Bohr’s Theory of the Hydrogen Atom

The great Danish physicist Niels Bohr (1885–1962) made immediate use of Rutherford’s planetary model of the atom. ([link]). Bohr became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on the planetary model of the atom. For decades, many questions had been asked about atomic characteristics. From their sizes to their spectra, much was known about atoms, but little had been explained in terms of the laws of physics. Bohr’s theory explained the atomic spectrum of hydrogen and established new and broadly applicable principles in quantum mechanics.

A photograph of Niels Bohr.

Mysteries of Atomic Spectra

As noted in Quantization of Energy , the energies of some small systems are quantized. Atomic and molecular emission and absorption spectra have been known for over a century to be discrete (or quantized). (See [link].) Maxwell and others had realized that there must be a connection between the spectrum of an atom and its structure, something like the resonant frequencies of musical instruments. But, in spite of years of efforts by many great minds, no one had a workable theory. (It was a running joke that any theory of atomic and molecular spectra could be destroyed by throwing a book of data at it, so complex were the spectra.) Following Einstein’s proposal of photons with quantized energies directly proportional to their wavelengths, it became even more evident that electrons in atoms can exist only in discrete orbits.

This figure has two parts. Part a shows a discharge tube at the extreme left. Light from the discharge tube passes through a rectangular slit and a grating, going from left to right. From the grating, light of different colors falls on a photographic film. Part b of the figure shows the emission line spectrum for iron.

In some cases, it had been possible to devise formulas that described the emission spectra. As you might expect, the simplest atom—hydrogen, with its single electron—has a relatively simple spectrum. The hydrogen spectrum had been observed in the infrared (IR), visible, and ultraviolet (UV), and several series of spectral lines had been observed. (See [link].) These series are named after early researchers who studied them in particular depth.

The observed hydrogen-spectrum wavelengths can be calculated using the following formula:

1λ=R1nf21ni2, size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

where λ size 12{λ} {}

is the wavelength of the emitted EM radiation and R size 12{R} {}

is the Rydberg constant, determined by the experiment to be

R = 1 . 097 × 10 7 / m ( or m 1 ) . size 12{R=1 "." "097" times "10" rSup { size 8{7} } `/"m " \( "or m" rSup { size 8{ - 1} } \) } {}

The constant nf

is a positive integer associated with a specific series. For the Lyman series, nf=1

; for the Balmer series, nf=2

; for the Paschen series, nf=3

; and so on. The Lyman series is entirely in the UV, while part of the Balmer series is visible with the remainder UV. The Paschen series and all the rest are entirely IR. There are apparently an unlimited number of series, although they lie progressively farther into the infrared and become difficult to observe as nf

increases. The constant ni

is a positive integer, but it must be greater than nf

. Thus, for the Balmer series, nf=2

and ni=3, 4, 5, 6, ...

. Note that ni size 12{n rSub { size 8{i} } } {}

can approach infinity. While the formula in the wavelengths equation was just a recipe designed to fit data and was not based on physical principles, it did imply a deeper meaning. Balmer first devised the formula for his series alone, and it was later found to describe all the other series by using different values of nf size 12{n rSub { size 8{f} } } {}

. Bohr was the first to comprehend the deeper meaning. Again, we see the interplay between experiment and theory in physics. Experimentally, the spectra were well established, an equation was found to fit the experimental data, but the theoretical foundation was missing.

The figure shows three horizontal lines at small distances from each other. Between the two lower lines, the Lyman series, with four vertical red bands in compact form, is shown. The value of the constant n sub f is 1 and the wavelengths are ninety-one nanometers to one hundred nanometers. The Balmer series is shown to the right side of this series. The value of the constant n sub f is two, and the range of wavelengths is from three hundred sixty five to six hundred fifty six nanometers. At the right side of this, the Paschen series bands are shown. The value of the constant n sub f is three, and the range of the wavelengths is from eight hundred twenty nanometers to one thousand eight hundred and seventy five nanometers.

