Phys.3
(lec.3)
Dr.ehab elfalaky
[5] THE COMPTON EFFECT
• Further justification for the photon nature of light came from an experiment
conducted by Arthur H. Compton in 1923. In his experiment, Compton directed an
x-ray beam of wavelength λ0 toward a block of graphite. He found that the
scattered x-rays had a slightly longer wavelength λ than the incident x-rays, and
hence the energies of the scattered rays were lower. The amount of energy reduction
depended on the angle at which the x-rays were scattered. The change in wavelength
∆λ between a scattered x-ray and an incident x-ray is called the Compton shift. In
order to explain this effect, Compton assumed that if a photon behaves like a
particle, its collision with other particles is similar to a collision between two billiard
balls. Hence, the x-ray photon carries both measurable energy and momentum, and
these two quantities must be conserved in a collision.
∆λ = λ – λ0 = h / me C ( 1 – cos θ )
Problem X-rays of wavelength λ0 = 0.200 000 nm are scattered from a block of material. The
scattered x-rays are observed at an angle of 45.0° to the incident beam. (a) Calculate the
wavelength of the x-rays scattered at this angle. (b) Compute the fractional change in the energy
of a photon in the collision.
Solution (a) Calculate the wavelength of the x-rays.
∆λ = h / me C ( 1 – cos θ )
= 6.63x10-34 J. s / (9.11x10-31 kg) (3.00 x108 m/s) (1 - cos 45.0)
= 7.11 x 10
-13
m = 0.000711 nm
λ = ∆λ + λ0 = 0.200711 nm
(b) Find the fraction of energy lost by the photon in the collision.
Rewrite the energy E in terms of wavelength, using c = f λ
E= hf = h c / λ
Compute ∆E/E using this expression:
∆E / E = Ef - Ei / Ei = ( hc / λ
f
- hc/ λ i ) / hc / λ
i
Cancel hc and rearrange terms:
∆E / E = ( 1 / λ
f
- 1/ λ i ) / 1 / λ
i
= (λ
i
/ λ i ) – 1= ( λ i - λ
∆E / E = - 0.000711 nm / 0.200711 nm = -3.54 x 10-3
f
)/ λ
f
= - ∆λ / λf
[6] THE DUAL NATURE OF LIGHT AND MATTER
Light has a dual nature, exhibiting both wave and particle characteristics.
• The Wave Properties of Particles
• In his doctoral dissertation in 1924, Louis de Broglie postulated that, because photons
have wave and particle characteristics, perhaps all forms of matter have both properties.
This was a highly revolutionary idea with no experimental confirmation at that time.
According to de Broglie, electrons, just like light, have a dual particle–wave nature. The
energy of a photon :-
• E = hf = hc / λ
• Consequently, the momentum of a photon can be expressed as
• P = E / C = hc / cλ = h / λ
[6] THE DUAL NATURE OF LIGHT AND MATTER
• From this equation, we see that the photon wavelength can be specified by its momentum,
or λ = h/p. De Broglie suggested that all material particles with momentum p should have a
characteristic wavelength λ = h/p. Because the momentum of a particle of mass m and
speed v is mv = p, the de Broglie wavelength of a particle is :
λ = h / p = h / mv
• Further, de Broglie postulated that the frequencies of matter waves (waves associated with
particles having nonzero rest energy) obey the Einstein relationship for photons, E = hf, so
that f = E / h
[6] THE DUAL NATURE OF LIGHT AND MATTER
Problem (a) Compare the de Broglie wavelength for an electron (me = 9.11x 10-31
kg) moving at a speed of 1.x107 m/s with that of a baseball of mass 0.145 kg pitched
at 45.0 m/s. (b) Compare these wavelengths with that of an electron traveling at
0.999c.
Solution (a) Compare the de Broglie wavelengths of the electron and the baseball.
λe = h / me v = 6.63 x 10-34 J.s
= 7.28 x 10
-11
/ (9.11 x 10-31 kg) (1.00 x 107 m/s)
m.
Repeat the calculation with the baseball data:
λb = h / mb v = 6.63 x 10-34 J.s
/ (0.145 kg) (45 m/s) = 1.02x 10
-34
m
(b) Find the wavelength for an electron traveling at 0.999c.Replace the momentum
in Equation with the relativistic momentum:
λe = h /
mev/√1 - v2/ c2 = h √1 - v2/c2
λe = (6.63 x 10
=
1.09x 10
-34
-13
2
J.s)√1 - (0.999c)
m
/ mev
/ c2 /
(9.11x 10-31 kg)(0.999 . 3.00 x108 m/s)
[8] The Uncertainty Principle
• According to Heisenberg’s uncertainty principle, it is impossible to measure
simultaneously the exact position and exact momentum of a particle. If ∆x is
the uncertainty in the measured position and ∆px the uncertainty in the
momentum, the product ∆x ∆px is given by :
∆x ∆px ≥ h / 4π Also,
∆E ∆t ≥ h / 4π
• where ∆E is the uncertainty in the energy of the particle and ∆t is the
uncertainty in the time it takes to measure the energy.
[8] The Uncertainty Principle
Problem The speed of an electron is measured to be 5 x 103 m/s to an accuracy of
0.003 00%. Find the minimum uncertainty in determining the position of this
electron.
Strategy
After
computing
the
momentum
and
its
uncertainty,
substitute
into
Heisenberg’s uncertainty principle.
Solution Calculate the momentum of the electron:
px= mev = (9.11x 10
-31
kg)(5x 103m/s) = 4.56 x10-27 kgm/s
The uncertainty in p is 0.003 00% of this value:
px = 0.0000300p = (0.0000300) (4.56 x10
-31
27 kgm/s) =1.37x10
-31
kgm/s
Now calculate the uncertainty in position using this value of ∆px and ∆x ∆px ≥ h /
4π
……∆x ∆px ≥ h / 4π ………∆x ≥ h / ∆px 4π
∆xx ≥ 6.626x10
-34
J.s / 4π ( 1.37x10
-31
kg.m/s ) = 0.384 mm
Chapter Two
[1] The Nature of Light
• An explanation of the photoelectric effect was proposed by Einstein in
1905 in a theory that used the concept of quantization developed by
Max Planck (1858-1947) in 1900. The quantization model assumes that
the energy of a light wave is present in particles called photons; hence,
the energy is said to be quantized. According to Einstein's theory, the
energy of a photon is proportional to the frequency of the
electromagnetic wave:
• E=hv
(1)
[2] Properties of light
• [2.1] Reflection of light
• when light travelling in a medium encounters a boundary leading to a
second medium, part of the incident light is returned to the first medium
from which it came. This phenomenon is called reflection.