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26-1 Photoelectric Effect
26-3 x- ray
T.Norah Ali Al moneef
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26-1Photoelectric Effect
• Scientists had noticed that when you shine light onto some types of metal, a
measurable voltage is produced
– The light seems to transfer energy to the metal which causes an electric
current
• But, not every kind of light produces the current
– And it doesn’t help to initiate the current by making the light brighter
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• 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 ΔV’s
• No current flows for voltages less than or
equal to –ΔV0, the stopping potential
• The stopping potential is independent of
the radiation intensity
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A specific value of V can be found at which the ammeter reading just drops to
zero
. This is called the stopping potential (Vstop).
• When the potential is at Vstop the most energetic electrons were turned back
just before hitting the collector.
• This indicates that the maximum kinetic energy of the photoelectrons,
Kmax= e Vstop where e is the elementary charge.
Interestingly, it was found that Kmax does not depend upon the intensity
of the incident light.
• It is difficult to explain this observation with classical physics where
light is viewed as a continuous wave.
• In classical physics, the electrons would be viewed as oscillating
under the influence of an alternating electric field. If the intensity of
the light is increased, so is the amplitude of the electric field, which
should make it easier for the electron to acquire enough energy to
break free of the surface of the metal plate. This, however, does
not agree with experiment
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Photo-electric effect observations
•The kinetic energy of the photoelectrons
is independent of the light intensity.
•The kinetic energy of the photoelectrons,
for a given emitting material, depends
only on the frequency of the light.
Note that there is no photoelectric effect if the
light is below a certain cutoff frequency, f0. This
occurs no matter how bright the incident light is.
The work function of each metal can be determined by taking
the negative y-intercept of each line.
– The cutoff frequency of each metal can be determined by
taking the x intercept of each line.
– Note that all three lines have the same slope. This slope is
Planck’s
constant
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Photo-electric effect observations
•When photoelectrons are produced,
their number (not their kinetic energy)
is proportional to the intensity of light.
(number of
electrons)
Also, the photoelectrons are emitted
almost instantly following illumination
of the photocathode, independent of
the intensity of the light.
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The Photoelectric Effect (cont’d)
Each photoemissive
material has a
characteristic threshold
frequency of light.
When light that is above
the threshold frequency
strikes the photoemissive
material, electrons are
ejected and current flows.
Light of low frequency
does not cause current
flow … at all.
As with line spectra, the
photoelectric effect cannot be
explained by classical physics.
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The Photoelectric Effect
• Albert Einstein won the 1921 Nobel Prize in Physics for
explaining the photoelectric effect.
• He applied Planck’s quantum theory: electromagnetic
energy occurs in little “packets” he called photons.
Energy of a photon (E) = hf
• The photoelectric effect arises when photons of light
striking a surface transfer their energy to surface electrons.
• The energized electrons can overcome their attraction for
the nucleus and escape from the surface …
• … but an electron can escape only if the photon provides
enough energy.
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The Photoelectric Effect
Explained
The electrons in a
photoemissive material
need a certain minimum
energy to be ejected.
Short wavelength (high
frequency, high energy)
photons have enough
energy per photon to
eject an electron.
A long wavelength—low
frequency—photon doesn’t
have enough energy to
eject an electron.
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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 f0 that is characteristic of the material being
illuminatedsince 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 kinetic energy kmax of the photoelectrons
is independent of the light intensity
• The maximum kinetic energy of the photoelectrons
increases with increasing light frequency, The maximum
KE depends only on the frequency and the work function,
• Electrons are emitted from the surface almost
instantaneously, even at low intensities
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Quantum Theory of The Atom
– In 1901, Max Planck suggested light was made up of
‘packets’ of energy:
–
E  nhf
E
v
n
h
(Energy of Radiation)
(Frequency)
(Quantum Number) = 1,2,3…
(Planck’s Constant, a Proportionality Constant)
6.626 x 10-34 Js)
6.626 x 10-34 kgm2/s
– Atoms, therefore, emit only certain quantities of energy
and the energy of an atom is described as being
“quantized”
– Thus, an atom changes its energy state by emitting (or
absorbing) one or more quanta
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h0
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 electron is considered to be in a well of height w
which is called the work function of the metal
• The maximum kinetic energy of the liberate photoelectron
is KEmax = hƒ –W
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Photons easily explain the results of both photoelectric effect
experiments.
• Light energy transferred as photons is independent of intensity as
shown in both experiments
• Furthermore, the existence of a cutoff frequency indicates that
below that frequency electrons are not receiving enough energy to
overcome the electric forces that bind them to the metal,
supporting the idea that the energy is proportional to the frequency
•The minimum energy that an electron needs to escape the metal
plate depends on the material plate is made out of. This minimum
energy is called the work function (w) of the material.
• Einstein was able to sum up the results of the two experiments
with the photoelectric equation
h f= Kmax +w.
Note that if this equation is rearranged to Kmax=h f - w
it is obviously linear in nature, agreeing with the photoelectric
effect graph seen earlier
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Photon:
A packet or bundle of energy is called a photon.
Energy of a photon is
E = hf =
hc
λ
where h is the Planck’s constant, f is the frequency of the
radiation or photon, c is the speed of light (e.m. wave)
and λ is the wavelength.
Properties of photons:
i) A photon travels at a speed of light c in vacuum. (3 x 108 m/s)
ii) It has zero rest mass. i.e. the photon can not exist at rest.
E
iii) The kinetic mass of a photon is,
m=
iv) The momentum of a photon is,
E
p=
c
h
=
c2
cλ
h
=
λ
v) Photons travel in a straight line.
vi) Energy of a photon depends upon frequency of the photon; so the
energy of the photon does not change when photon travels from one
medium to another.
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Cutoff Wavelength
• The cutoff wavelength is related to the work function
l
0
l
max

