Sections to arrive at a complete description of the hydrogen atom. This will culminate in the de nition of the hydrogen-atom orbitals and associated energies. From these functions, taken as a complete basis, we will be able to construct approximations to more complex wave functions. This explains why we need (why we use) three positions (states) to articulate hydrogen (a wave) (a function). Allowing us to ‘remove’ (understand) (articulate) the ‘duplicity’ (redundancy. A hydrogen atom can be described in terms of its wave function, probability density, total energy, and orbital angular momentum. The state of an electron in a hydrogen atom is specified by its quantum numbers ( n, l, m ). THE HYDROGEN ATOM; ATOMIC ORBITALS Atomic Spectra When gaseous hydrogen in a glass tube is excited by a 5000-volt electrical discharge, four lines are observed in the visible part of the emission spec-trum: red at 656.3 nm, blue-green at 486.1 nm, blue violet at 434.1 nm and violet at 410.2 nm: Figure 1. Visible spectrum of atomic hydrogen. The energy levels are degenerate, meaning that the electron in the hydrogen atom can be in different states, with different wave functions, labeled by different quantum numbers, and still have the same energy. The electron wave functions however are different for every different set of quantum numbers.
Learning Objectives
By the end of this section, you will be able to:
- Describe the hydrogen atom in terms of wave function, probability density, total energy, and orbital angular momentum
- Identify the physical significance of each of the quantum numbers (n, l, m) of the hydrogen atom
- Distinguish between the Bohr and Schrödinger models of the atom
- Use quantum numbers to calculate important information about the hydrogen atom
The hydrogen atom is the simplest atom in nature and, therefore, a good starting point to study atoms and atomic structure. The hydrogen atom consists of a single negatively charged electron that moves about a positively charged proton (Figure (PageIndex{1})). In Bohr’s model, the electron is pulled around the proton in a perfectly circular orbit by an attractive Coulomb force. The proton is approximately 1800 times more massive than the electron, so the proton moves very little in response to the force on the proton by the electron. (This is analogous to the Earth-Sun system, where the Sun moves very little in response to the force exerted on it by Earth.) An explanation of this effect using Newton’s laws is given in Photons and Matter Waves.
With the assumption of a fixed proton, we focus on the motion of the electron.
In the electric field of the proton, the potential energy of the electron is
[U(r) = -kfrac{e^2}{r},]
where (k = 1/4piepsilon_0) and (r) is the distance between the electron and the proton. As we saw earlier, the force on an object is equal to the negative of the gradient (or slope) of the potential energy function. For the special case of a hydrogen atom, the force between the electron and proton is an attractive Coulomb force.
Notice that the potential energy function (U(r)) does not vary in time. As a result, Schrödinger’s equation of the hydrogen atom reduces to two simpler equations: one that depends only on space (x, y, z) and another that depends only on time (t). (The separation of a wave function into space- and time-dependent parts for time-independent potential energy functions is discussed in Quantum Mechanics.) We are most interested in the space-dependent equation:
[frac{-hbar}{2m_e}left(frac{partial^2psi}{partial x^2} + frac{partial^2psi}{partial y^2} + frac{partial^2psi}{partial z^2}right) - kfrac{e^2}{r}psi = Epsi,]
where (psi = psi (x,y,z)) is the three-dimensional wave function of the electron, meme is the mass of the electron, and (E) is the total energy of the electron. Recall that the total wave function (Psi (x,y,z,t)), is the product of the space-dependent wave function (psi = psi(x,y,z)) and the time-dependent wave function (varphi = varphi(t)).
In addition to being time-independent, (U(r)) is also spherically symmetrical. This suggests that we may solve Schrödinger’s equation more easily if we express it in terms of the spherical coordinates ((r, theta, phi)) instead of rectangular coordinates ((x,y,z)). A spherical coordinate system is shown in Figure (PageIndex{2}). In spherical coordinates, the variable (r) is the radial coordinate, (theta) is the polar angle (relative to the vertical z-axis), and (phi) is the azimuthal angle (relative to the x-axis). The relationship between spherical and rectangular coordinates is (x = r , sin , theta , cos , phi), (y = r , sin theta , sin , phi), (z = r , cos , theta).
