Download energy - U of L Class Index

Light as a Particle
The photoelectric effect. In 1888, Heinrich Hertz discovered that electrons could be
ejected from a sample by shining light on it.
Note the effects of changing:
Intensity
- No matter how low the intensity there is still a current??
Frequency - Must have sufficient energy to eject an electron
Threshold Frequency, no
1
Black Body
Radiation
Matter will emitted radiation when heated.
How?
The amount of radiation, and its frequency,
depends on the temperature
The higher the T, the lower the l, i.e. higher n
Classical Physics predicts infinite
energy emission, which violates the law
of conservation of energy.
Plank was able to show that the light intensity
decays exponentially with wavelength
allowing for energy to be conserved, and in
so doing discovered the constant, h.
White Light
Prism:
disperses light into
its components
White light :
Continuous spectrum
Contains all frequencies in equal amounts
Light Emission From Atomic Gas
Atomic emissions
- Not continuous - Why?
- Result from changes in the electron motion
around the nucleus.
- Types of changes in motion are restricted
corresponding to specific frequencies – i.e.
quantized
Emission from the atoms
The electron remains in
a stable trajectory
around the nucleus
i.e its kinetic energy
is in balances with
the electron nuclear
potential energy.
Otherwise electron
will collapse into the
nucleus, losing energy
as radiation
The electron can change
to a lower orbit
A photon is emitted
when the electron
changes from higher
orbit to lower orbit
Energy Levels
0
Energy of free electron
5
Energy
5 4321
-ve
4
3
2
1
- ve energy change
=> more stable than free electron
Frequency
Atomic Spectra
Hydrogen Spectrum : Anders Ångström (1817-1874)
In 1885, Johann Balmer (1825-1898) showed that the
wavelengths of H could be described by:
1/l = (1.0974*107 m-1)*(1/4 -1/n2)
This equation was later generalized by, Johannes Rydberg
(1854-1919) to described all the spectral lines of H as:
Rydberg Equation
1/l = R*|1/n12 -1/n22|
R = 1.0974*107 m-1 = Rydberg Constant
n1 final
n2 initial
Exercise
Calculate the wavelength of a photon emitted when a
hydrogen atom changes to the n = 4 state from the n = 5
state. What type of electromagnetic radiation is this?
1/l = R*|1/n12 -1/n22|
Final n1 = 4; initial n2 = 5
1/l = (1.0974*107 m-1 )*|1/42 -1/52|
1/l = (1.0974*107 m-1 )*|1/16 -1/25|
1/l = (1.0974*107 m-1 )*|0.0225|
1/l = 246920 m-1
l = 0.0000040500 m = 4.0500*10-6 m = 4.0500 mm
Visible light
The Bohr Model of the Hydrogen Atom
In 1913, Neils Bohr (1885-1962) proposed
explanation H hydrogen based on three postulates:
an
1. The orbital angular momentum of electrons in an atom
is quantized.
Only those electrons whose orbitals
correspond to integer multiples of h/2π are “allowed”.
2. Electrons within an allowed orbital can move without
radiating (so that there is no net loss of energy).
3. The emission or absorption of light occurs when
electrons ‘jump’ from one orbital to another
Energy Level of Electrons
The energy of an electron in the n th
orbital of a hydrogen atom
E= - RhC/n2
n = principle quantum
number
Energy is
negative, i.e.
means its
stabilized
E/RhC = -1/n2 = -1, -1/4, -1/9, -1/16 ….
For any atomic system:
Z = atomic number
En=- Ry Z/n2
Ry =RhC= 2.179*10-18 J = Rydberg unit
Bohr calculated the radius of each orbital:
r = ao (n2/Z)
ao= Bohr radius
Absorption, Emission and Energy Levels
Lowest energy state : Ground state
Electrons cannot stand still therefore have an absolute
minimum energy
When a photon of the correct energy passes by it is
absorbed and the electron goes to a higher energy level.
i.e. An Excited state
Absorption, Emission and Energy Levels
The electron can relax back to the ground state.
Upon relaxation it releases a photon.
The energy of the photon absorbed or released has energy
matching the difference between the energy levels involved:
DE = Eex.s. – E g.s
Exercise
Calculate the energy of a photon absorbed by a hydrogen atom
when an electron jumps from the ground state to the n = 3
excited state?
DE = Eex.s. – E g.s
DE = E(n=3) – E(n=1)
E=-RhC/n2
DE = -RhC/32 – (-RhC/12)
DE = -(RhC)(1/9 –1)=-(Ry)(-8/9)
DE = (8/9) Ry
DE = (8/9) (2.179*10-18 J)
DE = 1.937*10-18 J
Ionization Limit
Notice that as you increase n the energy
approaches 0 but does not quite get there.
The corresponding orbital radius would
approach infinity.
What does this mean?
The electron is no longer in orbit
i.e. The atom has ionized
The ionization energy is therefore the limit of
DE as n → ∞
IE = -RhC/∞2 – (-RhC/ninitial2)
IE = RhC/n2 = -E(n) > 0