# Duel - Red Hook Central School District ```Quantum Nature of Matter &amp; Energy
• How is light produced?
• Vibration of charged particles.
• Is light a wave or a particle?
Cite 2 pieces of evidence for light as a wave.
• Diffraction
• Interference
What is the evidence for light as transverse.
• Polarization
• Max Planck studied how blackbodies emit radiation
(EM waves).
• A blackbody is an object that absorbs &amp; emits all
wavelengths of radiation. It does not reflect any light
incident on it.
• It was known that energy of a wave is related to its
amplitude - Intensity.
• As T goes up, total light emitted (I) goes up.
• Frequency goes up too!
Hot Objects Emit Waves
T vs. Intensity of a blackbody
Spectrum of wavelengths emitted, but more in certain range.
Classical Physics predicts E (I) becomes infinite as l
approaches 0.
As T , molc. vibration.
Planck proposed that the E is emitted by
vibrating molecules is quantized—E could only
take on certain values.
Energy come in little &quot;chunks&quot; of the frequency
multiplied by a constant now called Planck's
constant, h:
Matter absorbs &amp; emits energy in discrete
units called quanta or photons. Plank’s
formula gives the amount of energy based on
the frequency of waves.
E = hf.
h is Plank’s constant 6.63 x 10-34 Js.
E is energy in Joules
1 eV = 1.6 x 10-19 J
• Emitted E proportional to frequency.
• Proportionality Constant = h.
• Planck’s theory’s predictions agreed with
experimental observations.
Ex 1. Each photon of a certain
color light has an energy of 2.5
eV. What is the frequency of
and color of the light?
Solution:
E = hf
f = E/h
convert eV to Joules.
(2.5 eV)(1.6 x 10-19J/eV) = 6 x 1014 s-1 or Hz
6.626 x 10-34 J s
When you need to find # joules in a
wave/photon use
E = hf.
Or
E = hc/l.
2. What is the energy of light at
686 nm in Joules and eV?
•
•
•
•
E = hf
E = h c/l.
2.9 x 10-19 J
1.81 eV
waves come in discrete packets indicated
• More evidence for the particle nature of
light was later to come from work with the
photoelectric effect.
Photoelectric Effect
It's been known (previously) that when EM
waves shine on a metal surface, the surface can
emit e-.
You can start a current in a circuit just by
shining a light on a metal plate.
Materials that emit e- in this way are called
photoemissive. The ejected e- are called
photoelectrons.
Predictions
Classical
Photon
• EM waves absorbed by
metal shake e- in metal
until they overcome bond
E and escape.
• Longer exposure time
needed to eject e-.
• Higher A or f, will
decrease needed exposure
time.
• Higher A will increase E of
ejected e-.
• 1 on 1 collisions btw
photon &amp; e- cause ejection.
• Photon must have enough
E to free e-.
• No waiting time required.
• Since E ~ f, there is a cut
off f, below it, no e- ejected
no matter how long the
wait.
• High A, I = more photons,
more photoelectrons.
Observations
•
•
•
•
•
•
Cut off f, yes
Minimum fo, need to eject e- (threshold f)
Wo, F.
No time delay
Hi A, I (bright) = more e-.
Higher f, faster e-.
Easy Quantitative
•
•
•
•
•
•
If F for a metal = 7eV.
Red light has 6eV. What happens?
No e- ejected.
Blue light has 10 eV. What happens?
e- ejected but 3 extra eV E.
Could get lost in collisions in metal, but
could be KE of e-. Called KE max.
Example:
1. In the photoelectric effect, the following observations
• I. The kinetic energy of the emitted electrons increases
with increasing light frequency.
• II. The electrons are emitted without time delay.
Which of these observations, if any, can be explained in terms
of the wave theory of light?
• A. Neither I nor II
• B. I and II
• C. I only
• D. II only
2. Photoelectron emission only occurs if the light
incident on a metal surface is
•
•
•
•
A.
B.
C.
D.
coherent.
above a certain minimum intensity.
below a certain minimum frequency.
below a certain minimum wavelength.
3. Monochromatic light is incident on a metal surface
and electrons are released. The intensity of the incident
light is increased. What changes, if any, occur in the rate
of emission of electrons and the KE of the emitted
electrons?
Rate of emission of e-
KE of the emitted e-
A.
Increase
increase
B.
decrease
no change
C.
decrease
increase
D.
increase
no change
VDGgqw
Photoelectric Equation
• Photon E in = electron E out.
• hf
=
F + KE max.
• hc/l = F +
KE max.
Ex. What minimum f, is needed to eject e- if
F of metal is 4.3 x 10-19 J?
•
•
•
•
hf =
F + KE max.
hf =
F
(6.63x10-34) f = (4.3 x 10-19 J)
6.5 x 1014 Hz.
• Called threshold freq = fo.
Ex: UV light l = 285 nm, falls on a metal surface.
The maximum KE of photoelectrons is 1.40 eV.
What is the work function of the metal?
•
•
•
•
•
•
hc/l
= F
+ KE max.
(6.63x10-34)(3x108) = F
+
1.40 eV.
285 x 10-9 m
Change left side to eV (divide by e = 1.6 x 10-19)
4.36 eV =
F
+
1.40 eV.
3.06 eV = F .
The Experiment and Associated
Graph
Stopping Potential
Milliken’s experiment measured the KE of photo e- by
applying a stopping potential – a voltage that brought the
velocity of e- to 0.
