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=Ei−Ef.ΔE=hf=Ei−Ef. size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}
(7)Here, ΔEΔE size 12{ΔE} {} is the change in energy between the initial and final orbits, and hfhf 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 Figure 4.)
Figure 5 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.
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 LL size 12{L} {} of an electron in its orbit is quantized, that is, it has only specific, discrete values. The value for LL size 12{L} {} is given by the formula
L=mevrn=nh2πn=1, 2, 3,…,L=mevrn=nh2πn=1, 2, 3,…, size 12{ left (n=1,2,3, dotslow right )} {}
(8)where LL is the angular momentum, meme is the electron’s mass, rnrn is the radius of the nn th orbit, and hh is Planck’s constant. Note that angular momentum is L=IωL=Iω. For a small object at a radius r, I=mr2r, I=mr2 and ω=v/rω=v/r, so that L=mr2v/r=mvrL=mr2v/r=mvr. Quantization says that this value of mvrmvr can only be equal to h/2,2h/2,3h/2h/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 ZZ size 12{Z} {} protons (Z=1Z=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/rnmev2/rn size 12{m rSub { size 8{e} } v rSup { size 8{2} } /r rSub { size 8{n} } } {}, while the Coulomb force is kZqeqe/rn2kZqeqe/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).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} } } } } {}
(9)Angular momentum quantization is stated in an earlier equation. We solve that equation for vv 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 n=1,2,3,…,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 )} {}
(10)where aBaB size 12{a rSub { size 8{B} } } {} is defined to be the Bohr radius, since for the lowest orbit n=1n=1 size 12{ left (n=1 right )} {} and for hydrogen Z=1Z=1 size 12{ left (Z=1 right )} {}, r1=aBr1=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=h24π2mekqe2=0.529×10−10 m.aB=h24π2mekqe2=0.529×10−10 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"} {}
(11)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 n2n2 size 12{n rSup { size 8{2} } } {}, as illustrated in Figure 6.
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.En= KE + PE. size 12{E rSub { size 8{n} } =" KE "+" PE"} {}
(12)Kinetic energy is the familiar KE=1/2mev2KE=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=qeVPE=qeV size 12{ ital "PE"=q rSub { size 8{e} } V} {}, where VV size 12{V} {} is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge ZqeZqe size 12{ ital "Zq" rSub { size 8{e} } } {} ; thus,
V
=
kZq
e
/
r
n
V
=
kZq
e
/
r
n
, recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that PE=−kZqe/rnPE=−kZqe/rn size 12{ ital "PE"= - ital "kZq" rSub { size 8{e/r rSub { size 6{n} } } } } {}. Entering the expressions for KEKE size 12{ ital "KE"} {} and PEPE size 12{ ital "PE"} {}, we find
En=12mev2−kZqe2rn.En=12mev2−kZqe2rn. 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} } } } } {}
(13)Now we substitute rnrn size 12{r rSub { size 8{n} } } {} and vv size 12{v} {} from earlier equations into the above expression for energy. Algebraic manipulation yields
E
n
=
−
Z
2
n
2
E
0
(
n
=
1, 2, 3, ...
)
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} } \( n=1," 2, 3, " "." "." "." \) } {}
(14)for the orbital energies of hydrogen-like atoms. Here, E0E0 size 12{E rSub { size 8{0} } } {} is the ground-state energy n=1n=1 size 12{ left (n=1 right )} {} for hydrogen Z=1Z=1 size 12{ left (Z=1 right )} {} and is given by
E0=2π2qe4mek2h2=13.6 eV.E0=2π2qe4mek2h2=13.6 eV.
(15)Thus, for hydrogen,
En=−13.6 eVn2 (n=1, 2, 3, ...).En=−13.6 eVn2 (n=1, 2, 3, ...).
(16)Figure 7 shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.
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 nn size 12{n} {} approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since rnrn size 12{r rSub { size 8{n} } } {} gets very large for large nn 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=Ei−Ef.ΔE=hf=Ei−Ef. size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}
(17)Substituting En=(–13.6 eV/n2)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 eV1nf2−1ni2.hf=13.6 eV1nf2−1ni2. 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 )} {}
(18)Dividing both sides of this equation by hchc size 12{ ital "hc"} {} gives an expression for 1/λ1/λ size 12{1/λ} {}:
hfhc=fc=1λ=13.6 eVhc1nf2−1ni2.hfhc=fc=1λ=13.6 eVhc1nf2−1ni2. 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 )} {}
(19)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
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
(20)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
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 )} {}
(21)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/n21/n2 size 12{1/n rSup { size 8{2} } } {}, where nn size 12{n} {} is a non-negative integer. A downward transition releases energy, and so nini size 12{n rSub { size 8{i} } } {} must be greater than nfnf 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=1nf=1 size 12{n rSub { size 8{f} } =1} {} — that is, all the transitions end in the ground state (see also Figure 7). For the Balmer series, nf=2nf=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.
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