Physics 1C
Modern Physics Lecture II
"Splitting the atom is like trying to shoot a gnat in
the Albert Hall at night and using ten million
rounds of ammunition on the off chance of getting it.
That should convince you that the atom will always
be a sink of energy and never a reservoir of energy."
--Ernest Rutherford
Bohr Atom
In 1913, Neils Bohr explained atomic spectra by
utilizing Rutherford’s Planetary model and
quantization.
In Bohr’s postulate for
the hydrogen atom,
the electron moves in
circular orbits around
the proton.
The Coulomb force
provides the
centripetal
acceleration for
continued motion.
Bohr Atom
But the electron could only have certain orbits
that are stable.
In these orbits the
atom does not emit
energy in the form of
electromagnetic
radiation.
Radiation is only
emitted by the atom
when the electron
“jumps” between
stable orbits.
Bohr Atom
The electron will move from a more energetic
initial state to less energetic final state.
The frequency of the
light emitted in the
“jump” is related to
the change in the
atom’s energy.
E i − E f = hf
If the electron is not
“jumping” between
allowed orbitals, then
the energy of the
atom remains constant.
€
Bohr Atom
Bohr then turned to conservation of energy of
the atom in order to determine the allowed
electron orbitals.
The total energy of the atom will be:
2
E tot = KE + PE elec
1
e
2
= me v − ke
2
r
But the electron is undergoing centripetal
acceleration:
2
Fcent
2
e
v
= k e 2 = me = macent
r
r
2
e
2
me v = ke
r
Standing Waves
Recall that the wavelength of an electron is
given by the deBroglie formula:
h
h
λ= =
p mv
As the electron orbits the
hydrogen atom and creates
a stable€orbit, a standing
wave must form which
restricts the
circumference of the orbit
to an integer number of
deBroglie wavelengths.
Bohr Atom
Solving the previous two equations for
wavelength gives us:
we can define ħ to be h/2π.
nh
v=
me r
This gives us a velocity of:
Substituting into the last equation from two
slides before:
2
e
me v = ke
r
2
2
2
⎛
⎞
nh
e
€m
⎟ = k e
e ⎜
r
⎝ me r ⎠
Bohr Atom
Solving for radius gives us:
2 2
n h
rn =
2
me ke e
The integer values of n give you the quantized
Bohr orbits:
Electrons can only exist in certain allowed orbits
determined by €
the integer n.
When n = 1, the orbit has the smallest radius,
called the Bohr radius, ao.
ao = 0.0529nm
€
Bohr Atom
Recall, from centripetal acceleration we found
that:
2
e
2
me v = ke
r
Putting this back into energy we get:
2 ⎞
2
⎛
1
e
1
e
e
2
=€ me v − k e = ⎜ k e ⎟ − ke
2
r 2 ⎝ r ⎠
r
2
E tot
E tot
2 ⎞
⎛
1
e
= − ⎜ ke ⎟
2 ⎝ r ⎠
But we can go back to the result for the
radius (rn = n2ao) to get a numerical result.
Bohr Atom
E tot
2 ⎞
⎛
1
e
−13.6 eV
= − ⎜ ke 2 ⎟ =
2
2 ⎝ n ao ⎠
n
This is energy of any orbit. Please note the
negative sign in the equation.
€ When n = 1, the total energy is –13.6eV.
This is the lowest energy state and is called the
ground state.
The ionization energy is the energy needed to
completely remove the electron from the atom.
The ionization energy for hydrogen is 13.6eV.
Bohr Atom
So, a general
expression for the
radius of any orbit in
a hydrogen atom is:
rn = n2ao
The energy of any
orbit is:
If you would like to
completely remove
the electron from
the atom it requires
13.6eV of energy.
Bohr Atom
What are the first four energy levels for the
hydrogen atom?
When n = 1 => E1 = –13.6eV.
When n = 2 => E2 = -13.6eV/22 = -3.40eV.
When n = 3 => E3 = -13.6eV/32 = -1.51eV.
When n = 4 => E4 = -13.6eV/42 = -0.850eV.
Note that the energy levels get closer together
as n increases (similar to how the wavelengths got
closer in atomic spectra).
When the atom releases a photon it will move
from a final higher energy level (ni) to a final
lower energy level (nf).
Bohr Atom
The energies can be
compiled in an energy
level diagram.
As the atom is in a
higher energy state and
moves to a lower energy
state it will release
energy (in the form of a
photon).
The wavelength of this
photon will be determined
by the starting and
ending energy levels.
