PHYS 2053: Homework VI: Hydrogen Atom and Analytic Considerations
By: Alex Marshall
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
Wave Function vs. Probablility Density
Probability Density
0.25
0.2
0.15
y = 0.2595x + 0.0022
R² = 0.9991
0.1
0.05
0
0
0.2
0.4
0.6
Wave Function
0.8
1
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.
1
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.
2. 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 there is no chance for electron to be in existence at that
distance from the nucleus. This is basically where there is a gap in between the energy levels or
electron clouds. 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.
3
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.
4
b. 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.
5
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
6
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
7
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).
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.
8
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 balloonshaped symmetry as the last case, but this time it is symmetrical upon four planes: x-axis, z-axis, and two
diagonal symmetries.
Analytic Considerations
3. Prove that the most likely distance from the origin of an electron in the n = 2, l = 1 state is
4a0, where a0 is the Bohr radius.
a. The radial probability density for n=2 and l=1 is:
(EQ1)
𝑃2,1 (𝑟) = 𝑟 2 |𝑅2,1 (𝑟)|2 = 𝑟 2 ∗ (
1
3
𝑟
⁄ 𝑎
√3𝑎𝑜 2 𝑜
−𝑟
𝑒 ⁄2𝑎𝑜 )
2
9
𝑃2,1 (𝑟) = 𝑟 2
𝑃2,1 (𝑟) =
1 𝑟 2 −𝑟⁄𝑎
𝑜
𝑒
24𝑎𝑜3 𝑎𝑜2
𝑟4
−𝑟⁄
𝑎𝑜
5∙𝑒
24𝑎0
To find the maximum, find the first derivative:
𝑑𝑃
𝑟4
𝑑
−𝑟
=(
∙ 𝑒 ⁄𝑎𝑜 )
5
𝑑𝑟
𝑑𝑟
24𝑎0
Then, set it to equal to zero:
1
𝑟 4 −𝑟⁄𝑎
3 −𝑟⁄𝑎𝑜
𝑜)
0=
(4𝑟 𝑒
− 𝑒
𝑎0
24𝑎05
−𝑟
𝑒 ⁄𝑎𝑜
𝑟4
3
0=(
)
(4𝑟
−
)
𝑎0
24𝑎05
The left term will never equate to zero, therefore only solution is found by solving for r so that the right
term is zero shown by:
−𝑟
𝑒 ⁄𝑎𝑜
0=(
) (0)
24𝑎05
When plugging in r= 4a0:
(
𝑒 −4
(4𝑎0 )4
𝑒 −4
𝑒 −4
3
4 (𝑎 3
3 ))
)
(4(4𝑎
)
−
)
=
(
)
(4
−
𝑎
=
(
) (0) = 0
0
0
0
𝑎0
24𝑎05
24𝑎05
24𝑎05
This proves the mostly likely distance from the origin of an electron in the n=2 and l=1 state is 4a0.
a. Compare your answers with the graphical solution in the app of Problem 1.
The radius was not directly found because there is no graphical solution of the radial probability density;
instead, since the radial function for this state is 𝑅2,1 (𝑟) and radial probability function is
therefore 𝑃2,1 (𝑟),
𝑅2,1 (𝑟) =
1
𝑟
3⁄ 𝑎
√3𝑎𝑜 2 𝑜
𝑒
−𝑟⁄
2𝑎𝑜
2
𝑃(𝑟) = 𝑟 2 |𝑅2,1 (𝑟)|2 = 𝑟 2 ∗ (
1
𝑟 −𝑟⁄2𝑎
𝑜)
3⁄ 𝑎 𝑒
2 𝑜
√3𝑎𝑜
I have created a graph to visually represent the relationship between the two, the red line is the radial
function (same as Part 1) and the radial probability function (units are in Å):
10
Notice, the maximum point for the radial function is 1.058, which equates to 2*0.529 Å or 2 a0 (maximum
from Part 1 was ~2 a0). As for the radial probability function, the maximum point is at 2.116; therefore,
this is at 4*0.529 Å or 4 a0. This means that for the n=2 and l=1 case, the max for the radial probability
function is twice the radial wavefunction’s max:
rP(r),max = 2 * rR(r),max = 2 * 2 a0 = 4 a0
By this logic, Part 1 solidifies the equated max found in this problem.