The factor (r , sin , theta) is the magnitude of a vector formed by the projection of the polar vector onto the xy-plane. Also, the coordinates of x and y are obtained by projecting this vector onto the x- and y-axes, respectively. The inverse transformation gives
[begin{align*} r &= sqrt{x^2 + y^2 + z^2} [4pt] theta &= cos^{-1} left(frac{z}{r}right), [4pt] phi &= cos^{-1} left( frac{x}{sqrt{x^2 + y^2}}right) end{align*}]
Schrödinger’s wave equation for the hydrogen atom in spherical coordinates is discussed in more advanced courses in modern physics, so we do not consider it in detail here. However, due to the spherical symmetry of (U(r)), this equation reduces to three simpler equations: one for each of the three coordinates ((r), (θ), and (ϕ)). Solutions to the time-independent wave function are written as a product of three functions:
[psi (r, theta, phi) = R(r) Theta(theta) Phi (phi),]
where (R) is the radial function dependent on the radial coordinate (r) only; (Θ) is the polar function dependent on the polar coordinate (θ) only; and (Φ) is the phi function of (ϕ) only. Valid solutions to Schrödinger’s equation (ψ(r, θ, ϕ)) are labeled by the quantum numbers (n), (l), and (m).
- (n): principal quantum number
- (l): angular momentum quantum number
- (m): angular momentum projection quantum number
(The reasons for these names will be explained in the next section.) The radial function (R) depends only on (n) and (l); the polar function (Theta) depends only on (l) and (m); and the phi function (Phi) depends only on (m). The dependence of each function on quantum numbers is indicated with subscripts:
[psi_{nlm}(r, theta, phi) = R_{nl}(r)Theta_{lm}(theta)Phi_m(phi).]
Not all sets of quantum numbers ((n), (l), (m)) are possible. For example, the orbital angular quantum number (l) can never be greater or equal to the principal quantum number (n(l < n)). Specifically, we have
- (n = 1,2,3,...)
- (l = 0,1,2,...,(n-1))
- (m = -l, (-l+1), . . ., 0, . . ., (+l - 1), +l)
Notice that for the ground state, (n = 1), (l = 0), and (m = 0). In other words, there is only one quantum state with the wave function for (n = 1), and it is (psi_{100}). However, for (n = 2), we have
[l = 0, , m = 0 nonumber]
and
[l = 1, , m = -1, 0, 1. nonumber]
Therefore, the allowed states for the (n = 2) state are (psi_{200}), (psi_{21-1}), (psi_{210}), and (psi_{211}). Example wave functions for the hydrogen atom are given in Table (PageIndex{1}). Note that some of these expressions contain the letter (i), which represents (sqrt{-1}). When probabilities are calculated, these complex numbers do not appear in the final answer.
| (n = 1, , l = 0, , m_l = 0) | (displaystyle psi_{100} = frac{1}{sqrt{pi}} frac{1}{a_0^{3/2}}e^{-r/a_0}) |
| (n = 2, , l = 0, , m_l = 0) | (displaystylepsi_{200} = frac{1}{4sqrt{2pi}} frac{1}{a_0^{3/2}}(2 - frac{r}{a_0})e^{-r/2a_0}) |
| (n = 2, , l = 1, , m_l = -1) | (displaystylepsi_{21-1} = frac{1}{8sqrt{pi}} frac{1}{a_0^{3/2}}frac{r}{a_0}e^{-r/2a_0}sin , theta e^{-iphi}) |
| (n = 2, , l = 1, , m_l = 0) | ( displaystyle psi_{210} = frac{1}{4sqrt{2pi}} frac{1}{a_0^{3/2}}frac{r}{a_0}e^{-r/2a_0}cos , theta) |
| (n = 2, , l = 1, , m_l = 1) | ( displaystylepsi_{211} = frac{1}{8sqrt{pi}} frac{1}{a_0^{3/2}}frac{r}{a_0}e^{-r/2a_0}sin , theta e^{iphi}) |
Physical Significance of the Quantum Numbers
Each of the three quantum numbers of the hydrogen atom ((n), (l), (m)) is associated with a different physical quantity.
