Download Kvantfysik Lecture Notes No. 4x

Kvantfysik Lecture Notes No. 4x
In this lecture we consider the semi-classical approximation and apply
it to the hydrogen atom. Here we will find Bohr’s result for the spectrum
of energies of the hydrogen atom. It turns out that this was somewhat
accidental, but it is still considered one of the most important milestones in
the development of quantum mechanics.
1
Quantization of angular momentum (handwaving version)
Recall our result for a particle on a circle with circumference L. The wave
function is single valued on the circle, satisfying ψ(x+L) = ψ(x). This forces
quantization of the momentum p. The corresponding wave-functions are
1
ψ(x) = √ eipx/h̄ ,
L
where single-valuedness sets p = 2π`h̄/L = `h̄/R. ` is any integer and we
have written the last expression in terms of the radius of the circle R.
Now imagine that our circle is in the x − y plane and that positive p
corresponds to a momentum that is in the counter-clockwise direction (see
figure). The particle then has angular momentum pointing in the ẑ direction
given by Lz = pR = `h̄. Hence, the angular momentum is also quantized, and
in fact comes in integer multiples of h̄. We will call ` the angular momentum
quantum number. In a later lecture we will show more precisely how the
quantization of the angular momentum works.
1
2
The classical hydrogen atom
The hydrogen atom consists of an electron with charge q = −e =≈ −1.602 ×
10−19 coulombs bound to a proton with the opposite charge. We can generalize this to a “hydrogenic” atom by replacing the proton with a nucleus of
charge +Ze. The electron mass is me ≈ 9.109 × 10−31 kg, and the proton
mass is mp ≈ 1.67 × 10−27 kg. Since the nuclear mass is at least 2000 times
the mass of the electron, we can assume that the nucleus is fixed and only
consider the electron motion. In the later installment of the notes we will
take the small motion of the nucleus into account and replace the electron
mass me with the reduced mass.
There is an electrostatic force between the electron and the nucleus, whose
potential is
V (r) = −
Ze2
,
4π0 r
where 0 is the vacuum permittivity which has the value
0 ≈ 8.854 × 10−12 A2 s4 kg−1 m−3
≈ 0.05526 eV V−2 nm−1 .
(It is also convenient to use (4π0 )−1 = 9 × 109 ohm-m/sec.) r is the distance
between the electron and the nucleus. Since the potential only depends on
r and not on any angles the angular momentum of the electron around the
nucleus is conserved. We call such a potential a central force potential.
Without any loss of generality, we assume that the electron’s motion is
restricted to the x − y plane. If the electron is bound to the nucleus, then
it will follow an elliptical orbit. We will now give up some generality and
assume that the motion is circular with the radius of the orbit at r. In this
case the energy of the electron is given by
E=
~ ·L
~
L
L2z
Ze2
+
V
(r)
=
−
,
2me r2
2me r2 4π0 r
2
(1)
where Lz is the component of angular momentum in the z direction. For a
given Lz we can find the r that minimizes the energy by taking the derivative
of E in (1) with respect to r and setting it to zero. This gives us
dE
L2
Ze2
=− z3 +
=0
dr
me r
4π0 r2
⇒r=
4π0 L2z
.
me Ze2
(2)
Substituting this expression for r back into (1), we get
E=−
3
Z 2 e4 me
.
2(4π0 )2 L2z
(3)
The Bohr atom
Everything done in the last section is classical. We now make what is known
as the semi-classical approximation, where we substitute the quantized values
for Lz into (3). We then find that the allowed values for E are
E=−
Z 2 e4 me
.
2`2 (4π0 h̄)2
(4)
This is an expression first derived by Bohr — and it is not quite correct.
Notice that ` = 0 is an allowed quantum number, but with this value the
hydrogen atom would have −∞ energy. However, Bohr was somewhat lucky.
He noticed that if he threw out the ` = 0 result then he could reproduce the
energy spectrum completely. Then the lowest allowed energy is
E1 = −
Z 2 e4 me
.
2(4π0 h̄)2
(5)
Using (2) we can see that the radius of the electron for this lowest level in
hydrogen, where Z = 1, is
4π0 h̄2
r=
≡ aµ .
me e2
(6)
The constant aµ is called the Bohr radius.
We can naively understand why we cannot have Bohr’s ` = 0 state. This
would correspond to an electron at rest sitting on top of the proton. If this
were the case, then we would know both the position and momentum of the
electron exactly, violating Heisenberg uncertainty.
It turns out that the ` in (4) should be replaced by a different quantum
number called the principle quantum number, n, where in this case n = |`|+1
3
(We will derive the principle quantum number in the Lecture 11 notes).
