- Hydrogen Atom Levels
- Energy Of Hydrogen Atom
- Energy Of Hydrogen Atom At N 3
- Energy Of Hydrogen Atom In Its Ground State
- Energy Of Hydrogen Atom In First Excited State
1 day ago In many textbooks, energy changes of the hydrogen atom are attributed to the electron transitioning between energy levels. However, the energy itself is that of the whole system (proton+electron) s.
Next:- A muonic hydrogen is a bound state of a proton to a muon (instead of a proton and an electron as in the usual hydrogen). I) Use Bohr’s theory to calculate the energy levels of the muonic hydrogen. What is the energy of the ground state (in eV)? We proceed as we (well, Bohr) did in the regular hydrogen atom. We use Newton’s law e2 4ˇ 0r2.
- Recommended ionization energy: Δ f H (+) ion,0K: Enthalpy of formation of positive ion at 0K: Δ r G° Free energy of reaction at standard conditions: Δ r H° Enthalpy of reaction at standard conditions.
- 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.
A hydrogen atom consists of an electron, of charge and mass ,and a proton, of charge and mass , moving in the Coulombpotential
where is the position vector of the electron with respect to theproton. Now, according to the analysis in Sect. 6.4, this two-bodyproblem can be converted into an equivalent one-body problem. In thelatter problem, a particle of mass
| (662) |
moves in the central potential
Note, however, that since the differencebetween and is very small. Hence, in the following,we shall write neglect this difference entirely.
Writing the wavefunction in the usual form,
| (664) |
it follows from Sect. 9.2 that the radial function satisfies
Let , with
| (666) |
where and are defined in Eqs. (678) and (679),respectively.Here, it is assumed that , since we are only interested in bound-states of the hydrogen atom. The above differential equation transformsto
where
| (668) |
Suppose that . It follows that
We now need to solve the above differential equation in the domain to , subject to the constraint that be square-integrable.
Let us look for a power-law solution of the form
| (670) |
Substituting this solution into Eq. (669), we obtain
Equating the coefficients of gives the recursion relation
| (672) |
Now, the power series (670) must terminate at small , atsome positive value of , otherwise behaves unphysically as [i.e., it yields an that is not square-integrableas ]. From the above recursion relation, this is only possible if , where the first term in the series is . There are two possibilities: or . However, the former possibility predicts unphysical behaviour of at . Thus, we conclude that . Note that, since at small , there is a finiteprobability of finding the electron at the nucleus for an state, whereasthere is zero probability of finding the electron at the nucleus foran state [i.e., at , except when].
For large values of , the ratio of successive coefficients in the power series (670)is
| (674) |
which converges to . We conclude that as . It thus follows that as . This does not correspond to physically acceptable behaviour of the wavefunction, since must be finite.The only way in which we can avoid this unphysical behaviour is if thepower series (670) terminates at some maximum value of . According tothe recursion relation (672), this is only possible if
where is an integer, and the last term in the series is . Since thefirst term in the series is , it follows that mustbe greater than , otherwise there are no terms in the series at all. Finally, it is clear from Eqs. (666), (668), and (675) that
| (676) |
and
where
| (678) |
and
Here, is the energy of so-called ground-state (or lowest energy state) of thehydrogen atom, and the length is known as the Bohr radius. Note that , where is the dimensionless fine-structure constant. The fact that is the ultimate justification for our non-relativistic treatment of the hydrogen atom.
We conclude that the wavefunction of a hydrogen atom takes the form
| (680) |
Here, the are the spherical harmonics (see Sect 8.7), and is the solution of
which varies as at small .Furthermore, the quantum numbers , , and can only take valueswhich satisfy the inequality
| (682) |
where is a positive integer, a non-negative integer, and an integer.
Now, we expect the stationary states of the hydrogen atom to be orthonormal: i.e.,
| (684) |
It, thus, follows that the radial wavefunction satisfies theorthonormality constraint
The first few radial wavefunctions for the hydrogen atom are listed below:
| (686) |
| (687) |
| (688) |
| (689) |
| (690) |
| (691) |
These functions are illustrated in Figs. 21 and 22.