Principal Quantum Number
The principal quantum number (n) is associated with the total energy of the electron, (E_n). According to Schrödinger’s equation:
[E_n = - left(frac{m_ek^2e^4}{2hbar^2}right)left(frac{1}{n^2}right) = - E_0 left(frac{1}{n^2}right), label{8.3}]
where (E_0 = -13.6 , eV). Notice that this expression is identical to that of Bohr’s model. As in the Bohr model, the electron in a particular state of energy does not radiate.
Example (PageIndex{1}): How Many Possible States?
For the hydrogen atom, how many possible quantum states correspond to the principal number (n = 3)? What are the energies of these states?
Strategy
For a hydrogen atom of a given energy, the number of allowed states depends on its orbital angular momentum. We can count these states for each value of the principal quantum number, (n = 1,2,3). However, the total energy depends on the principal quantum number only, which means that we can use Equation ref{8.3} and the number of states counted.
Solution
If (n = 3), the allowed values of (l) are 0, 1, and 2. If (l = 0), (m = 0) (1 state). If (l = 1), (m = -1, 0, 1) (3 states); and if (l = 2), (m = -2, -1, 0, 1, 2) (5 states). In total, there are 1 + 3 + 5 = 9 allowed states. Because the total energy depends only on the principal quantum number, (n = 3), the energy of each of these states is
[E_{n3} = -E_0 left(frac{1}{n^2}right) = frac{-13.6 , eV}{9} = - 1.51 , eV. nonumber]
Significance
An electron in a hydrogen atom can occupy many different angular momentum states with the very same energy. As the orbital angular momentum increases, the number of the allowed states with the same energy increases.
Angular Momentum Orbital Quantum Number
The angular momentum orbital quantum number (l) is associated with the orbital angular momentum of the electron in a hydrogen atom. Quantum theory tells us that when the hydrogen atom is in the state (psi_{nlm}), the magnitude of its orbital angular momentum is
[L = sqrt{l(l + 1)}hbar,]
where (l = 0, 1, 2, . . . , (n - 1)).
This result is slightly different from that found with Bohr’s theory, which quantizes angular momentum according to the rule (L = n), where (n = 1,2,3, ...)
Spectroscopic Notation
Quantum states with different values of orbital angular momentum are distinguished using spectroscopic notation (Table (PageIndex{2})). The designations s, p, d, and f result from early historical attempts to classify atomic spectral lines. (The letters stand for sharp, principal, diffuse, and fundamental, respectively.) After f, the letters continue alphabetically.
The ground state of hydrogen is designated as the 1s state, where “1” indicates the energy level ((n = 1)) and “s” indicates the orbital angular momentum state ((l = 0)). When (n = 2), (l) can be either 0 or 1. The (n = 2), (l = 0) state is designated “2s.” The (n = 2), (l = 1) state is designated “2p.” When (n = 3), (l) can be 0, 1, or 2, and the states are 3s, 3p, and 3d, respectively. Notation for other quantum states is given in Table (PageIndex{3}).
| Orbital Quantum Number (l) | Angular Momentum | State | Spectroscopic Name |
|---|---|---|---|
| 0 | 0 | s | Sharp |
| 1 | (sqrt{2}h) | p | Principal |
| 2 | (sqrt{6}h) | d | Diffuse |
| 3 | (sqrt{12}h) | f | Fundamental |
| 4 | (sqrt{20}h) | g | |
| 5 | (sqrt{30}h) | h |
Angular Momentum Projection Quantum Number
The angular momentum projection quantum number (m) is associated with the azimuthal angle (phi) (see Figure (PageIndex{2})) and is related to the z-component of orbital angular momentum of an electron in a hydrogen atom. This component is given by
[L_z = mhbar,]
where (m = -l, -l + 1, ..., 0, ..., +l - 1, l).
