| 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. | |
Hydrogen Atom: A hydrogen atom contains only two subatomic particles: (1) an electron and (2) a proton. It is the only atom that exists in the universe with no neutron in the atomic nucleus.
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.
Mechanical nature of the hydrogen atom helps us understand how these lines arise. Series of lines in the hydrogen spectrum, named after the scien-tists who observed and characterized them, can be related to the energies associated with transsitions from the various energy lev-els of the hydrogen atom. The relation, simple enough as it is. Using Balmer-Rydberg equation to solve for photon energy for n=3 to 2 transition. Solving for wavelength of a line in UV region of hydrogen emission spectrum. According to the Bohr model, the wavelength of the light emitted by a hydrogen atom when the electron falls from a high energy (n = 4) orbit into a lower energy (n = 2) orbit.Substituting the appropriate values of R H, n 1, and n 2 into the equation shown above gives the following result. Solving for the wavelength of this light gives a value of 486.3 nm, which agrees with the experimental.
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.
|
Energy Of Hydrogen Atom In Its Ground State
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?
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: |
Hydrogen, like all atoms, provides very distinctive lines when the frequency of its electromagnetic waves are measured as its electron changes between orbitals. These electromagnetic waves come in the form of packets, or photons, that are absorbed or emitted by an electron. Photons are absorbed if the electron takes energy to move away from the hydrogen nucleus and emitted if the electron moves closer to the hydrogen nucleus.
Despite the electron’s probability distribution in an atomic orbit, the energy transition states are very distinct when they move between orbitals. The photon energy and its wavelength are measured by spectral analysis such as hydrogen’s spectral lines below:
In wave theory, these transitions are described as a conversion from transverse to longitudinal wave energy and vice versa. Particles produce spherical, longitudinal waves that decrease with the square of the distance from the particle core. Two or more particles will produce waves that are amplified, their longitudinal wave amplitude is constructively or destructively added as a result of this interaction. Thus, this energy changes as the distance between the particles change. However, energy is always conserved. It transitions from one wave form to another.
Longitudinal wave energy is spherical, where each shell (n) in the wave equations is a longitudinal wavelength. Each spherical shell has an energy level that can be computed with the Longitudinal Energy equation. This energy can be transferred. Transverse waves are a secondary wave as a result of particle vibration. Whether it is particle annihilation or orbital transition, the electron will vibrate as it settles into its final position and this vibration creates a transverse wave. The difference in longitudinal energy is then transferred to vibrational energy in a transverse wave (photon). A photon will travel until reaching another particle, transferring its energy to longitudinal energy, once again affecting the wave amplitude between particles. The photon will be absorbed in this case and a particle once again moves to minimize its amplitude. It moves further from the atom’s core once the photon is absorbed.
Photons are transverse waves which we know as electromagnetic waves responsible for light, radio waves, microwaves and more. They are a result of short bursts of electron vibrations, creating a secondary transverse wave. Energy is conserved and we can calculate this as a difference in longitudinal energy between two or more particles.
Evidence
The evidence is an exact match for the hydrogen atom for both ionization and orbital transitions when using the wave equations to calculate photon energies and wavelengths. When an electron leaves the atom, or when the electron transitions orbitals, a photon is emitted or absorbed and this can be calculated using the wave equations. The calculations and results for each are shown below.
In addition to the wave equations and constants that are used elsewhere on this site in other applications, two important variables that are required to determine transverse energies and wavelengths are:
Distance – The distance between two particles, from the initial state to the final state. In the wave equations, this is expressed in terms of wavelengths (n) from the nucleus core to the electron.
Amplitude Difference – The energy released is also dependent on the amount of constructive or destructive wave interference between two particles. Amplitude factor (δ) is used in the wave equations to reflect this difference. A positron-electron interaction is set as the baseline of one, thus δ=1 for hydrogen energy calculations. However, other atomic elements will have different amplitude factors that will need to be considered. For this section, δ=1.
Hydrogen Ionization
When a particle experiences an amplitude difference, the transverse equation is modified to be the difference in energy from its initial position ni to its final position nf. In the case of ionization, an electron is ejected from an atom and experiences an amplitude difference, as it is no longer attracted to the atom’s nucleus. Once ejected, its distance (n wavelengths) is essentially infinity, so nf is set to infinity.
Using the Transverse Energy Equation, the final position is set to infinity, and the equation simplifies to:
A negative sign in this equation means that energy (a photon) needs to be absorbed. A positive value means that a photon is created.
A similar derivation can be used to calculate the photon’s wavelength required for ionization, starting with the Transverse Wavelength Equation and inserting infinity in place of nf.
Transverse Wavelength – Ionization
Energy Of Hydrogen Atom In Ground State In Joules
Using the orbital distances (n, in wavelengths) and amplitude factor of 1 (δ=1, for a single positron-electron interaction), the wave equations can now be used to calculate energies and wavelengths of particle interaction. Beginning with hydrogen, this example calculates the wavelength of a photon absorbed to eject an electron (K=10) from the first orbital (shell N1, or n=187,1789). Starting with the Transverse Wavelength – Ionization equation from above:
The equation above solves to be -9.113 x 10-8 meters, or –91.13 nanometers, a match of known hydrogen ionization energy data. In fact, the calculated values for each of the remaining orbitals (displayed in red in the table below) exactly match the known ionization energies and wavelengths of hydrogen data, shown in italics in the table.
In shell n=5, the ionization energy and wavelength of the electron-positron annihilation also match, showing a relation between ionization and annihilation energies as one equation; simply a difference of amplitude based on the distance between the particles.
Hydrogen Orbital Transitions
Another set of data demonstrating proof of the wave equations is the transitional energies and wavelengths for an electron in a hydrogen atom that moves between orbitals. In this case, it is a difference in energy between two positions relative to the nucleus, where ni is the initial orbital and nf is the final orbital.
For example, the wavelength distances for hydrogen for the third orbital (N3) is 1,690,098 wavelengths from the proton, and 751,155 wavelengths for the second orbital (N2). Thus for an electron that transitions from N=3 to N=2 (3->2) the wavelength values are: nf = 751,155 and ni = 1,690,098. Inserting these values into the Transverse Energy Interaction equation yields:
Note a positive value in the above energy calculation, meaning a photon is created. And using the Transverse Wavelength Interaction equation, the wavelength of this photon emitted for hydrogen orbital transition 3->2 is:
The above equation is solved to be 6.56 x 10-7 meters, or 656 nanometers. This is an exact match of photon wavelengths measured in experiments. The table below shows the calculations of the above transition (3->2) through to an electron in the ninth orbital transitioning to the second orbital (9->2). The calculations using the above equations are marked in red and the measured data from the hydrogen spectral series in italics.
CLICK TO ENLARGE
The above table is an exact match of the hydrogen spectrum, without the use of the Rydberg constant. It’s accurate to at least three digits and can be used not only for orbital transitions, but also for ionization energies and annihilation properties.