Given the (properly normalized) hydrogen wavefunction (680),plus our interpretation of as a probability density, we can calculate
| (693) |
| (694) |
| (695) |
| (696) |
| (697) |
According to Eq. (676), the energy levels of the bound-states of a hydrogen atomonly depend on the radial quantum number . It turns out that this is a specialproperty of a potential. For a general central potential, , thequantized energy levels of a bound-state depend on both and (see Sect. 9.3).
The fact that the energy levels of a hydrogen atom only depend on ,and not on and , implies that the energy spectrum of a hydrogenatom is highly degenerate: i.e., there are many differentstates which possess the same energy. According to the inequality(682) (and the fact that , , and are integers), fora given value of , there are different allowed values of (i.e., ). Likewise, for a given value of ,there are different allowed values of (i.e., ). Now,all states possessing the same value of have the same energy (i.e., they are degenerate). Hence, the total number ofdegenerate states corresponding to a given value of is
Next:Rydberg Formula Up:Central Potentials Previous:Infinite Spherical Potential WellRichard Fitzpatrick2010-07-20
Learning Objectives
By the end of this section, you will be able to:
- Describe the physical meaning of the position-momentum uncertainty relation
- Explain the origins of the uncertainty principle in quantum theory
- Describe the physical meaning of the energy-time uncertainty relation
Heisenberg’s uncertainty principle is a key principle in quantum mechanics. Very roughly, it states that if we know everything about where a particle is located (the uncertainty of position is small), we know nothing about its momentum (the uncertainty of momentum is large), and vice versa. Versions of the uncertainty principle also exist for other quantities as well, such as energy and time. We discuss the momentum-position and energy-time uncertainty principles separately.
Momentum and Position
To illustrate the momentum-position uncertainty principle, consider a free particle that moves along the x-direction. The particle moves with a constant velocity (u) and momentum (p = mu). According to de Broglie’s relations, (p = hbar k) and (E = hbar omega). As discussed in the previous section, the wavefunction for a free particle is given by
[ begin{align*} psi_k(x,t) &= A[cos , (omega t - kx) - i , sin , (omega t - kx)] [4pt] &= A , e^{-i(omega t - kx)} [4pt] &= A , e^{-i(omega t - kx)} [4pt] &=A, e^{-iomega t} e^{ikx} end{align*} ]
and the probability density (|psi_k (x,t)|^2 = A^2) is uniform and independent of time. The particle is equally likely to be found anywhere along the x-axis but has definite values of wavelength and wave number, and therefore momentum. The uncertainty of position is infinite (we are completely uncertain about position) and the uncertainty of the momentum is zero (we are completely certain about momentum). This account of a free particle is consistent with Heisenberg’s uncertainty principle.
Similar statements can be made of localized particles. In quantum theory, a localized particle is modeled by a linear superposition of free-particle (or plane-wave) states called a wave packet. An example of a wave packet is shown in Figure (PageIndex{1}). A wave packet contains many wavelengths and therefore by de Broglie’s relations many momenta—possible in quantum mechanics! This particle also has many values of position, although the particle is confined mostly to the interval (Delta x). The particle can be better localized ((Delta x) can be decreased) if more plane-wave states of different wavelengths or momenta are added together in the right way ((Delta p) is increased). According to Heisenberg, these uncertainties obey the following relation.
Definition: The Heisenberg’s Uncertainty Principle
The product of the uncertainty in position of a particle and the uncertainty in its momentum can never be less than one-half of the reduced Planck constant:
[Delta x Delta p geq dfrac{hbar}{2}. label{Heisen}]
This relation expresses Heisenberg’s uncertainty principle. It places limits on what we can know about a particle from simultaneous measurements of position and momentum. If (Delta x) is large, (Delta p) is small, and vice versa. Equation ref{Heisen} can be derived in a more advanced course in modern physics. Reflecting on this relation in his work The Physical Principles of the Quantum Theory, Heisenberg wrote “Any use of the words ‘position’ and ‘velocity’ with accuracy exceeding that given by [the relation] is just as meaningless as the use of words whose sense is not defined.”