Hence, the lowest allowed energy level has ` = 0 and not ` = 1. You can
see that for |`| 1 the difference between using ` and n in (4) is not very
important since their separation is small compared to their values. In general
the semi-classical approximation works best when one is dealing with large
quantum numbers. Writing the energy in terms of the Bohr radius and using
the principle quantum number, the allowed energy levels are
Z 2 e2
En = − 2
2n (4π0 aµ )
4
n ≥ 1.
(7)
Spectra
In the first lecture we pointed out that a classical hydrogenic atom should
radiate away all of its energy as the electron spirals into the nucleus. Quantum mechanically this cannot happen – an electron can only be in one of its
allowed states. Electrons can still radiate away energy, but only by jumping
from one allowed energy level to another. This leads to discrete emission
lines as the electron transfers the energy difference to a photon.
where ν is the frequency and
The photon energy is given by E = hν = hc
λ
λ is the wavelength. Hence if an electron in a hydrogen atom (Z = 1) jumps
from an energy level with principle quantum number n2 to an energy level
with principle quantum number n1 , the energy of the photon is
e2
E = hν =
8π0 aµ
1
1
− 2
2
n1 n2
!
= hcRH
1
1
− 2
2
n1 n2
!
where RH is known as the “Rydberg constant”
RH =
e2
= 1.10 × 107 m−1 ,
8π0 hcaµ
which has units of inverse wavelength. Therefore the wavelength is
λ=
1
n2 n2
c
=
· 21 22.
ν
RH n2 − n1
The spectral lines are grouped into series based on the final energy level.
In the case of hydrogen, the series of spectral lines with n1 = 2 is the “Balmer
series”, while the series of lines with n1 = 1 is the “Lyman series”. The
shortest wavelength in either series is found by taking the limit n2 = ∞. For
the Balmer series this minimum wavelength is
λmin =
4n22
1
4
·
→
≈ 3600 Å .
2
n2 − 4 RH
RH
4
The maximum wavelength for the Balmer series occurs when n2 = 3. Hence,
λmax =
4·9 1
9 4
=
= 6500 Å .
9 − 4 RH
5 RH
Both the minimum and the maximum are in the visible, so the Balmer series
was observed first. The Lyman series is in the ultraviolet, which you can
quickly verify.
5
X-ray emission from atoms.
The Bohr model strictly speaking only works for hydrogen or ions with a
single electron. However, the Bohr model can still lead to very accurate
predictions for certain experiments using multi-electron atoms. In particular,
it leads to beautiful predictions for the x-ray spectra of atoms with Z > 15.
For now we make the following assumptions
1. The electrons are in discrete energy shells labeled by n = 1, 2, 3 . . ..
2. The lowest energy shell at n = 1 holds at most two electrons.
3. The atom can be excited by knocking one of the electrons out of the
n = 1 shell.
When the atom is excited as such, the electrons in the next to lowest
energy shell (n = 2) see one electron inside of their orbit and the nucleus
with charge +Ze. The electrons in the higher energy shells (n > 2) do not
play much of role and we will ignore them. The inner electron acts as a screen
to the nuclear charge and effectively reduces the nuclear charge to +(Z − 1)e.
Therefore, if an electron jumps from the n = 2 shell to the n = 1 shell, we
expect the frequency of the emitted light to be
2
ν = (Z − 1) cRH
1
1
− 2
1 2
3
= (Z − 1)2 cRH
4
Such emissions coming from electrons jumping from the n = 2 to the n = 1
shells are known as Kα emission lines. For example, calcium, with Z = 20,
would have a Kα line with wave-length
λ=
c
4
−1
=
RH
= 3.3 Å .
2
ν
3(19)
Wave-lengths this short are in the x-ray spectrum.
5
(8)
We could also consider transitions from the n = 3 shell to the n = 2 shell,
which can occur if the atom is excited by knocking an electron out of the
n = 2 shell. For an electron in the n = 3 shell, there are many more inner
electrons that screen the nuclear charge. In this case the effective charge seen
by the electron is experimentally measured to be +(Z − 7.4)e, and so the
frequency of the radiation is
2
ν = (Z − 7.4) cRH
1
1
−
22 32
=
5
(Z − 7.4)2 cRH .
36
The emission lines coming from this transition are called Lα lines.
These results were experimentally
verified by Moseley in 1913 by
bombarding many different elements of the periodic table with
electrons to excite the atoms, and
observing the resulting spectra.
What he found in his original experiment is shown in the figure.
Here he√presents the data as Z
versus ν, so that the graphs
are linear. It is interesting to
note that the relation of the periodic table to the atomic number
Z, that is the charge of the nucleus, was unknown before Moseley’s work. In fact, Moseley could
see gaps along the lines at Z = 43,
61 and 75, allowing him to predict
the existence of these elements.
(Note that L and K in the figure refer to the Lα line and the
Kα line. The Kα and Lα are actually part of multiple lines called
the K and the L series. The presence of multiple lines is caused by
the electron’s spin. )
6