Calculating Wave Interference of a Hydrogen Line

What is the distance between the slits of a grating that produces a first-order maximum for the second Balmer line at an angle of 15º size 12{"15"°} {}

?

Strategy and Concept

For an Integrated Concept problem, we must first identify the physical principles involved. In this example, we need to know (a) the wavelength of light as well as (b) conditions for an interference maximum for the pattern from a double slit. Part (a) deals with a topic of the present chapter, while part (b) considers the wave interference material of Wave Optics.

Solution for (a)

Hydrogen spectrum wavelength. The Balmer series requires that nf=2 size 12{n rSub { size 8{f} } =2} {}

. The first line in the series is taken to be for ni=3 size 12{n rSub { size 8{i} } =3} {}

, and so the second would have ni=4 size 12{n rSub { size 8{i} } =4} {}

.

The calculation is a straightforward application of the wavelength equation. Entering the determined values for nf size 12{n rSub { size 8{f} } } {}

and ni size 12{n rSub { size 8{i} } } {}

yields

1 λ = R 1 n f 2 1 n i 2 = 1.097 × 10 7 m –1 1 2 2 1 4 2 = 2.057 × 10 6 m –1 .

Inverting to find λ size 12{λ} {}

gives

λ = 1 2.057 × 10 6 m –1 = 486 × 10 9 m = 486 nm. alignl { stack { size 12{λ= { {1} over {2 "." "057" times "10" rSup { size 8{6} } " /m"} } ="486" times "10" rSup { size 8{ - 9} } " m"} {} # ="486"`"nm" "." {} } } {}

Discussion for (a)

This is indeed the experimentally observed wavelength, corresponding to the second (blue-green) line in the Balmer series. More impressive is the fact that the same simple recipe predicts all of the hydrogen spectrum lines, including new ones observed in subsequent experiments. What is nature telling us?

Solution for (b)

Double-slit interference (Wave Optics). To obtain constructive interference for a double slit, the path length difference from two slits must be an integral multiple of the wavelength. This condition was expressed by the equation

dsinθ=, size 12{d"sin"θ=mλ} {}

where d size 12{d} {}

is the distance between slits and θ size 12{θ} {}

is the angle from the original direction of the beam. The number m size 12{m} {}

is the order of the interference; m=1 size 12{m=1} {}

in this example. Solving for d size 12{d} {}

and entering known values yields

d=1486 nmsin 15º=1.88×106 m. size 12{d= { { left (1 right ) left ("486"" nm" right )} over {"sin""15"°} } =1 "." "88" times "10" rSup { size 8{ - 6} } " m"} {}

Discussion for (b)

This number is similar to those used in the interference examples of Introduction to Quantum Physics (and is close to the spacing between slits in commonly used diffraction glasses).

Bohr’s Solution for Hydrogen

Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new proposals. His first proposal is that only certain orbits are allowed: we say that the orbits of electrons in atoms are quantized. Each orbit has a different energy, and electrons can move to a higher orbit by absorbing energy and drop to a lower orbit by emitting energy. If the orbits are quantized, the amount of energy absorbed or emitted is also quantized, producing discrete spectra. Photon absorption and emission are among the primary methods of transferring energy into and out of atoms. The energies of the photons are quantized, and their energy is explained as being equal to the change in energy of the electron when it moves from one orbit to another. In equation form, this is

ΔE=hf=EiEf. size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}

Here, ΔE size 12{ΔE} {}

is the change in energy between the initial and final orbits, and hf size 12{ ital "hf"} {}

is the energy of the absorbed or emitted photon. It is quite logical (that is, expected from our everyday experience) that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets, which can have any orbit given the proper energy. (See [link].)

The orbits of Bohr’s planetary model of an atom; five concentric circles are shown. The radii of the circles increase from innermost to outermost circles. On the circles, labels E sub one, E sub two, up to E sub i are marked.