hc
w
hc

w
• Wavelengths greater than l0 incident on a material with a work
function w don’t result in the emission of photoelectrons
Photoemission
• The number of electrons emitted by a surface is proportional
to the number of incident photons
• An electron is emitted as soon as a photon reaches the surface
hf = W + ½ mV2max
W = work function
½ mV2max = max kinetic energy
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Threshold Frequency
( f0)
hfo = W
hc = W
λo
c = speed of light
Work Function (W )
λo = wavelength
The minimum amount of energy that has to be given to an
electron to release it from the surface of the material and varies
depending on the material
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1 2
E photon  hf  2 mvmax  w
hf  eVo  hf o
eVo  h f  f o 
h
Vo 
e
f
Slope = h/e
 fo 
• The stopping potential doesn’t depend on the incident light
intensity.
• The stopping potential depends on the incident frequency.
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(a) What are the energy and momentum of a photon of red
light of wavelength 650nm?
(b) What is the wavelength of a photon of energy 2.40 eV?
(a) E = hc/l
= (6.62x10-34 Js)X(3x108 m/s) /650 nm
= 3.110-19J
( b ) p = E/c = 3.110-19J / (3x108 m/s) =
(c) l = hc/E = (6.62x10-34 Js)X(3x108 m/s) / 2.40 X (1.6x10-19)
= 517 nm
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Another solution using eV instead of J
1 eV = 1.6x10-19 J
hc= (6.62x10-34 Js)·(3x108 m/s) = [6.62x10-34 ·(1.6x10-19)-1eV·s]·(3x108 m/s)
= 1.24eV·10-6m = 1240eV·nm
1 eV/c = (1.6x10-19)J/ (3x108 m/s) = 5.3x10-28 Ns
(a) E = hc/l
= 1240 eVnm /650 nm
= 1.91 eV (= 3.110-19J)
(b) p = E/c = 1.91 eV/c (= 1x10-27 Ns)
(c) l = hc/E
= 1240eV·nm /2.40 eV
= 517 nm
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Example
The work function of silver is 4.73eV. EM radiation with
a frequency of 1.20x1015Hz strikes a piece of pure silver.
What is the speed of the electrons that are emitted?
hf = K + W …
Calculate the maximum K first…
K= hf – W
= (4.14x10-15eVs)(1.20x1015Hz) – 4.73eV
= 0.238 eV = 3.81 x 10-20J
Then calculate the speed of the electron…
K = ½ mv2
1
K  9.11  1031kg( v 2 )
2
2  3,81  1020 j
2
v 
9  1031 kg
v = 2.89 x 105m/s
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(II) When 230-nm light falls on a metal, the current through a
photoelectric circuit) is brought to zero at a stopping voltage
of 1.64 V. What is the work function of the metal
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What is threshold frequency of a material with a
work function of 10eV?
Since the value for the work function is given in
electron volts, we might as well use the value for
Planck’s constant that is in eV s.
W = h fo
fo = W / h
= (10eV) / (4.14 x 10-15eVs)
fo = 4.14 x 10-14 Hz
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What is the maximum wavelength of electromagnetic radiation
which can eject electrons from a metal having a work function
of 3 eV? (3 points)
Answer
The minimum photon energy needed to knock out an electron
is 3 eV. We just need to convert that to wavelength.
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Find the frequency for EM radiation with wavelength
equal to 1 inch ~ 2.5 cm =2.5x10-2 m
l f c
f 
c
l
3 108
10
f 