The z-component of angular momentum is related to the magnitude of angular momentum by
[L_z = L , cos theta,]
where (theta) is the angle between the angular momentum vector and the z-axis. Note that the direction of the z-axis is determined by experiment - that is, along any direction, the experimenter decides to measure the angular momentum. For example, the z-direction might correspond to the direction of an external magnetic field. The relationship between (L_z) and (L) is given in Figure (PageIndex{3}).
| (l = 0) | (l = 1) | (l = 2) | (l = 3) | (l = 4) | (l = 5) | |
| (n = 1) | 1s | |||||
| (n = 2) | 2s | 2p | ||||
| (n = 3) | 3s | 3p | 3d | |||
| (n = 4) | 4s | 4p | 4d | 4f | ||
| (n = 5) | 5s | 5p | 5d | 5f | 5g | |
| (n = 6) | 6s | 6p | 6d | 6f | 6g | 6h |
The quantization of (L_z) is equivalent to the quantization of (theta). Substituting (sqrt{l(l + 1)}hbar) for (L) and (m) for (L_z) into this equation, we find
[mhbar = sqrt{l(l + 1)}hbar , cos , theta.]
Thus, the angle (theta) is quantized with the particular values
[theta = cos^{-1}left(frac{m}{sqrt{l(l + 1)}}right).]
Notice that both the polar angle ((θ)) and the projection of the angular momentum vector onto an arbitrary z-axis ((L_z)) are quantized.
The quantization of the polar angle for the (l = 3) state is shown in Figure (PageIndex{4}). The orbital angular momentum vector lies somewhere on the surface of a cone with an opening angle (theta) relative to the z-axis (unless (m = 0), in which case (θ = 90^o) and the vector points are perpendicular to the z-axis).
A detailed study of angular momentum reveals that we cannot know all three components simultaneously. In the previous section, the z-component of orbital angular momentum has definite values that depend on the quantum number (m). This implies that we cannot know both x- and y-components of angular momentum, (L_x) and (L_y), with certainty. As a result, the precise direction of the orbital angular momentum vector is unknown.
Example (PageIndex{2}): What Are the Allowed Directions?
Calculate the angles that the angular momentum vector (vec{L}) can make with the z-axis for (l = 1), as shown in Figure (PageIndex{5}).
Strategy
The vectors (vec{L}) and (vec{L_z}) (in the z-direction) form a right triangle, where (vec{L}) is the hypotenuse and (vec{L_z}) is the adjacent side. The ratio of (L_z) to |(vec{L})| is the cosine of the angle of interest. The magnitudes (L = |vec{L}|) and (L_z) are given by
[L = sqrt{l(l + 1)} hbar nonumber]
and
[L_z = mhbar. nonumber]
Solution
We are given (l = 1), so (m) can be +1, 0, or +1. Thus, (L) has the value given by
[L = sqrt{l(l + 1)}hbar = sqrt{2}hbar. nonumber]
The quantity (L_z) can have three values, given by (L_z = m_lhbar).
