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General Physics (PHY 2140)
Lecture 28
¾ Modern Physics
9Quantum Physics
9Photoelectric effect
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 27
11/10/03
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11/10/03
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Lightning Review
Last lecture:
1. Quantum physics
9 Blackbody radiation
9 Planck’s hypothesis
λmaxT = 0.2898 ×10−2 m ⋅ K
En = nhf ,
n = 1, 2,3,...
Review Problem: All objects radiate energy. Why, then, are we not able to
see all objects in a dark room?
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Reminder (for those who don’t read syllabus)
Reading Quizzes (bonus 5%):
It is important for you to come to class prepared, i.e. be familiar with the
material to be presented. To test your preparedness, a simple five-minute
quiz, testing your qualitative familiarity with the material to be discussed in
class, will be given at the beginning of some of the classes. No make-up
reading quizzes will be given.
There could be one today…
… but then again…
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QUICK QUIZ
A photon (quantum of light) is reflected from a mirror. True or false:
(a) Because a photon has a zero mass, it does not exert a force on
the mirror.
(b) Although the photon has energy, it cannot transfer any energy to
the surface because it has zero mass.
(c) The photon carries momentum, and when it reflects off the mirror,
it undergoes a change in momentum and exerts a force on
the mirror.
(d) Although the photon carries momentum, its change in momentum
is zero when it reflects from the mirror, so it cannot exert a
force on the mirror.
(a)
(b)
(c)
(d)
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False
False
True
False
∆p = Ft
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Problem: a lightbulb
Assuming that the tungsten filament of a lightbulb is a blackbody,
determine its peak wavelength if its temperature is 2 900 K.
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Assuming that the tungsten filament of a lightbulb is a blackbody, determine its
peak wavelength if its temperature is 2 900 K.
Given:
Recall Wein;s law: λmax T = 0.2898 x 10-2 m • K. It can
be used to solve for λmax:
T = 2900 K
λmax
0.2898 ×10−2 m ⋅ K
=
T
Thus,
Find:
λmax = ?
λmax
0.2898 ×10
(
=
−2
2900 K
m⋅ K
) = 9.99 ×10
−7
m = 999nm
This is infrared, so most of the electric energy
goes into heat!
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27.2 Photoelectric Effect
When light is incident on certain metallic surfaces, electrons are
emitted from the surface
„
„
This is called the photoelectric effect
The emitted electrons are called photoelectrons
The effect was first discovered by Hertz
The successful explanation of the effect was given by Einstein in
1905
„
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Received Nobel Prize in 1921 for paper on electromagnetic radiation, of
which the photoelectric effect was a part
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Photoelectric Effect Schematic
When light strikes E,
photoelectrons are emitted
Electrons collected at C and
passing through the ammeter
are a current in the circuit
C is maintained at a positive
potential by the power supply
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Photoelectric Current/Voltage Graph
The current increases with
intensity, but reaches a
saturation level for large ∆V’s
No current flows for voltages
less than or equal to –∆Vs, the
stopping potential
„
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The stopping potential is
independent of the radiation
intensity
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Features Not Explained by Classical
Physics/Wave Theory
No electrons are emitted if the incident light frequency is
below some cutoff frequency that is characteristic of the
material being illuminated
The maximum kinetic energy of the photoelectrons is
independent of the light intensity
The maximum kinetic energy of the photoelectrons
increases with increasing light frequency
Electrons are emitted from the surface almost
instantaneously, even at low intensities
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Einstein’s Explanation
A tiny packet of light energy, called a photon, would be emitted
when a quantized oscillator jumped from one energy level to the
next lower one
„ Extended Planck’s idea of quantization to electromagnetic
radiation
The photon’s energy would be E = hƒ
Each photon can give all its energy to an electron in the metal
The maximum kinetic energy of the liberated photoelectron is
KE = hƒ – Φ
Φ is called the work function of the metal
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Explanation of Classical “Problems”
The effect is not observed below a certain cutoff
frequency since the photon energy must be greater than
or equal to the work function
„
Without this, electrons are not emitted, regardless of the intensity
of the light
The maximum KE depends only on the frequency and
the work function, not on the intensity
The maximum KE increases with increasing frequency
The effect is instantaneous since there is a one-to-one
interaction between the photon and the electron
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Verification of Einstein’s Theory
Experimental observations of a
linear relationship between KE
and frequency confirm
Einstein’s theory
The x-intercept is the cutoff
frequency
Φ
fc =
h
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Problem: photoeffect
Electrons are ejected from a metallic surface with speeds ranging
up to 4.6 × 105 m/s when light with a wavelength of λ = 625 nm
is used.
(a) What is the work function of the surface?
(b) What is the cutoff frequency for this surface?
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Electrons are ejected from a metallic surface with speeds ranging up to 4.6 × 105 m/s
when light with a wavelength of λ = 625 nm is used.
(a) What is the work function of the surface?
(b) What is the cutoff frequency for this surface?
Given:
v = 4.6x105 m/s
λ = 625 nm
Recall that KEmax=hf - φ. This can be used to solve
for φ. First find the kinetic energy
KEmax
2
mvmax
1
=
= 9.11×10−31 kg 4.6 ×105 m s = 9.6 ×10−20 J
2
2
(
Thus, Φ = hf − KEmax =
Find:
φ=?
ν =?
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)(
hc
λ
)
− KEmax = 2.2 ×10−19 J
It equals to 1.4 eV
Cutoff frequency is
Φ
2.2 ×10−19 J
14
=
×
3.3
10
fc = =
Hz
−34
h 6.63 ×10 J ⋅ s
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27.3 Application: Photocells
Photocells are an application of the photoelectric effect
When light of sufficiently high frequency falls on the cell,
a current is produced
Examples
„
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Streetlights, garage door openers, elevators
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27.4 X-Rays
Electromagnetic radiation with short wavelengths
„
„
„
Wavelengths less than for ultraviolet
Wavelengths are typically about 0.1 nm
X-rays have the ability to penetrate most materials with relative
ease
Discovered and named by Roentgen in 1895
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Production of X-rays
X-rays are produced when highspeed electrons are suddenly
slowed down
„
Can be caused by the electron
striking a metal target
A current in the filament causes
electrons to be emitted
These freed electrons are
accelerated toward a dense
metal target
The target is held at a higher
potential than the filament
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Production of X-rays
An electron passes near a
target nucleus
The electron is deflected from
its path by its attraction to the
nucleus
„
This produces an acceleration
It will emit electromagnetic
radiation when it is accelerated
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Diffraction of X-rays by Crystals
For diffraction to occur, the spacing between the lines
must be approximately equal to the wavelength of the
radiation to be measured
For X-rays, the regular array of atoms in a crystal can act
as a three-dimensional grating for diffracting X-rays
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Schematic for X-ray Diffraction
A continuous beam of X-rays
is incident on the crystal
The diffracted radiation is very
intense in certain directions
„
These directions correspond
to constructive interference
from waves reflected from the
layers of the crystal
The diffraction pattern is
detected by photographic film
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Photo of X-ray Diffraction Pattern
The array of spots is called a Laue
pattern
The crystal structure is determined by
analyzing the positions and intensities
of the various spots
This is for NaCl
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Bragg’s Law
The beam reflected from the lower
surface travels farther than the
one reflected from the upper
surface
If the path difference equals some
integral multiple of the wavelength,
constructive interference occurs
Bragg’s Law gives the conditions
for constructive interference
2 d sin θ = m λ, m = 1, 2, 3…
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If you want to know your progress so far, please
send me an email request at
[email protected]
11/10/03
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