31 Atomic Physics
Outline
31-1
Early Models of the Atom
31 2
31-2
The Spectrum of Atomic Hydrogen
31-3
Bohr’s Model of the Hydrogen Atom
31-3 Bohr’s Model of the Hydrogen Atom
Bohr′s model of Hydrogen Atom
n = 1, 2, 3, …..
Figure 31-7. The First Three Bohr Orbits
31-3 Bohr’s Model of the Hydrogen Atom
The spectrum of Hydrogen
How to create spectrum?
• Electron jumps from one orbit to the another: it gives up or
absorbs photons (as a result of energy difference)
difference).
• Different orbit has different energy level.
•
The p
photon wavelength
g is determined by
y hf.
Electron Energies include:
• Electric Potential energy.
• Kinetic energy.
31-3 Bohr’s Model of the Hydrogen Atom
Assumptions of the Bohr model:
•
The electron in a hydrogen atom moves in a circular orbit
around the nucleus.
• Only certain orbits are allowed, where the angular momentum in
the nth allowed orbit is
• Electrons in allowed orbits do not radiate photons. Radiation is
emitted when an electron changes from one orbit to another, with
frequency given by the energy difference:
31-3 Bohr’s Model of the Hydrogen Atom
In order for an electron to move in a circle of radius
r at speed v, the electrostatic force must provide the
2
2
mv
e
required centripetal force. Hence: ( = k 2 )
r
2
e
mv 2 = k
rn
r
31 − 3
Adding the angular momentum requirement gives:
According Eq.11 − 12 L = rmv,
mrn vn = nh / 2π ,
nh
vn =
, n = 1, 2,3,...
2π mrn
31 − 4
31-3 Bohr’s Model of the Hydrogen Atom
Solving for rn and vn from these two equations, the
allowed radii are:
h2
2
rn = ( 2
n
)
, n = 1,, 2,3,...
, ,
2
4π mke
k
31 − 5
The speeds at the allowed radii are:
2π ke
vn =
, n = 1, 2,3,...
nh
2
31 − 6
Example
p 1
Find the radius of the Hydrogen at the smallest orbit (Bohr Orbit).
Solution:
h2
2
)
,
rn = ( 2
n
2
4π mke
with n = 1
(6.626 ×10−34 ) 2
2
)
1
=( 2
×
4π (9.109 ×10−31 kg )(8.988 ×109 N .m 2 / C 2 )(1.602 ×10−19 ) 2
= 5.29 × 10−11 m
Examples
Find the speed and kinetic energy of the electron in the Second Bohr orbit
(n=2).
Solution:
2π ke
k
vn =
with
n=2
,
nh
2π (8.99 ×109 N ⋅ m 2 / C 2 )(1.60 ×10−19 ) 2
v2 =
2 × (6.63 ×10−34 J .s )
2
= 1.09 × 10−6 m / s
1) Speed
2π ke
vn =
,
with
n=2
nh
2π (8.99 ×109 N ⋅ m 2 / C 2 )(1.60 ×10−19 ) 2
v2 =
2 × (6.63
(6 63 ×10−34 J .s )
2
= 1.09 × 10−6 m / s
2) Kinetic energy
1 2
K 2 = mv2
2
1
= (9.11
(9 11×10−31 kg
k )(1.09
)(1 09 × 106 m / s ) 2 = 55.4
4 × 10−19 J
2
31-3 Bohr’s Model of the Hydrogen Atom
The total mechanical energy of an electron in a Bohr orbit is the
sum of its kinetic and potential energies.
After some algebraic manipulation, and substituting known
values of constants, we find for hydrogen atom:
1
En = −(13.6 ⋅ eV ) 2 , n = 1, 2,3,...
n
1 eV = 1.60x10-19 Joule
The lowest energy is called the ground state. Most atoms at room
temperature are in
i the
h groundd state.
The Origin of Spectral Series in Hydrogen: Lyman Series
Figure 31-9a
The Origin of Spectral Series in Hydrogen: Balmer Series
Figure 31-9b
The Origin of Spectral Series in Hydrogen: Paschen Series
Figure 31-9c
Active Exam 31-2 Absorbing a photon and what is the
wavelength?
Find the wavelength of a photon that will be absorded, if a electron
i tto raise
is
i iin a h
hydrogen
d
atom
t
ffrom th
the n=3
3 state
t t tto the
th n=5
5 state.
t t
Solution:
1) The energy at n=5 state is
1
1
=
−
(13
(13.6
6
⋅
eV
)
= −0.544
0 544 eV
2
2
n
5
= −0.544 ×1.6 × 10−19 J = −0.87 × 10−20 J
E5 = −(13.6
(13 6 ⋅ eV )
2) Th
The energy att n=3
3 state
t t is
i
E3 == −(13.6 ⋅ eV )
1
= −2.42 × 10−19 J
2
5
3) The energy difference is
ΔE5,3 = 1.55 × 10−19 J
4) The wavelength is
hf = ΔE5,3 = 1.55 × 10−19 J ⇒ f = 2.34 × 1014 Hz
f λ = c ⇒ λ = 1.28 × 10−6 m
Chapter Summary
The total mechanical energy of an electron in a Bohr orbit is
the sum of its kinetic and potential energies. For Hydrogen
atom, the total energy is
1
En = −(13.6
(13 6 ⋅ eV ) 2 , n = 11, 22,3,...
3
n