[link] shows an energy-level diagram, a convenient way to display energy states. In the present discussion, we take these to be the allowed energy levels of the electron. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. Given the energies of the lines in an atomic spectrum, it is possible (although sometimes very difficult) to determine the energy levels of an atom. Energy-level diagrams are used for many systems, including molecules and nuclei. A theory of the atom or any other system must predict its energies based on the physics of the system.

The energy level diagram is shown. A number of horizontal lines are shown. The lines are labeled from bottom to top as n is equal to one, n is equal to two and so on up to n equals infinity; the energy levels increase from bottom to top. The distance between the lines decreases from the bottom line to the top line. A vertical arrow shows an electron transitioning from n equals four to n equals two.

Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. This was an important first step that has been improved upon, but it is well worth repeating here, because it does correctly describe many characteristics of hydrogen. Assuming circular orbits, Bohr proposed that the **angular momentum L size 12{L} {}

of an electron in its orbit is quantized**, that is, it has only specific, discrete values. The value for L size 12{L} {}

is given by the formula

L=mevrn=nh2πn=1, 2, 3,, size 12{ left (n=1,2,3, dotslow right )} {}

where L

is the angular momentum, me

is the electron’s mass, rn

is the radius of the n

th orbit, and h

is Planck’s constant. Note that angular momentum is L=

. For a small object at a radius r, I=mr2

and ω=v/r

, so that L=mr2v/r=mvr

. Quantization says that this value of mvr

can only be equal to h/2,2h/2,3h/2 size 12{h/2,` 2h/2, `3h/2} {}

, etc. At the time, Bohr himself did not know why angular momentum should be quantized, but using this assumption he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

From Bohr’s assumptions, we will now derive a number of important properties of the hydrogen atom from the classical physics we have covered in the text. We start by noting the centripetal force causing the electron to follow a circular path is supplied by the Coulomb force. To be more general, we note that this analysis is valid for any single-electron atom. So, if a nucleus has Z size 12{Z} {}

protons (Z=1 size 12{Z=1} {}

for hydrogen, 2 for helium, etc.) and only one electron, that atom is called a hydrogen-like atom. The spectra of hydrogen-like ions are similar to hydrogen, but shifted to higher energy by the greater attractive force between the electron and nucleus. The magnitude of the centripetal force is mev2/rn size 12{m rSub { size 8{e} } v rSup { size 8{2} } /r rSub { size 8{n} } } {}

, while the Coulomb force is kZqeqe/rn2 size 12{k left ( ital "Zq" rSub { size 8{e} } right ) left (q rSub { size 8{e} } right )/r rSub { size 8{n} } rSup { size 8{2} } } {}

. The tacit assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. This is consistent with the planetary model of the atom. Equating these,

kZqe2rn2=mev2rn(Coulomb = centripetal). size 12{k { { ital "Zq" rSub { size 8{e} } rSup { size 8{2} } } over {r rSub { size 8{n} } rSup { size 8{2} } } } = { {m rSub { size 8{e} } v rSup { size 8{2} } } over {r rSub { size 8{n} } } } } {}

Angular momentum quantization is stated in an earlier equation. We solve that equation for v size 12{v} {}

, substitute it into the above, and rearrange the expression to obtain the radius of the orbit. This yields:

rn=n2ZaB, for allowed orbits size 12{r rSub { size 8{n} } = { {n rSup { size 8{2} } } over {Z} } a rSub { size 8{B} } } {}n=1,2,3,, size 12{ left (n=1, 2, 3, dotslow right )} {}

where aB size 12{a rSub { size 8{B} } } {}

is defined to be the Bohr radius, since for the lowest orbit n=1 size 12{ left (n=1 right )} {}

and for hydrogen Z=1 size 12{ left (Z=1 right )} {}

, r1=aB size 12{r rSub { size 8{1} } =a rSub { size 8{B} } } {}

. It is left for this chapter’s Problems and Exercises to show that the Bohr radius is

aB=h22mekqe2=0.529×1010 m. size 12{a rSub { size 8{B} } = { {h rSup { size 8{2} } } over {4π rSup { size 8{2} } m rSub { size 8{e} } ital "kq" rSub { size 8{e} } rSup { size 8{2} } } } =0 "." "529" times "10" rSup { size 8{ - "10"} } " m"} {}

