angular function, which also creates the symmetrical circles.

4. Look at the cases n = 3, m = 0 for the l values 0, 1, 2. Drag the mouse around in the probability density widow to determine approximately where the probability for finding the electron goes to zero.

a. Case 1: n = 3, m = 0, l = 0

The black rings are along the Bohr radius of ~2.3 and ~7.5, then falls off to zero at ~27.5 in all directions.

b. Case 1: n = 3, m = 0, l = 1

The black ring is at ~6.3 radii. There is also a zero probability band that runs vertical and becomes thicker after the black ring just described. Also, the probability reduces to zero in a symmetrical fashion about the vertical black band and falls off radially at ~27.5. The overall shape looks like two nuclear mushroom clouds (the top part of a nuclear explosion) and the shape that the the mushroom cloud as it actively curl inwards and towards the steam, is where the probability density’s amplitude also goes to zero.

Once again, it seems as if the angular function determines the shape, but this time, there are internal orbits that it does not correlate with the angular function.

c. Case 1: n = 3, m = 0, l = 2

The positive amplitude clouds fall off to zero when moving vertical in either direction from the nucleus at ~29 radii, but instead of falling off in a circular way (equal radii in all directions), the positive clouds go to zero horizontally at ~20 radii (blimp shapes like the angular function graph).

For the negative, it is just smaller balloon shapes:

Moving from the center, horizontally in either direction, these go to zero at ~24 radii, then the zeros for the vertical symmetry of these clouds is around ~18 radii.

d. How does the size of the atom for this energy level with different l values compare?

The size of the overall shape is very similar for l = 0 and l = 1, but then the l=2 is elongated vertically, so it goes to zero further out moving vertically in either direction, but falls to zero quick when moving horizontally away from the nucleus.

e. How does the size of the atom with electrons in the n = 3 energy level compare with the size of the atom with electrons in the n = 2 and n = 1 levels?

For the following comparison, l = m = 0: when n=1 the overall radius was ~4; when n=2 the overall radius was ~16; then, when n=3, the overall radius was ~27.5.

5. Explain the probability density pattern for the case n = 4, l = 2, m =1 in terms of the radial and angular wave functions.

Radial function: from the center, there is a rapid increase to the maximum value (antinode) at ~5 radii, which is the center of each internal probability cloud; then with a similar rate of change as this function increased, it decreases to zero (node) at ~13 radii, which is where there is a black band on the probability density graph. From there, the magnitude of the rate of change in the radial function reduces, and the negative antinode is at ~18 radii and this is where each outer cloud of the probability density is at their maximum magnitudes of ~0.005. Finally, the function goes back to zero, but there seems to be a limit at 0. It reminds me of an extremely damped spring wave function from classical mechanics.

Angular function: This once again determines the overall shape of the atom. There is another balloon-shaped symmetry as the last case, but this time it is symmetrical upon four planes: x-axis, z-axis, and two diagonal symmetries.

3)

P (r )dr=r 2∨R2,1(r )¿2dr

¿∨ r√24 a0∙ a0

∙ e−r2a0 ¿2 ∙4 πr

2

¿ π6a0

5 ∙ e−ra0

To find the maximum, set the derivative to zero

dPdr

=0= π6 a0

5 ∙ e−ra0 ddr

¿ π6a0

5 (4 r3e−ra0 − r

4

a0e

−ra0 )

¿( π r36 a05 ∙ e

−ra0 )(4− r

a0 )When r is solved for zero,

r = 4a0

4)

For n = 2, l = 0

P (r )dr=r 2∨R2,0 (r ) ¿2dr= r 2

8a03 ¿

P=∫0

a0

P (r )dr=¿ 18a0

3∫0

a0

(4 r2−4 r3

a0+ r

4

a02 )e

−ra0 dr¿

x= r/a0

P=18∫0

1

(4 x2−4 x3+ x4 )e−x dx=.034

For n = 2, l = 1

P (r )dr=r 2∨R2,1 (r )¿2dr= r 2

24a03r2

a02 dr

P=∫0

a0

P (r )dr=¿ r2

24 a03∫0

a0 r 4

a02 e

−ra0 dr¿

x= r/a0

P= 124∫0

1

x4 e−x dx=.0037

5)For l=1 ml=0 we obtain the function

P (θ ,φ )=¿Θ2,0 (θ )Φ0 (∅ ) ¿2= 34 πcos2θ

To find the maximum, set the derivative to zerodPdθ

= 34 π

(−2cosθsinθ )=0

The solutions to this equation are cos = 0 where = /2 and sin = 0 where = o θ θ π θ θor πThe second derivative shows cos =0 is a minumum and sin =0 is a maximum. θ θTherefore the direction with maximum probability is along the positive z-axis ( =0) θand the negative z-axis ( = )θ πFor l=1 ml=±1 we obtain the function

P (θ ,φ )=¿Θ2 , ±1 (θ )Φ± (∅ )¿2= 38πsin2θ

dPdθ

= 34 π

( cosθsinθ )=0

The solutions to this equation are cos = 0 where = /2 and sin = 0 where = o θ θ π θ θor π

The second derivative shows cos =0 is a maximum and sin =0 is a minimum. θ θTherefore the direction with maximum probability is along the xy plane

6)

The Bohr formula is En=−1n2me2h2

¿where n is the priciple quantum number. n is

defined as an eigenvalue function of n= jmax+ l+1.From the definition of n, is can be determined that l is within the range of 0 to n-1In each state, there are 2l + 1 values. There one, is for a state involving no spin, and the are two l values, one each for positive and negative spin which is determined by m.

Using the above information, it can clearly be seen that ∑l=0

n=1

2 (2 l+1 )=2n2