Note that the uncertainty principle has nothing to do with the precision of an experimental apparatus. Even for perfect measuring devices, these uncertainties would remain because they originate in the wave-like nature of matter. The precise value of the product (Delta x Delta p) depends on the specific form of the wavefunction. Interestingly, the Gaussian function (or bell-curve distribution) gives the minimum value of the uncertainty product:
[Delta x Delta p = dfrac{hbar}{2} ]
Example (PageIndex{1}): The Uncertainty Principle Large and Small
Determine the minimum uncertainties in the positions of the following objects if their speeds are known with a precision of (1.0 times 10^{-3} m/s):
- an electron and
- a bowling ball of mass 6.0 kg.
Strategy
Given the uncertainty in speed (Delta u = 1.0 times 10^{-3} m/s), we have to first determine the uncertainty in momentum (Delta p = mDelta u) and then invert Equation ref{Heisen} to find the uncertainty in position
[Delta x = dfrac{hbar}{2Delta p}. nonumber]
Solution
- For the electron: [begin{align*} Delta p &= mDelta u [4pt] &= (9.1 times 10^{-31} kg)(1.0 times 10^{-3}m/s) [4pt] &= 9.1 times 10^{-34} kg cdot m/s,end{align*} ] [begin{align*} Delta x &= frac{hbar}{2Delta p} [4pt] &= 5.8 , cm. end{align*} ]
- For the bowling ball: [begin{align*} Delta p &= mDelta u [4pt] &= (6.0 , kg)(1.0 times 10^{-3}m/s) [4pt] &= 6.0 times 10^{-3} kg cdot m/s, end{align*}] [begin{align*} Delta x &= frac{hbar}{2Delta p} [4pt] &= 8.8 times 10^{-33}m. end{align*} ]
Significance
Unlike the position uncertainty for the electron, the position uncertainty for the bowling ball is immeasurably small. Planck’s constant is very small, so the limitations imposed by the uncertainty principle are not noticeable in macroscopic systems such as a bowling ball.
Example (PageIndex{2}): Uncertainty and the Hydrogen Atom
Hydrogen Atom Levels
Estimate the ground-state energy of a hydrogen atom using Heisenberg’s uncertainty principle. (Hint: According to early experiments, the size of a hydrogen atom is approximately 0.1 nm.)
Strategy
An electron bound to a hydrogen atom can be modeled by a particle bound to a one-dimensional box of length (L = 0.1 , nm). The ground-state wavefunction of this system is a half wave. This is the largest wavelength that can “fit” in the box, so the wavefunction corresponds to the lowest energy state. Note that this function is very similar in shape to a Gaussian (bell curve) function. We can take the average energy of a particle described by this function (E) as a good estimate of the ground state energy ((E_0)). This average energy of a particle is related to its average of the momentum squared, which is related to its momentum uncertainty.
Solution
To solve this problem, we must be specific about what is meant by “uncertainty of position” and “uncertainty of momentum.” We identify the uncertainty of position (Δx) with the standard deviation of position ((σ_x)), and the uncertainty of momentum ((Δp)) with the standard deviation of momentum ((σ_p)). For the Gaussian function, the uncertainty product is
Energy Of Hydrogen Atom
[sigma_xsigma_p = frac{hbar}{2}, nonumber]
where
[sigma_x^2 = x^2 - overline{x}^2 nonumber]
and
[sigma_p^2 = p^2 - overline{p}^2.nonumber]
The particle is equally likely to be moving left as moving right, so (overline{p}^2 = 0). Also, the uncertainty of position is comparable to the size of the box, so (sigma_x = L). The estimated ground state energy is therefore
[begin{align*} E_0 &= E_{Gaussian} [4pt] &= dfrac{overline{p}^2}{m} [4pt] &= frac{sigma_p^2}{2m} [4pt] &= frac{1}{2m} left(frac{hbar}{2sigma_x}right)^2 [4pt] &= frac{1}{2m} left(frac{hbar}{2L}right)^2 [4pt] &= frac{hbar^2}{8mL^2}. end{align*}]
[begin{align*}E_0 &= dfrac{(hbar c)^2}{8(mc^2)L^2} [4pt] &= frac{(197.3 , eV cdot nm)^2}{8(0.511 cdot 10^6 eV)(0.1 , nm)^2} [4pt] &= 0.952 , eV approx 1 , eV. end{align*}]
Multiplying numerator and denominator by (c^2) gives
Significance
Based on early estimates of the size of a hydrogen atom and the uncertainty principle, the ground-state energy of a hydrogen atom is in the eV range. The ionization energy of an electron in the ground-state energy is approximately 10 eV, so this prediction is roughly confirmed. (Note: The product ℏcℏc is often a useful value in performing calculations in quantum mechanics.)