These last two equations can be used to calculate the radii of the allowed (quantized) electron orbits in any hydrogen-like atom. It is impressive that the formula gives the correct size of hydrogen, which is measured experimentally to be very close to the Bohr radius. The earlier equation also tells us that the orbital radius is proportional to n2 size 12{n rSup { size 8{2} } } {}

, as illustrated in [link].

The electron orbits are shown in the form of four concentric circles. The radius of each circle is marked as r sub one, r sub two, up to r sub four.

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy:

En= KE + PE. size 12{E rSub { size 8{n} } =" KE "+" PE"} {}

Kinetic energy is the familiar KE=1/2mev2 size 12{ ital "KE"= left (1/2 right )m rSub { size 8{e} } v rSup { size 8{2} } } {}

, assuming the electron is not moving at relativistic speeds. Potential energy for the electron is electrical, or PE=qeV size 12{ ital "PE"=q rSub { size 8{e} } V} {}

, where *V size 12{V} {}

, recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that *PE=kZqe/rn size 12{ ital "PE"= - ital "kZq" rSub { size 8{e/r rSub { size 6{n} } } } } {}

*. Entering the expressions for KE size 12{ ital "KE"} {}

and PE size 12{ ital "PE"} {}

, we find

En=12mev2kZqe2rn. size 12{E rSub { size 8{n} } = { {1} over {2} } m rSub { size 8{e} } v rSup { size 8{2} } - k { { ital "Zq" rSub { size 8{e} } rSup { size 8{2} } } over {r rSub { size 8{n} } } } } {}

Now we substitute rn size 12{r rSub { size 8{n} } } {}

and *v size 12{v} {}

E n = Z 2 n 2 E 0 ( n = 1, 2, 3, ... ) size 12{E rSub { size 8{n} } = - { {Z rSup { size 8{2} } } over {n rSup { size 8{2} } } } E rSub { size 8{0} } \( n=1," 2, 3, " "." "." "." \) } {}

for the orbital energies of hydrogen-like atoms. Here, *E0 size 12{E rSub { size 8{0} } } {}

for hydrogen Z=1 size 12{ left (Z=1 right )} {}

and is given by

E0=2π2qe4mek2h2=13.6 eV.

Thus, for hydrogen,

En=13.6 eVn2 (n=1, 2, 3, ...).

[link] shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

An energy level diagram is shown. At the left, there is a vertical arrow showing the energy levels increasing from bottom to top. At the bottom, there is a horizontal line showing the energy levels of Lyman series, n is one. The energy is marked as negative thirteen point six electron volt. Then, in the upper half of the figure, another horizontal line showing Balmer series is shown when the value of n is two. The energy level is labeled as negative three point four zero electron volt. Above it there is another horizontal line showing Paschen series. The energy level is marked as negative one point five one electron volt. Above this line, some more lines are shown in a small area to show energy levels of other values of n.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As n size 12{n} {}

approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since rn size 12{r rSub { size 8{n} } } {}

gets very large for large n size 12{n} {}

, and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

ΔE=hf=EiEf. size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}

Substituting En=(13.6 eV/n2) size 12{E rSub { size 8{n} } = - "13" "." 6``"eV"/n rSup { size 8{2} } } {}

, we see that

hf=13.6 eV1nf21ni2. size 12{ ital "hf"= left ("13" "." 6" eV" right ) left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

