Bohr model of hydrogen atom. Bohr Model of hydrogen atom was adopted by Neils Bohr in 1913 for the explanation of the Rutherford model and the atomic energy levels of the hydrogen spectrum. According to classical mechanics, when a charged electron is subjected to acceleration, it emits radiation and energy released to hit the nucleus of an atom. Hydrogen is a colorless, odorless gas. It is easily ignited. Once ignited it burns with a pale blue, almost invisible flame. The vapors are lighter than air. It is flammable over a wide range of vapor/air concentrations.
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.
Bohr Model Of Hydrogen Atom
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
[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).
| In 1897 J. J. Thomson discovered the electron, a negatively charged particle more than two thousand times lighter than a hydrogen atom. In 1906 Thomson suggested that each atom contained a number of electrons roughly equal to its atomic number. Since atoms are neutral, the charge of these electrons must be balanced by some kind of positive charge. Thomson proposed a 'plum pudding' model, with positive and negative charge filling a sphere only one ten billionth of a meter across. This plum pudding model was generally accepted. Even Thomson's student Rutherford, who would later prove the model incorrect, believed in it at the time. | |
| In 1911 Ernest Rutherford proposed that each atom has a massive nucleus containing all of its positive charge, and that the much lighter electrons are outside this nucleus. The nucleus has a radius about ten to one hundred thousand times smaller than the radius of the atom. Rutherford arrived at this model by doing experiments. He scattered alpha particles off fixed targets and observed some of them scattering through very large angles. Scattering at large angles occurs when the alpha particles come close to a nucleus. The reason that most alpha particles are not scattered at all is that they are passing through the relatively large 'gaps' between nuclei. Links: The Rutherford Experiment | |
| Rutherford revised Thomson's 'plum pudding' model, proposing that electrons orbit a positively charged nucleus, like planets orbit a star. But orbiting particles continuously accelerate, and accelerating charges produce electromagnetic radiation. According to classical physics the planetary atom cannot exist. Electrons quickly radiate away their energy and spiral into the nucleus. In 1915 Niels Bohr adapted Rutherford's model by saying that the orbits of the electrons were quantized, meaning that they could exist only at certain distances from the nucleus. Bohr proposed that electrons did not emit EM radiation when moving in those quantized orbits. | |
Quantum mechanics now predicts what measurements can reveal about atoms. The hydrogen atom represents the simplest possible atom, since it consists of only one proton and one electron. The electron is bound, or confined. Its potential energy function U(r) expresses its electrostatic potential energy as a function of its distance r from the proton.
U(r) = -q2/(4πε0r) = -e2/r. Here e2 is defined as q2/(4πε0). In SI unit 1/(4πε0) = 9*109 Nm2/C2, and q = 1.6*10-19 C. |
The energy levels in a hydrogen atom can be obtained by solving Schrödinger’s equation in three dimensions. We have to solve the radial equation
(-ħ2/(2m))∂2(rR))/∂r2 + (l(l + 1)ħ2/(2mr2))(rR) - (E - e2/r)(rR) = 0
or
∂2(rR))/∂r2 + [(2m/ħ2)(E + e2/r) - l(l + 1)/r2](rR) = 0,
or
∂2(rR))/∂r2 + k2(r)(rR) = 0,
with k2(r) = (2m/ħ2)(E + e2/r) - l(l + 1)/r2.
This equation can be integrated using the Numerov method. Click on the linked spreadsheet to find the allowed electron energies in the hydrogen atom numerically for states with zero angular momentum. All distance are measured in Å (10-10 m) and all energies in eV. (Note: We are solving the differential equation for the function rR(r), not for the function R(r).) Because we cannot integrate from infinity, the program assumes that rR(r) = 0 at r = 30Å. It integrates inward towards the origin. The radial functions R(r) have to be finite at the origin, and therefore the functions rR(r) have to be zero at the origin for a solution that fulfills the boundary conditions.
A spreadsheet macro increments the trial energies in small steps. When rR(0) changes sign the program records an eigenvalue. Only eigenvalues associated with radial functions, which rapidly decrease as r increases beyond a few Å, are physically reasonable solutions. Confinement leads to energy quantization.
The electron energies in the hydrogen atom do nor depend on the quantum numbers m and l which characterize the dependence of the wave function on the angles θ and φ. The allowed energies are
En = -me4/(2ħ2n2) = -13.6 eV/n2.
