Download Bohr`s atomic model

Bohr’s atomic model
Bohr tried to modify Rutherford’s planetary model in such a way that it could
fit the experimentally observed spectrum of hydrogen, described by Rydberg’s
empirical formula.
1. In Rutherford’s model, electrons move around the nucleus in circular orbits, like planets around the Sun. There are no restrictions on the radii,
r, of these orbits. Since the energy of an electron depends on the radius
of its orbit, this means that all energies are allowed.
Note: Why does the energy of an electron depend on the radius of its
orbit?
• The potential energy of the electron is the Coulomb energy of its
interaction with the nucleus. This is inversely proportional to the
distance between the electron and the nucleus (the radius of the orbit,
r):
Ze2
,
(1)
U =−
4πε0 r
where Z is the atomic number and e is the elementary charge (the
absolute value of the electron charge) – thus, Ze is the charge of the
nucleus (for hydrogen, Z = 1, so Ze is simply e).
• The kinetic energy of the electron is defined by its velocity. From
classical mechanics we know the expression for the magnitude of the
centripetal force (that holds the electron in orbit): me v 2 /r, where me
is the electron mass and v is its velocity. In the hydrogen atom, the
electron interacts with the nucleus by means of Coulomb attraction,
and there are no other forces acting on it. Thus, the centripetal force
is just the Coulomb force between the electron and the nucleus:
me v 2
Ze2
=
.
r
4πε0 r2
(2)
Multiplying the right and left hand side by r/2, we get the expression
for the kinetic energy:
Ze2
me v 2
=
.
2
8πε0 r
T =
(3)
• The total energy of the electron is the sum of the potential (Eq. 1)
and kinetic (Eq. 3) energy:
E =U +T =−
Ze2
Ze2
Ze2
+
=−
,
4πε0 r 8πε0 r
8πε0 r
and we see that this depends on r.
1
(4)
Notice that so far we have treated the electron purely classically. The
radius r could assume any value, and so could the energy E, given by
Eq. 4.
2. To restrict the values that the energy of the electron can assume, Bohr
added the angular momentum quantization condition to Rutherford’s planetary model. This condition is:
me vr = nh̄,
(5)
where h̄ is the reduced Planck’s constant and n can take positive integer
values (1, 2, 3, ...). The allowed values of r are then:
r=
nh̄
.
me v
(6)
These values are discrete.
3. Now it is time to do some math. Squaring the left and right hand side of
Eq. 6, we get:
n2 h̄2
n2 h̄2
r2 = 2 2 =
.
(7)
me v
me (me v 2 )
From Eq. 2 we can express me v 2 as:
me v 2 =
Ze2
.
4πε0 r
(8)
Plugging this last expression into the denominator on the right hand side
of Eq. 7, we get:
n2 h̄2 4πε0 r
r2 =
.
(9)
2
me | Ze
{z }
= m 1v2
e
Dividing the left and right hand side of Eq. 9 by r, we get:
r=
4πε0 h̄2 n2
.
me e2 Z
}
| {z
(10)
=rB
Notice that the first term contains constants only, and is thus itself a
constant, rB , known as the “Bohr radius”. Its value is rB ≈ 0.529 Å (1 Å=
10−10 m).
4. Eq. 10 tells us that the allowed values of the electron orbits in the Bohr
model of the hydrogen atom, r, are:
r = rB
2
n2
.
Z
For hydrogen, the atomic number Z = 1, so the smallest allowed orbit,
corresponding to n = 1, has a radius equal to the Bohr radius, r1 = rB , the
second smallest orbit, corresponding to n = 2, has a radius of r2 = 4rB ,
and so on.
5. From Eq. 4 we see that the energy of an electron is inversely proportional
to the radius of its orbit. Plugging in Eq. 10 into Eq. 4, we find the allowed
energy values:
E=−
m e e4
Z2
Ze2 me e2 Z
Ze2
=
−
=
=−
8πε0 r
8πε0 4πε0 h̄2 n2
8ε20 4π 2 h̄2 n2
| {z }
| {z }
= r1
4
=h2
2
(11)
Z
me e
,
=−
8ε20 h2 n2
| {z }
=const
where we have used the fact that h = 2πh̄, by definition.
6. The energy of the light quantum (photon) absorbed or emitted by an
atom has to be the same as the difference between the energy that the
atom posesses after and before it absorbs or emits the photon. This is
because energy must be conserved, so whatever energy the photon carries
is passed to the atom when the photon is absorbed, or taken away from
the atom when the photon is emitted:
∆E = Efinal − Einitial = hν.
(12)
In Eq. 12, the initial and the final energy are both given by Eq. 11 (with
different values of n), and ν is the frequency of the photon. Using this
relation together with Eq. 11 allows calculating the frequencies at which
the hydrogen atom can absorb and emit light. Since the frequency ν is
related to the wavelength λ by the relation c = λν (where c is the speed
of light), we can also calculate the wavelengths at which hydrogen absorbs
and emits light (that is, derive Rydberg’s formula).
7. You might be wondering why our equations include the atomic number
Z, if we are only talking about hydrogen here. The reason is that this
reasoning also applies to any ion with arbitrary charge of the nucleus, but
only one electron: He+ , Li2 +, etc. For He+ , for instance, Z = 2, so the
radius of the smallest allowed orbit is rB /2 instead of rB , and the energy
levels are adjusted accordingly. Note that the model does not work for
any atom with more than one electron (we will talk about some of the
reasons for this later in this course).
8. So, Bohr’s model describes the energy levels of the hydrogen atom correctly. What is the problem with it, then? The problem is exactly the
same as with Rutherford’s original planetary model.
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• An electron moving in a circular orbit is an accelerated charged particle. Even if the absolute value of the electron’s velocity is constant,
the direction of the velocity is changing all the time. Thus, there
is an acceleration directed towards the center of the electron’s orbit
(the nucleus) that prevents the electron from moving along a straight
line.
• From electrodynamics we know that an accelerated charged particle
must continuously emit electromagnetic waves. Since electromagnetic waves carry energy, the electron would lose energy, it’s orbit
would become smaller and smaller (see Eq. 4), and it would eventually fall onto the nucleus. We know that in reality this does not
happen: atoms remain in stable states.
• Bohr postulated, in an ad hoc manner, that electrons remain in circular orbits and do not fall onto the nucleus, and that they do not
emit energy while in orbit (only when transitions between orbits of
different radii occur). This assumption is inconsistent with the previous point: it violates the laws of classical physics. Thus, Bohr’s
model was useful for calculating the energy levels and the spectrum
of the hydrogen atom, but there was no clear physical reasoning to
support his assumption of angular momentum quantization.
9. “New” quantum theory considers the electron to be a wave. It discards
the notion of orbits and replaces it with the notion of orbitals. Orbitals
are stationary (standing) waves, with no spatial redistribution of charge
over time. Thus, since there is no net movement of charge, there is also no
requirement for electrons to constantly emit electromagnetic waves. The
atom can thus remain in a stable state. Like orbits, orbitals correspond
to specific energy states. We will learn how to calculate these states in
upcoming lectures.
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