Dividing both sides of this equation by hc size 12{ ital "hc"} {}

gives an expression for 1/λ size 12{1/λ} {}

:

hfhc=fc=1λ=13.6 eVhc1nf21ni2. size 12{ { { ital "hf"} over { ital "hc"} } = { {f} over {c} } = { {1} over {λ} } = { { left ("13" "." 6" eV" right )} over { ital "hc"} } left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

It can be shown that

13.6 eV hc = 13.6 eV 1.602 × 10 −19 J/eV 6.626 × 10 −34 J·s 2.998 × 10 8 m/s = 1.097 × 10 7 m –1 = R

is the Rydberg constant. Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

1 λ = R 1 n f 2 1 n i 2 size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to 1/n2 size 12{1/n rSup { size 8{2} } } {}

, where *n size 12{n} {}

must be greater than nf size 12{n rSub { size 8{f} } } {}

. The various series are those where the transitions end on a certain level. For the Lyman series, nf=1 size 12{n rSub { size 8{f} } =1} {}

— that is, all the transitions end in the ground state (see also [link]). For the Balmer series, nf=2 size 12{n rSub { size 8{f} } =2} {}

, or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and Limits of the Bohr Theory

Bohr did what no one had been able to do before. Not only did he explain the spectrum of hydrogen, he correctly calculated the size of the atom from basic physics. Some of his ideas are broadly applicable. Electron orbital energies are quantized in all atoms and molecules. Angular momentum is quantized. The electrons do not spiral into the nucleus, as expected classically (accelerated charges radiate, so that the electron orbits classically would decay quickly, and the electrons would sit on the nucleus—matter would collapse). These are major triumphs.

But there are limits to Bohr’s theory. It cannot be applied to multielectron atoms, even one as simple as a two-electron helium atom. Bohr’s model is what we call semiclassical. The orbits are quantized (nonclassical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there are clouds of probability. Bohr’s theory also did not explain that some spectral lines are doublets (split into two) when examined closely. We shall examine many of these aspects of quantum mechanics in more detail, but it should be kept in mind that Bohr did not fail. Rather, he made very important steps along the path to greater knowledge and laid the foundation for all of atomic physics that has since evolved.

PhET Explorations: Models of the Hydrogen Atom

How did scientists figure out the structure of atoms without looking at them? Try out different models by shooting light at the atom. Check how the prediction of the model matches the experimental results.* * *


Section Summary

Conceptual Questions

How do the allowed orbits for electrons in atoms differ from the allowed orbits for planets around the sun? Explain how the correspondence principle applies here.

Explain how Bohr’s rule for the quantization of electron orbital angular momentum differs from the actual rule.

What is a hydrogen-like atom, and how are the energies and radii of its electron orbits related to those in hydrogen?

Problems & Exercises

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

1λ=R1nf21ni2λ=1R(ninf)2ni2nf2;ni=2,nf=1, size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right ) drarrow λ= { {1} over {R} } left [ { { \( n rSub { size 8{i} } cdot n rSub { size 8{f} } \) rSup { size 8{2} } } over {n rSub { size 8{i} } rSup { size 8{2} } - n rSub { size 8{f} } rSup { size 8{2} } } } right ]; n rSub { size 8{i} } =2" , "n rSub { size 8{f} } =1" ,"} {}

so that

λ = m 1.097 × 10 7 ( 2 × 1 ) 2 2 2 1 2 =1.22×107 m=122 nm size 12{λ= left ( { {m} over {1 "." "097" times "10" rSup { size 8{7} } } } right ) left [ { { \( 2 "." 1 \) rSup { size 8{2} } } over {4 - 1} } right ]=1 "." "22" times "10" rSup { size 8{ - 7} } " m"="122"" nm"} {}

, which is UV radiation.

Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.