Energy and Time
Another kind of uncertainty principle concerns uncertainties in simultaneous measurements of the energy of a quantum state and its lifetime,
[Delta E Delta t geq frac{hbar}{2} label{H2}]
where (Delta E) is the uncertainty in the energy measurement and (Delta t) is the uncertainty in the lifetime measurement. The energy-timeuncertainty principle does not result from a relation of the type expressed by Equation ref{Heisen} for technical reasons beyond this discussion. Nevertheless, the general meaning of the energy-time principle is that a quantum state that exists for only a short time cannot have a definite energy. The reason is that the frequency of a state is inversely proportional to time and the frequency connects with the energy of the state, so to measure the energy with good precision, the state must be observed for many cycles.
To illustrate, consider the excited states of an atom. The finite lifetimes of these states can be deduced from the shapes of spectral lines observed in atomic emission spectra. Each time an excited state decays, the emitted energy is slightly different and, therefore, the emission line is characterized by a distribution of spectral frequencies (or wavelengths) of the emitted photons. As a result, all spectral lines are characterized by spectral widths. The average energy of the emitted photon corresponds to the theoretical energy of the excited state and gives the spectral location of the peak of the emission line. Short-lived states have broad spectral widths and long-lived states have narrow spectral widths.
Example (PageIndex{3}): Atomic Transitions
An atom typically exists in an excited state for about (Delta t = 10^{-8} s). Estimate the uncertainty (Delta f) in the frequency of emitted photons when an atom makes a transition from an excited state with the simultaneous emission of a photon with an average frequency of (f = 7.1 times 10^{14} Hz). Is the emitted radiation monochromatic?
Energy Of Hydrogen Atom At N 3
Strategy
We invert Equation ref{H2} to obtain the energy uncertainty (Delta E approx hbar /2Delta t) and combine it with the photon energy (E = hf) to obtain (Delta f). To estimate whether or not the emission is monochromatic, we evaluate (Delta f/f).
Solution
The spread in photon energies is (Delta E = h Delta f). Therefore,
[Delta E approx frac{hbar}{2 Delta t} Rightarrow h Delta t approx frac{hbar}{2 Delta t} Rightarrow Delta f approx frac{1}{4pi Delta t} = frac{1}{4pi (10^{-8}s)} = 8.0 times 10^6 , Hz,]
[frac{Delta f}{f} = frac{8.0 times 10^6 , Hz}{7.1 times 10^{14} , Hz} = 1.1 times 10^{-8}.]
Significance
Because the emitted photons have their frequencies within (1.1 times 10^{-6}) percent of the average frequency, the emitted radiation can be considered monochromatic.
Energy Of Hydrogen Atom In Its Ground State
Exercise (PageIndex{1})
A sodium atom makes a transition from the first excited state to the ground state, emitting a 589.0-nm photon with energy 2.105 eV. If the lifetime of this excited state is (1.6 times 10^{-8} s), what is the uncertainty in energy of this excited state? What is the width of the corresponding spectral line?
(4.1 times 10^{-8} eV); (1.1 times 10^{-5} nm)
Contributors and Attributions
Energy Of Hydrogen Atom In First Excited State
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).