Here n is called the principle quantum number. The values En are the possible value for the total electron energy (kinetic and potential energy) in the hydrogen atom. The average potential energy is -me4/(ħ2n2) and the average kinetic energy is me4/(2ħ2n2).
The wave functions ψnlm(r,θ,φ) = Rnl(r)Ylm(θ,φ) are products of functions Rnl(r), which depend only on the coordinate r, and the spherical harmonics Ylm(θ,φ), which depend only on the angular coordinates. They are characterized by three quantum numbers, n, l, and m.
| The electron has three spatial degrees of freedom. To completely determine its initial wave function, we, in general, have to make three compatible measurements. Some observables that are compatible with energy measurements and compatible with each other are
We can know the values of these observables labeled by n, l, and m, simultaneously. For the hydrogen atom, the energy levels only depend on the principal quantum number n. 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.
Below is a link to plots of the square of the wave functions or the probability densities for the electron in the hydrogen atom for different sets of quantum numbers n, l, and m. Links:
Note: Energy eigenfuctions characterize stationary state. We cannot track the electron and know its energy at the same time. If we know its energy, we can only predict probabilities for where we might find it if we tried to measure its position. If we determine the position of the electron, we lose the energy information. | Examples of hydrogen atom |
The probability of finding the electron in a small volume ∆V about the point (r,θ,φ) is |ψnlm(r,θ,φ)|2∆V. |ψnlm(r,θ,φ)|2 is the probability density, the probability per unit volume in three dimensions. |ψnlm(r,θ,φ)|2 is zero at the origin unless l = 0. But if l = 0, then the electron has zero orbital angular momentum, and there is a finite probability of finding it at the same position as the nucleus. |ψn00(r = 0)|2 is not equal to zero. This can lead to a special type of nuclear decay. Certain nuclei can de-excite by internal conversion, which is a process whereby the excitation energy is transferred directly to one of the atomic electrons, causing it to be ejected from the atom. This process competes with de-excitation by photon emission, which is called gamma decay. The probability of de-excitation by internal conversion is directly proportional to the probability of an electron being at the nucleus, and therefore only electrons with zero orbital angular momentum are involved.
The hydrogen-atom wave function for n = 1, 2, and 3 are given below. The constant a0 appearing in these functions has the value a0 = 52.92 pm.
The probability of finding the electron in a small volume ∆V about the point (r,θ,φ) is |ψnlm(r,θ,φ)|2∆V. The probability of finding the electron whose wave function depends only on the coordinate r a distance r from the nucleus is |ψ(r)|2 4πr2∆r. [The volume ∆V a distance r from the nucleus is a spherical shell with radius r and thickness ∆r.] Only electrons in state with l = 0 have spherically symmetric wave functions.
Problem:
Find the probability per unit length of finding an electron in the ground state of hydrogen a distance r from the nucleus. At what value of r does this probability have its maximum value?
Solution:
Given the ground state wave function ψ100(r,θ,φ) = ψ100(r) = [1/(π1/2a03/2)]exp(-r/a0), we find the probability per unit length,
P100(r) = |ψ100(r)|2 4πr2 = (4/a03) r2 exp(-2r/a0). We can plot P100(r) versus r. Let us measure r in units of a0. Open the linked spreadsheet to view the plot.
The plot shows that P100(r) has its maximum value at r = 1 (in units of a0), i.e at r = a0.
Suggestion: Change the spreadsheet to plot P200(r) = |ψ200(r)|2 4πr2 = (1(/4a03)) r2 (2 - r/a0)2exp(-r/a0). At what value of r does this probability have its maximum value? Note: Because we measuring distances in units of a0, a0 in units of a0 is equal to 1, and you need to plot P200(r) = (1/4) r2 (2 - r)2exp(-r).
Spectroscopic notation
Often texts use a different (spectroscopic) notation to refer to the energy levels of the hydrogen atom.