4. For the n = 2 states (l = 0 and l = 1), compare the probabilities of the electron being found
inside the Bohr radius.
a. For the n = 2 and l = 0 case, I used EQ1 above:
2
𝑃(𝑟) = 𝑟 2 |𝑅2,1 (𝑟)|2 = 𝑟 2 ∗ (
1
𝑟 −𝑟⁄2𝑎
𝑜)
3⁄ 𝑎 𝑒
2 𝑜
√3𝑎𝑜
In order to isolate the probability inside a Bohr radius, the integral must be taken from 0 to a0:
𝑃(𝑟)𝑑𝑟 = 𝑟 2 |𝑅2,0 (𝑟)|2 𝑑𝑟 =
𝑎0
𝑃(0: 𝑎0 ) = ∫ 𝑃(𝑟)𝑑𝑟 =
0
𝑟2
𝑟 2 −𝑟
(2
−
) 𝑒 𝑎0 𝑑𝑟
𝑎0
8𝑎03
−𝑟
1 𝑎0
4𝑟 3 𝑟 4
2
𝑎0 𝑑𝑟
∫
(4𝑟
−
+
)
𝑒
𝑎0 𝑎02
8𝑎03 0
𝑃(0: 𝑎0 ) = .034
b. For n = 2, l = 1, the same process was employed:
2
𝑃(𝑟)𝑑𝑟 = 𝑟 |𝑅2,1
(𝑟)|2
𝑟2 𝑟2
𝑑𝑟 =
𝑑𝑟
24𝑎03 𝑎02
11
𝑎0
𝑎0 4 −𝑟
𝑟2
𝑟
𝑃(0: 𝑎0 ) = ∫ 𝑃(𝑟)𝑑𝑟 =
∫
𝑒 𝑎0 𝑑𝑟
24𝑎03 0 𝑎02
0
𝑃(0: 𝑎0 ) = .0037
c. Compare your answers with the graphical solution in the app of Problem 1.
These results show that the l = 0 state has a multiplicity of 10 relative to the l = 1 state, which is because
in the l = 0 state, the electrons are more likely to be in the first orbital and also far away from the nucleus.
This supports my results in part 1 because the l = 0 state maintained two spherical orbits and had high
values for the wave function close to the nucleus (then quickly decreases) and then after the node, the
return to 0 is much slower. For l = 1, the probability density shows two distinct dumb-bell shaped clouds,
but most of all, there is the black band through the nucleus; meaning for this state, there the density is
reduced close to the nucleus and the wave function shows the same relationship, and even that the
highest values of the wave function soon after the initial radii, then the decrease begins gradual, but falls
off quicker than l = 0.
4. For the n = 2, l = 1 wave functions, find the direction in space at which the maximum
probability occurs when ml = 0 and when ml = ±1.
a. For the n=2, l = 1 and ml = 0 state, the angular component of the wave function was
used:
𝑃(𝜃, 𝜑) = |𝛩2,0 (𝜃)𝛷0 (∅)|2 =
3
𝑐𝑜𝑠 2 𝜃
4𝜋
Then, just like before, the maximum is found by setting the first derivative of the probability to zero:
𝑑𝑃
3
(−2𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃) = 0
=
𝑑𝜃
4𝜋
For the equation above, the only solutions that satisfy a maximum are when either cos(θ) = 0 or sin(θ) =
0, which gives solutions of θ = +/- π/2, for the cosine and θ = 0 or π for the sine. The second derivative
then is used to find minimums and maximums:
𝑑2𝑃
3
=
cos(2𝜃)
𝑑𝜃 2 −2𝜋
This results in relative maximums at 0 and 𝜋 shown by the negative value resulting in the shape being
concave downward at that point, meaning that it is the maximum point:
𝑑2𝑃
3
π
3
3
=
cos (2 ∗ ) =
cos(𝜋) =
2
𝑑𝜃
−2𝜋
2
−2𝜋
2𝜋
𝑑2 𝑃
3
π
3
3
=
cos (2 ∗ − ) =
cos(−𝜋) =
2
𝑑𝜃
−2𝜋
2
−2𝜋
2𝜋
12
𝑑2𝑃
3
3
=
cos(2 ∗ 0) =
cos(0) = −3/( 2𝜋)
2
𝑑𝜃
−2𝜋
−2𝜋
𝑑2𝑃
3
3
=
cos(2 ∗ π) =
cos(2𝜋) = −3/( 2𝜋)
2
𝑑𝜃
−2𝜋
−2𝜋
Therefore the direction with maximum angular wave component is along the z-axis, which is exactly what
my results were for part 1 with n=2, l = 1 and ml = 0.