[L_z = begin{cases} hbar, & text{if } m_l=+1 0, & text{if } m_l=0 hbar, & text{if } m_l=-1 end{cases} nonumber]
As you can see in Figure (PageIndex{5}), (cosθ=Lz/L), so for (m=+1), we have
[cos , theta_1 = frac{L_z}{L} = frac{hbar}{sqrt{2}hbar} = frac{1}{sqrt{2}} = 0.707 nonumber]
Thus,
[theta_1 = cos^{-1}0.707 = 45.0°. nonumber]
Similarly, for (m = 0), we find (cos , theta_2 = 0); this gives
[theta_2 = cos^{-1}0 = 90.0°. nonumber]
Then for (m_l = -1):
[cos , theta_3 = frac{L_Z}{L} = frac{-hbar}{sqrt{2}hbar} = -frac{1}{sqrt{2}} = -0.707, nonumber ]
so that
Hydrogen Wave Function
[theta_3 = cos^{-1}(-0.707) = 135.0°. nonumber]
Significance
The angles are consistent with the figure. Only the angle relative to the z-axis is quantized. (L) can point in any direction as long as it makes the proper angle with the z-axis. Thus, the angular momentum vectors lie on cones, as illustrated. To see how the correspondence principle holds here, consider that the smallest angle ((theta_1) in the example) is for the maximum value of (m_l), namely (m_l = l). For that smallest angle,
[cos , theta = dfrac{L_z}{L} = dfrac{l}{sqrt{l(l + 1)}}, nonumber]
which approaches 1 as (l) becomes very large. If (cos , theta = 1), then (theta = 0º). Furthermore, for large (l), there are many values of (m_l), so that all angles become possible as (l) gets very large.
Exercise (PageIndex{1})
Can the magnitude (L_z) ever be equal to (L)?
No. The quantum number (m = -l, -l + l, ..., 0, ..., l -1, l). Thus, the magnitude of (L_z) is always less than (L) because (<sqrt{l(l + 1)})
Using the Wave Function to Make Predictions
As we saw earlier, we can use quantum mechanics to make predictions about physical events by the use of probability statements. It is therefore proper to state, “An electron is located within this volume with this probability at this time,” but not, “An electron is located at the position (x, y, z) at this time.” To determine the probability of finding an electron in a hydrogen atom in a particular region of space, it is necessary to integrate the probability density (|ψ_{nlm}|^2)_ over that region:
[text{Probability} = int_{volume} |psi_{nlm}|^2 dV,]
where (dV) is an infinitesimal volume element. If this integral is computed for all space, the result is 1, because the probability of the particle to be located somewhere is 100% (the normalization condition). In a more advanced course on modern physics, you will find that (|psi_{nlm}|^2 = psi_{nlm}^* psi_{nlm}), where (psi_{nlm}^*) is the complex conjugate. This eliminates the occurrences (i = sqrt{-1}) in the above calculation.
Consider an electron in a state of zero angular momentum ((l = 0)). In this case, the electron’s wave function depends only on the radial coordinate (r). (Refer to the states (psi_{100}) and (psi_{200}) in Table (PageIndex{1}).) The infinitesimal volume element corresponds to a spherical shell of radius (r) and infinitesimal thickness (dr), written as
[dV = 4pi r^2dr.]
The probability of finding the electron in the region (r) to (r + dr) (“at approximately r”) is
[P(r)dr = |psi_{n00}|^2 4pi r^2 dr.]
Here (P(r)) is called the radial probability density function (a probability per unit length). For an electron in the ground state of hydrogen, the probability of finding an electron in the region (r) to (r + dr) is
[|psi_{n00}|^2 4pi r^2 dr = (4/a_)^3)r^2 exp(-2r/a_0)dr,]
where (a_0 = 0.5) angstroms. The radial probability density function (P(r)) is plotted in Figure (PageIndex{6}). The area under the curve between any two radial positions, say (r_1) and (r_2), gives the probability of finding the electron in that radial range. To find the most probable radial position, we set the first derivative of this function to zero ((dP/dr = 0)) and solve for (r). The most probable radial position is not equal to the average or expectation value of the radial position because (|psi_{n00}|^2) is not symmetrical about its peak value.
