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CHEM-E4110
Quantum mechanics and Spectroscopy
Exercise I, Solutions
Exercise 1 – The lifespan of a classical atom
This exercise can be done using pen and paper.
Lets assume that an electron is a point-wise particle, orbiting the nucleus in a
spherical trajectory with radius a0 = 0.529 Å. In a spherical orbit the electron
2
experiences only the centripetal acceleration, ~a = ~vr . A charged particle in an
accelerated motion emits energy according to the Larmor equation:
e2 |~a|2
dE
=−
dt
6π0 c3
where ~a is the acceleration of the particle. How long does it take for a classical,
point-wise electron to collapse into the nucleus?
Hint: The acceleration is caused by the electrostatic interaction between the
electron and the nucleus. Write the total energy as the sum of kinetic and
potential energy:
e2
1
2
E = m|v| −
2
4π0 r
Find a way to express the total energy as a function of the radius r. After this
you can use the Larmor equation and the chain rule
df (x)
df (x) dx
=
dt
dx dt
and integrate the radius from a0 → 0 and time t : 0 → tend .
Solution:
In the Bohr model for hydrogen atom, an electron is considered to be a
classical particle that can orbit the nucleus only in some allowed trajectories.
The term classical means that it has well-defined position and momentum
at all times (this is not case in quantum mechanics). The total energy of
1
the electron depends on the potential and kinetic energy. Potential energy
is determined by the Coulomb potential (electro-static interaction between
the electron and the nucleus). Kinetic energy depends on the velocity of the
electron.
e2
1
E = m|v|2 −
2
4π0 r
The kinetic energy can be expressed using the centripetal acceleration:
|F~ |
e2
v2
=
=
r
m
4π0 mr2
e2
⇒ v2 =
4π0 mr
|~a| =
where the force F~ can be obtained by differentiating the potential with respect
to r:
dV (r)
d e2
=
dr
dr 4π0 r
e2
=
4π0 r2
Now the total energy can be written as a function of r:
|F~ | =
1
e2
e2
E= m
−
2 4π0 mr 4π0 r
e2
=−
8π0 r
Using the chain rule the change in total energy can be expressed as a time
derivative:
dE
dE dr
e2 dr
e2 |~a|2
=
=
=
−
dt
dr dt
8π0 r2 dt
6π0 c3
e2
3c3 dr
,
|~
a
|
=
dt = −
4|~a|2 r2
4π0 mr2
12π 2 20 m2 c3 2
dt = −
r dr
e4
The lifespan of the classical atom can be calculated when the above differential
is integrated from t = 0 to tend when the electron collapses into the nucleus.
During this time the radius changes from r = a0 , when t = 0 to r = 0 when
t = tend .
For the final calculation one requires the natural constants: c = 299 792 458
m s-1 is the speed of light, 0 = 8.854 · 10-12 C2 N-1 m-2 is vacuum permittivity,
e = 1.602 · 10-19 C is an elementary charge and m = 9.109 · 10-31 kg is the mass
of the electron.
2
t
0
12π 2 20 m2 c3 2
r dr
e4
aH
0
4π 2 20 m2 c3 a3H
∆t =
≈ 15.6 ps
e4
Thus, according to classical physics, the hydrogen atom should collapse in 16
pico seconds. The instability of the classical atom is one of the reasons a more
comprehensive model for our physical reality was needed and what eventually
lead to the development of quantum mechanics.
Z
Z
dt =
−
3