b. For the n=2, l = 1 and ml = 0 state, the same methodology was used:
3
𝑃(𝜃, 𝜑) = |𝛩2,±1 (𝜃)𝛷±(∅)|2 =
𝑠𝑖𝑛2 𝜃
8𝜋
𝑑𝑃
3
(𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃) = 0
=
𝑑𝜃
4𝜋
𝑑2𝑃
3
=
cos(2𝜃)
𝑑𝜃 2 4𝜋
𝑑2𝑃
3
π
3
=
cos
(2
∗
)
=
cos(𝜋) = −3/( 4π)
𝑑𝜃 2 4𝜋
2
4𝜋
𝑑2𝑃
3
π
3
=
cos (2 ∗ − ) =
cos(−𝜋) = −3/( 4π)
2
𝑑𝜃
4𝜋
2
4𝜋
𝑑2 𝑃
3
3
=
cos(2 ∗ 0) =
cos(0) = 3/( 4π)
2
𝑑𝜃
4𝜋
4𝜋
𝑑2 𝑃
3
3
=
cos(2
∗
π)
=
cos(−2𝜋) = 3/( 4π)
𝑑𝜃 2 4𝜋
4𝜋
c. Compare your solution with the graphical solution in the app of Problem 1:
Therefore the direction with maximum angular wave component is along the xy-axis, which is exactly
what my results were for part 1 with n=2, l = 1 and ml = 0; but, since the y-plane is not shown graphically
in the applet, I determined the x-axis as the highest probability point.
5. For each l value, the number of possible states is 2(2l + 1). Show explicitly that the total
number of states for each principal quantum number is
𝒏−𝟏
∑ 𝟐(𝟐𝒍 + 𝟏) = 𝟐𝒏𝟐
𝒍=𝟎
First, we know for each value of n that l has value range from [0, n-1]. Secondly, we know that each value
13
of l there are 2(2l+1) possible states.
Therefore, the total number of states per principle quantum number is (degeneracy per n):
𝑛−1
∑ 2(2𝑙 + 1)
Then, multiplying terms leads to:
𝑛−1
𝑙=0
𝑛−1
𝑙−1
𝑙−1
∑ 2(2𝑙 + 1) = ∑(4𝑙 + 2) = 4 ∑ 𝑙 + ∑ 2
𝑙=0
After expansion:
𝑙−1
𝑙=0
𝑙=0
𝑙=0
𝑙−1
4 ∑ 𝑙 + ∑ 2 = {((4 ∗ 0) + 2) + ((4 ∗ 1) + 2) + ((4 ∗ 2) + 2). . }
𝑙=0
𝑙=0
Obvious by the pattern, the summation will result in a series final term:
= {((4 ∗ 0) + 2) + ((4 ∗ 1) + 2) + ((4 ∗ 2) + 2). . ((𝑛 − 1) + 2)}
After separating like terms:
= {(4(0 + 1 + 2 + ⋯ (𝑛 − 1)) + (2 + 2 + 2 + ⋯ + 2)}
In order to reach this summation and to achieve the pattern, the first half can be represented as:
(𝑛 − 1)(𝑛 + 1 − 1)
(4(0 + 1 + 2 + ⋯ (𝑛 − 1)) = 4
2
Then, the second half can be represented as:
(2 + 2 + 2 + ⋯ + 2) = 2𝑛
Therefore:
= {((4 ∗ 0) + 2) + ((4 ∗ 1) + 2) + ((4 ∗ 2) + 2). . ((𝑛 − 1) + 2)} = 4
Finally, the proof for the degeneracy of the nth level is concluded:
(𝑛 − 1)(𝑛 + 1 − 1)
4
+ 2𝑛 = 2𝑛2
2
(𝑛 − 1)(𝑛 + 1 − 1)
+ 2𝑛
2
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