If the electron has orbital angular momentum ((l neq 0)), then the wave functions representing the electron depend on the angles (theta) and (phi); that is, (psi_{nlm} = psi_{nlm}(r, theta, phi)). Atomic orbitals for three states with (n = 2) and (l = 1) are shown in Figure (PageIndex{7}). An atomic orbital is a region in space that encloses a certain percentage (usually 90%) of the electron probability. (Sometimes atomic orbitals are referred to as “clouds” of probability.) Notice that these distributions are pronounced in certain directions. This directionality is important to chemists when they analyze how atoms are bound together to form molecules.
A slightly different representation of the wave function is given in Figure (PageIndex{8}). In this case, light and dark regions indicate locations of relatively high and low probability, respectively. In contrast to the Bohr model of the hydrogen atom, the electron does not move around the proton nucleus in a well-defined path. Indeed, the uncertainty principle makes it impossible to know how the electron gets from one place to another.
Contributors and Attributions
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
Solving the Schrödinger equation for hydrogen-like atoms
Consider the Schrödinger (time independent) Wave Equationwhich is expanded as
| (2) |
in Cartesian co-ordinates.
When applied to the hydrogen atom, the wave function should describe the behaviour of both the nucleus and the electron, . This means we have a two body problem, which is very difficult to solve.We can fortunately convert this two-body problem to an effective one-bodyproblem by transforming from the Laboratory System to the Centre of Mass System.
The appropriate potential is of course the simple radial electrostatic potentialfor a point charge nucleus of charge . (Note that if we are not dealing with hydrogen, then we are dealing with hydrogen-like atoms which are fully stripped of all but one electron.)
| (4) |
has spherical symmetry.One could write and solve the wave equation in Cartesian co-ordinates.This would work but it would be very tedious, as the mathematics does not display the symmetry of the physics.Accordingly we rather exploit the spherical symmetry of the potential, and perform a co-ordinate transformation from Cartesian Co-ordinates (efficient for rectangle shapes)to Spherical Polar Co-ordinates (efficient for spherical shapes)
These new co-ordinates are defined in figure 1.Under this co-ordinate transformation, the wave equation takes the form,
| (6) |
Exercise 1Writing the double differential operator in spherical polar co-ordinates is not trivial.Check Electromagnetic Fields and Waves by Lorrain and Carson, Chapter 1, if you requireinsight into the last step. You are not expected to be able to do this transformation.However, make sure, using a sketch, that you can show how an infinitesimal volume elementbehaves under this transformation.
Exercise 2
Write each of the variables in terms of the variables , also perform the inverse mapping. You may use figure 1.Note that the potential is radial, , which meansit depends only on , and not on or .
The wave function necessarily is separable into radial, polar and azimuthal factors under a radial potential as follows : .(Once again, can you think why ?)Substituting the above expression and the potential into the spherical polar representation of the wave equation, we find, after some manipulation,
| (8) |
Exercise 3
Ensure you can achieve the last result with your own pencil and paper. Doing these exercises andtutorials properly helps to make you familiarwith quantum mechanics, via your fingertips, into the marrow of your bones. You may use your textbooksthe first time you do this.
This equation has on the left functions of only, and on the right a function of only. Accordingly, it can only be satisfied for all values of the independent variables by requiring that both sides are equal to the same constant value. Hence it follows the the left hand side, which iscalled the azimuthal equation, is equal to a constant, which is called the azimuthal constant.We give the azimuthal constant the symbol , for reasons which will become clear withhindsight.
The solution for the azimuthal equation (by the ansatz method) is
| (10) |
Exercise 4
Check this with your own pencil and paper....
Imposing the boundary conditions, we must have single valued, and . Therefore, we require the constant to be quantisedas follows.
We call the magnetic quantum number. You may imagine that the quantity measures the change of the azimuthal part of the wave functionwith change of the azimuthal co-ordinate. Thus this quantity can sensedifferences under a rotation about the -axis. Intuitively, if the wave function of theelectron changed under such a rotation, one would be able to discern it, and classically,a rotating charge has a magnetic moment. This is a heuristic justification of the appellationof ``magnetic quantum number' for .