Bohr Atom
The photon will have a
wavelength λ and a
frequency f:
Ei − E f
f =
h
To find the wavelengths
between any energy
jump (nf and ni) we can
generalize the equation
discovered by Balmer to:
⎛ 1
1
1 ⎞
= RH ⎜⎜ 2 − 2 ⎟⎟
λ
⎝ nf ni ⎠
Bohr Atom
The wavelength will be
represented by a different
series depending on your
final energy level (nf).
For nf = 1 it is called the
Lyman series (ni = 2,3,4...).
For nf = 2 it is called the
Balmer series (ni = 3,4,5...).
For nf = 3 it is called the
Paschen series (ni = 4,5,6...).
Atomic Spectra
Example
What are the first four wavelengths for the
Lyman, Balmer, and Paschen series for the
hydrogen atom?
Answer
The final energy level for either series will
be nf = 1 (Lyman), nf = 2 (Balmer), and
nf = 3 (Paschen).
Atomic Spectra
Answer
Turn to the generalized Balmer equation:
⎛ 1
1
1 ⎞
= RH ⎜⎜ 2 − 2 ⎟⎟
λ
⎝ nf ni ⎠
For the Lyman series we have:
⎛ 1 1 ⎞
⎛ n i2 1 ⎞
1
= R H ⎜ − 2 ⎟ = R H ⎜ 2 − 2 ⎟
λ
⎝ 1 n i ⎠
⎝ n i n i ⎠
2
⎛
1
n i −1⎞
= R H ⎜ 2 ⎟
λ€
⎝ n i ⎠
⎛ n i2 ⎞
1 ⎛ n i2 ⎞
1
λn =
⎜ 2 ⎟ =
⎟
7
-1 ⎜ 2
R H ⎝ n i −1⎠ 1.097 ×10 m ⎝ n i −1⎠
Atomic Spectra
Answer
Finally for the Lyman series:
⎛ 2 2 ⎞
1
λ1 =
⎟ = 121 nm
7
-1 ⎜ 2
1.097 ×10 m ⎝ 2 −1⎠
€
€
€
⎛ 32 ⎞
1
λ2 =
⎟ = 103 nm
7
-1 ⎜ 2
1.097 ×10 m ⎝ 3 −1⎠
⎛ 4 2 ⎞
1
λ3 =
⎟ = 97.2 nm
7
-1 ⎜ 2
1.097 ×10 m ⎝ 4 −1⎠
⎛ 5 2 ⎞
1
λ4 =
⎟ = 95.0 nm
7
-1 ⎜ 2
1.097 ×10 m ⎝ 5 −1⎠
For the Balmer series we have from before:
λ1 = 656nm, λ2 = 486nm, λ3 = 434nm, λ4 = 410nm.
€
Atomic Spectra
Answer
For the Paschen series we have:
⎛ 1 1 ⎞
⎛ n i2
1
9 ⎞
= R H ⎜ − 2 ⎟ = R H ⎜ 2 − 2 ⎟
λ
⎝ 9 n i ⎠
⎝ 9n i 9n i ⎠
€
⎛ 9n i2 ⎞
1
λn =
⎟
7
-1 ⎜ 2
1.097 ×10 m ⎝ n i − 9 ⎠
2 ⎞
2 ⎞
⎛
⎛
9
4
9
5
( ) ⎟ = 1880 nm λ =
( ) ⎟ = 1280 nm
1
1
⎜
⎜
λ1 =
2
1.097 ×10 7 m-1 ⎜⎝ 4 2 − 9 ⎟⎠
1.097 ×10 7 m-1 ⎜⎝ 5 2 − 9 ⎟⎠
€
⎛ 9(6 2 ) ⎞
1
⎜
⎟ = 1090 nm
λ3 =
7
-1 ⎜ 2
1.097 ×10 m ⎝ 6 − 9 ⎟⎠
€
⎛ 9( 7 2 ) ⎞
1
⎜
⎟ = 1010 nm
λ4 =
7
-1 ⎜ 2
1.097 ×10 m ⎝ 7 − 9 ⎟⎠
Atomic Spectra
So, the only series that lies in the visible range
is the Balmer series.
The Lyman series lies in the ultraviolet range and
the Paschen series lies in the infrared range.
We can extend the Bohr hydrogen atom to fully
describe atoms that are “close” to hydrogen.
These “hydrogen-like” atoms are those that only
contain one electron.
In those cases, when you have Z as the atomic
number of the element (number of protons in the
atom), you substitute Ze2 into where e2 was in
the equations.
Quantum Numbers
n is considered the principle quantum number. n
can range from 1 to infinity in integer steps.
But other quantum numbers were added later on
to explain other subatomic effects (such as
elliptical orbits).