How to measure the values?
Use stopping/retarding voltage until no
photocurrent.
Phet
•
•
•
PE gained = KE lost by electron.
qV
=
KE max.
eV
=
KE max.
• http://www.walter-fendt.de/ph6en/
IB Data Booklet
Work
function.
Stopping
Potential.
Graphing E vs. f.
• hf =
F + KE max.
• Rearrange to solve for KE max.
• KE max. = hf
• eV (st pot) = hf
-
F.
hfo .
• Linear Eq. KE Y axis, h = slope, F = Y intercept.
• Metals have work functions a few eV.
• Energy in J/e = energy in eV.
Photoelectric Experiment Applet
create graph
• http://www.walter-fendt.de/ph6en/
• eV =
hf
hfo .
• Slope is Planck’s constant h/ e
(1.6 x 10-19).
Ex 1: Ultra-violet light is shone on a zinc surface and
photoelectrons are emitted. The sketch graph shows how the
stopping potential Vs varies with frequency f.
Planck’s constant may be determined from the charge of an
electron e multiplied by
• A. the x-intercept.
• B. the y-intercept.
• D. the area under the graph.
KE max =
hf
-
F.
As f of EM wave increases, KE increases, slope = h.
F min E J, needed to eject e.
Ex 2: Light of frequency f is incident on a metal surface.
The work function of the metal is φ. Which of the
following is the maximum kinetic energy of the electrons
emitted from the surface?
A.
hf – 
B.
h
( f )
e
C.
 – hf
D.
h
(  f )
e
3. Which of the following is a correct statement
associated with the photoelectric effect?
• A. Electron emission is instantaneous.
• B. Electrons are only emitted if the incident light is
above a certain minimum wavelength.
• C. The energy of the emitted electrons depends on
the light intensity.
• D. The energy of the emitted electrons does not
depend on the frequency of the incident light.
Summery
• Can measure KEmax using stopping V needed to bring
fastest e- to stop.
• Energy of e- is PEelc qV, for the retarding V.
• Graph of V vs. f, the Y intercept is the Wo, X intercept
= fo, and slope is h/e.
• Graph of KE vs. f, slope = h.
Hist of Quantum pt 1
British 15 min Max Planck and E= hf.
TbqOgdfEY
IB Set 3 Photoelectric Questions.
Hwk. Hamper pg 261 #2,3 Write out all.
v=B7pACq_xWyw&amp;feature=rela
ted
Watch this. Good summery of light. 9:45 min.
Hwk Holt Photo Elect and E = hf. 23 – 1
and 23 – 2
Do page 833 all and
page 856 # 2, 4, 7, 10, 11
Models of The Atom Rutherford
Equivalence of Mass
&amp; Energy
Einstein realized that matter contains
energy. There is an equivalence of mass
&amp; energy.
Energy is stored in the nucleus of atoms.
The energy stored any mass obeys
Einstein’s equation:
E = energy in J.
E = mc2.
m = mass kg
c = vel of light
Ex 2: How much energy is
produced when 2.5 kg of matter
are completely converted to
energy?
How much energy is that in eV?
E = mc2.
=(2.5 kg )(3x108 m/s)2. = 2.25 x 1017 J
in eV
(2.25 x 1017 J)(1 eV / 1.6 x 10 –19 J) =
1.4 x 1036 eV.
Atomic Mass Units:
amu or u
Mass of atoms very small so they are
measured in amu or u.
Since mass is equivalent to energy,
1 amu = 931 MeV or 931 x 106 eV.
Ex 3: One universal atomic mass
unit is equivalent to an energy of
931 MeV. Calculate the mass in kg
of one universal mass unit.
Hint: Use E = mc2 where energy is
known in eV.
Don’t forget to convert MeV to eV.
(1 u) x (931 MeV/u) x (106eV/MeV) x (1.6 x
10 –19 J / eV) =
1.49 x 1010 J
E = mc2 so
m = E/c2.
(1.49 x 1010 J) / (3x108 m/s)2 =
1.66 x 10 –27 kg
The mass units are based on
1
the mass of a proton or H.
(A hydrogen nucleus)
Particle Properties of Waves
extend to conservation of
energy and momentum.
Photons may give up all or part of their
energy in collisions, but the sum of the
momentums and energy before must
equal the sum after.
Compton Effect
If light behaves like a particle, then a
collision btw photon &amp; e- should be similar
to billiard balls colliding. Photons must
have momentum (p), &amp; energy.
In collision of photons with particles (like
e-), conservation of energy &amp; conservation
of momentum apply.
If the photon gives only part of its energy &amp;
momentum to an e-, its momentum decreases
after the collision by the same amount as
absorbed by the electron.
Therefore, the frequency or energy of the
photon decreases. The wavelength increases.
pbefore = pafter.
E photon before = KEelc after. + E photon after
hfi = KEelc after + hff photon after
pphoton
= hf/c =
h/l. The
wavelength of the photon increases after
collision.
Matter has wave-like properties.
1924 Louis DeBroglie suggested that
matter might have wave properties.
It turns out that matter does have wave
properties which are inversely related to
the momentum of the particle.
For matter:
l =h/p
or
l = h/mv.
Since the mass of most objects is so large,
the wavelengths would be very small &amp;
not measurable.
Electrons, however, do show diffraction &amp;
other wave characteristics.
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