Clearly, not only is the separation constant quantised, but also the azimuthal wave function becomespart of a family of wave functions labelled by the quantum number .
| (12) |
Congratulations !
You have just seen that the quantisation of the magnetic quantum number arises naturally from the condition that the wave function must be single valued and satisfy its boundary conditions.The imposition of boundary conditions also lead to quantisation of the wave function in the previousexamples we have seen (particle in a box).This is just a mathematical way of saying we have successfullytrapped a wave function in an attractive potential.The value of the prefactor will be set later via a normalisation condition.Thus fortified, we proceed to the polar wave function.Substituting the constant, (known as a separation constant), back into the wave equation above and re-arranging terms
Again, we have an equation which has on the left functions of only, and on the right functions of only. Accordingly, it can only be satisfied for all values of the independent variables by requiring that both sides are equal to the same constant value. Hence it follows,
| (14) |
We have achieved so far separated equations for the last two wave functions, viz. theradial wave equation for and the polar wave equation for. The solution of these two equations is beyond the scope of this course.Rest assured, it proceeds as in the case for the azimuthal wave function. That is, imposing the boundary conditions causes the separation constant to become quantised and also the radial wave function and the polar wave function to become part of a family labelled by theappropriate quantum number.
(The angle dependent wave functions are known as the Spherical Harmonic Functions, and the radialwave functions as Laguerre polynomials. These solutions are tabulated below in figure 2.)
Hence we find, on solving the polar wave equation for that
| (16) |
so that
We call the orbital quantum number. In a later section, we will evidence the relationship of this quantum number to orbital angular momentum. This is then the heuristic justification of its appellation.
Exercise 5
What do you think the boundary conditions are for the polar wave equation ?
In the introductory course on Quantum Mechanics, we saw that was also a separation constant.It arose when we separated the time and space parts of the Time dependent wave equation to arrive at the Time independent wave equation, which we have presented at the top of this section.So it will come as no surprise now to find that becomes quantised on requiring thatthe solutions of the radial wave equation above also obey boundary conditions. In the case of the radial wave equation, we obviously require that is finite and .We get ,
| (18) |
with the ancillary condition that
We call the principal quantum number.
Exercise 6
This is the same expression as in the Bohr model. Muse on what this correspondence with the Bohr model implies.
Exercise 7
How would these results change if we were dealing with positronium and not hydrogen ?
We may summarise our results so far :
The Schrödinger Wave Equation for hydrogen like atoms in three dimensions is best treated in spherical polar co-ordinates
| (20) |
because the Coulomb potential for this case
is spherically symmetric.Consequently, the wave function will be separable
| (22) |
and the Schrödinger Wave Equation reduces to the three equations
| (24) |
and
Solving these equations subject to the appropriate boundary conditions leads to the three sets of quantum numbers :
In this process, we also find a family of wave functions, labelled by the quantum numbers :
We can think of the set of quantum numbers as identifying a wave functionfor a particular state . It is typical that quantum numbers appear naturally when quantum particles are trappedin a particular region of space by an attractive potential.A selection of the lowest energy wave functions have been collected in the table above.These wave functions are normalised so that the probability density for finding an electron in a particular state represented by is unity when integrated over all space.
| (27) |
Observation
Schrödinger sure deserved the Nobel Prize !
Example
Suppose we want to verify the energy of the ground state wave function of thehydrogen atom, and . We note that it is only the radial waveequation which contains , the energy of the state. The appropriate radial wave functionis
| (29) |
Using and , we simplify to find
Each parenthesis must equal nought for the entire equation to equal nought.We therefore find an expression for , the Bohr radius,
| (31) |
and the ground state energy
Next:
Hydrogen Wave Function Hyperphysics
Quantum numbers Up:Quantum Mechanics of Atoms Previous:A full Quantum MechanicalSimon Connell2004-10-04