Look up the values of the quantities in aB=h24π2mekqe2

, and verify that the Bohr radius aB

is 0.529×1010 m size 12{0 "." "529" times "10" rSup { size 8{ - "10"} } " m"} {}

.

a B = h 2 4π 2 m e kZq e 2 = ( 6.626 × 10 34 J·s ) 2 4π 2 ( 9.109 × 10 31 kg ) ( 8.988 × 10 9 N · m 2 / C 2 ) ( 1 ) ( 1.602 × 10 19 C ) 2 = 0.529 × 10 10 m

Verify that the ground state energy E0 size 12{E rSub { size 8{0} } } {}

is 13.6 eV by using E0=2π2qe4mek2h2. size 12{E rSub { size 8{o} } = { {2π rSup { size 8{2} } q rSub { size 8{e} } rSup { size 8{4} } m rSub { size 8{e} } k rSup { size 8{2} } } over {h rSup { size 8{2} } } } "." } {}

If a hydrogen atom has its electron in the n=4 size 12{n=4} {}

state, how much energy in eV is needed to ionize it?

0.850 eV

A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is *n size 12{n} {}

Find the radius of a hydrogen atom in the n=2 size 12{n=2} {}

state according to Bohr’s theory.

2.12 × 10 –10 m size 12{2 "." "12" times "10" rSup { size 8{"-10"} } " m"} {}

Show that 13.6 eV/hc=1.097×107 m=R

(Rydberg’s constant), as discussed in the text.

What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?

365 nm

It is in the ultraviolet.

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.

Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

No overlap

365 nm

122 nm

(a) Which line in the Balmer series is the first one in the UV part of the spectrum?

(b) How many Balmer series lines are in the visible part of the spectrum?

(c) How many are in the UV?

A wavelength of 4.653 μm size 12{4 "." "653 μm"} {}

is observed in a hydrogen spectrum for a transition that ends in the nf=5 size 12{n rSub { size 8{f} } =5} {}

level. What was ni size 12{n rSub { size 8{i} } } {}

for the initial level of the electron?

7

A singly ionized helium ion has only one electron and is denoted He+ size 12{"He" rSup { size 8{+{}} } } {}

. What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

A beryllium ion with a single electron (denoted Be3+ size 12{"Be" rSup { size 8{3+{}} } } {}

) is in an excited state with radius the same as that of the ground state of hydrogen.

(a) What is *n size 12{n} {}

ion?

(b) How much energy in eV is needed to ionize the ion from this excited state?

(a) 2

(b) 54.4 eV

Atoms can be ionized by thermal collisions, such as at the high temperatures found in the solar corona. One such ion is C+5 size 12{C rSup { size 8{+5} } } {}

, a carbon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

(b) What is the wavelength of the first line in this ion’s Paschen series?

(c) What type of EM radiation is this?

Verify Equations rn=n2ZaB

and aB=h24π2mekqe2=0.529×1010 m

using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.

kZqe2rn2=meV2rn ,

so that rn=kZqe2meV2=kZqe2me1V2.

From the equation mevrn=nh2π ,

we can substitute for the velocity, giving: rn=kZqe2me4π2me2rn2n2h2

so that rn=n2Zh24π2mekqe2=n2ZaB ,

where aB=h22mekqe2

.

The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by 1λ=R1nf21ni2 size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

? It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

Glossary

hydrogen spectrum wavelengths
the wavelengths of visible light from hydrogen; can be calculated by 1λ=R1nf21ni2 size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}
Rydberg constant
a physical constant related to the atomic spectra with an established value of 1.097 × 10 7 m −1
double-slit interference
an experiment in which waves or particles from a single source impinge upon two slits so that the resulting interference pattern may be observed
energy-level diagram
a diagram used to analyze the energy level of electrons in the orbits of an atom
Bohr radius
the mean radius of the orbit of an electron around the nucleus of a hydrogen atom in its ground state
hydrogen-like atom
any atom with only a single electron
energies of hydrogen-like atoms
Bohr formula for energies of electron states in hydrogen-like atoms: E n = Z 2 n 2 E 0 ( n = 1, 2, 3, )

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