Letters of the alphabet are associated with various values of l.
|
The hydrogen line spectrum:
| When an electron changes from one energy level to another, the energy of the atom must change as well. It requires energy to promote an electron from one energy level to a higher one. This energy can be supplied by a photon whose energy E is given in terms of its frequency E = hf or wavelength E = hc/λ. Since the energy levels are quantized, only certain photon wavelengths can be absorbed. If a photon is absorbed, the electrons will be promoted to a higher energy level and will then fall back down into the lowest energy state (ground state) in a cascade of transitions. Each time the energy level of the electron changes, a photon will be emitted and the energy (wavelength) of the photon will be characteristic of the energy difference between the initial and final energy levels of the atom in the transition. The energy of the emitted photon is just the difference between the energy levels of the initial (ni) and final (nf ) states. The set of spectral lines for a given final state nf are generally close together. In the hydrogen atom they are given special names. The lines for which nf = 1 are called the Lyman series. These transitions frequencies correspond to spectral lines in the ultraviolet region of the electromagnetic spectrum. The lines for which nf = 2 are called the Balmer series and many of these spectral lines are visible. The spectrum of hydrogen is particularly important in astronomy because most of the Universe is made mostly of hydrogen. |
| The Balmer series, which is the only hydrogen series with lines in the visible region of the electromagnetic spectrum, is shown in the right in more detail. The Balmer lines are designated by H with a Greek subscript in order of decreasing wavelength. Thus the longest wavelength Balmer transition is designated H with a subscript alpha, the second longest H with a subscript beta, and so on. |
Problem:
What is the wavelength of the least energetic line in the Balmer series?
Hydrogen Atomic Emission Spectrum
Solution:
The transition from ni = 3 to nf = 2 is the lowest energy, longest wavelength transition in the Balmer series.
∆E = -13.6 eV(1/9 - 1/4) = 1.89 eV = 3*10-19 J. λ = hc/∆E = 658 nm.
Problem:
What is the shortest wavelength in the Balmer series?
Solution
The transition from ni = ∞ to nf = 2 is the highest energy, shortest wavelength transition in the Balmer series.
∆E = -13.6 eV(1/∞ - 1/4) = 13.6/ 4 eV = 3.4 eV = 5,44*10-19 J. λ = hc/∆E = 365 nm.
Hydrogenic atoms
Atoms with all but one electron removed are called hydrogenic atoms.
If the charge of the nucleus is Z times the proton charge, then U(r) = -Ze2/r.
The solutions to the Schroedinger equation of such atoms are obtained by simply scaling the the solutions for the hydrogen atom.
The energy levels scale with Z2, i.e. En = -Z2*13.6 eV/n2. It takes more energy to remove an electron from the nucleus, because the attractive force that must be overcome is stronger.
The average size of the wave functions scales as 1/Z, i.e. the electron, on average, stays closer to the nucleus, because the attraction is stronger. In the wave functions we replace a0 by a0/Z.
The Bohr Atom
In 1913 Bohr's model of the atom revolutionized atomic physics. The Bohr model consists of four principles:
- Electrons assume only certain orbits around the nucleus. These orbits are stable and called 'stationary' orbits. Electrons in these orbits do not radiate their energy away.
- Each orbit is associated with a definite value of the energy and the angular momentum. Bohr assumed that the angular momentum could only take on values that were integer multiples of ħ.
Angular momentum = mr2ω = mrv = nħ, n = 1, 2, 3, ... .
A classical electron with a definite angular momentum in an orbit about a proton also has a definite energy E.
If angular momentum = mrv = nħ, then En = -me4/(2ħ2n2) = -13.6 eV/n2.
The orbit closest to the nucleus has an energy E1, the next closest E2 and so on.
A definite angular momentum also implies a definite orbital radius.
If angular momentum = mrv = nħ, then rn = n2ħ2/(me2) = n2a0 = n2 * (52.92 pm).
a0 is called the Bohr radius. - A photon is emitted when an electron jumps from a higher energy orbit to a lower energy orbit and absorbed when it jumps from a lower energy orbit to higher energy orbit. The photon energy is equal to the energy difference ∆E = hf = Ehigh - Elow.
| With these conditions Bohr was able to explain the stability of atoms, as well as the emission spectrum of hydrogen. According to Bohr's model only certain orbits were allowed which means only certain energies are possible. These energies naturally lead to the explanation of the hydrogen atom spectrum. Bohr's model was so successful that he immediately received world-wide fame. Unfortunately, Bohr's model worked only for hydrogen and hydrogenic atoms, such as any atom with all but one electron removed. The Bohr model is easy to picture, but we now know that it is wrong. Any planetary model of the atom, so often seen in pictures and so easy to picture, is wrong. Links: |