1
.
2

10
Hz  12 GHz
2
2.5 10
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In a photoelectric-effect experiment it is observed that no current flows unless the
wavelength is less than 570 nm. (a) What is the work function of this material? (b)
What is the stopping voltage required if light of wavelength 400 nm is used?
At the threshold wavelength, the kinetic energy of the photoelectrons
is zero, so we have
The stopping voltage is the voltage that gives a potential energy change equal
to the maximum kinetic energy
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What is the energy associated with a light
quantum of wavelength 5.0 x 10-7 m? (h =
6.63 x 10-34 J-s)
4.0 x 10-19 J
3.3 x 10-19 J
1.5 x 10-19 J
1.7 x 10-19 J
hc 6.63  10 34 J. s  3  108 m s1
19
E


4

J
10
7
l
5  10
1.
2.
3.
4.
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If a monochromatic light beam with quantum energy value of
3.0 eV incident upon a photocell where the work function of
the target metal is 1.60 eV, what is the maximum kinetic
energy of ejected electrons?
1.
2.
3.
4.
4.6 eV
4.8 eV
1.4 eV
2.4eV
kmax = hf – W
= (3eV) – 1.6eV
kmax=1.4 eV
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Calculating the Energy of Radiation from Its Wavelength
PROBLEM: A cook uses a microwave oven to heat a meal. The
wavelength of the radiation is 1.20cm. What is the
energy of one photon of this microwave radiation?
PLAN: After converting cm to m, we can use the energy
equation, E = h combined with  = c/l to find the
energy.
SOLUTION:
E=
E = hc/l
6.626X10-34J-s x 3x108m/s = 1.66x10-23J
-2m
10
1.20cm
cm
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A photon at 300 nm will kick out an electron with an amount of
kinetic energy, KE300. If the wavelength is halved to 150 nm and
the photon hits an electron in the metal with same energy as the
previous electron, the energy of the electron coming out is
a. less than ½ KE300.
b. ½ KE300
c. = KE300
d. 2 x KE300
e. more than 2 x KE300
KE = photon energy-energy to get out
= hf – energy to get out
if l is ½ then, f twice as big, Ephot =2hf300
New KEnew= 2hf300- energy to get out
Old KE300 =hf300- energy to get out
so KEnew is more than twiceT.Norah
as big.
Ali Al moneef
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: Shine in light of 300 nm, most energetic electrons come out with
kinetic energy, KE300. A voltage diff of 1.8 V is required to stop these
electrons. What is the work function  for this plate? (e.g. the
minimum amount of energy needed to kick e out of metal?)
a. 1.2 eV
b. 2.9 eV
c. 6.4 eV
d. 11.3 eV
e. none
Energy is conserved so:
Ephot= energy need to exit (w) + electron’s left over
energy
so
w= Ephot – electron’s energy
When electron stops, all of initial KE has been converted
to electrostatic potential energy:
electron energy = qDV = e x 1.8V = 1.8 eV, and
Ephot = 1240 eV nm/300 nm = 4.1 eV.
So w = 4.1eV - 1.8 eV = 2.3 eV
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• (a) Lithium, beryllium and mercury have work
functions of 2.3 eV, 3.9 eV and 4.5 eV, respectively. If a
400-nm light is incident on each of these metals,
determine
• (i) which metals exhibit the photoelectric effect, and
• (ii) the maximum kinetic energy for the
photoelectron in each case (in eV)
The energy of a 400 nm photon is E = hc/l = 3.11 eV
The effect will occur only in lithium
For lithium,
Kmax = hf – W
= 3.11 eV – 2.30 eV
= 0.81 eV
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• (b) Molybdenum has a work function of 4.2 eV.
• (i) Find the cut-off wavelength (in nm) and threshold
frequency for the photoelectric effect.
• (ii) Calculate the stopping potential if the incident
radiation has a wavelength of 180 nm.
Known hfcutoff = W0
Cut-off wavelength = l cutoff = l max=c/fcutoff
= hc/W = 1240 nm eV / 4.2 eV of= 295 nm
Cut-off frequency (or threshold frequency), f cutoff
= c / l cutoff = 1.01 x 1015 Hz
Stopping potential Vstop = (hc/l – W0) / e
= (1240 nmeV/180 nm – 4.2 eV)/e = 2. 7 V
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• Light of wavelength 400 nm is incident upon lithium (W0 = 2.9 eV).
Calculate
• (a) the photon energy and
• (b) the stopping potential, Vs
• (c) What frequency of light is needed to produce electrons of
kinetic energy 3 eV from illumination of lithium?
• (a) E= h = hc/l = 1240eV·nm/400 nm = 3.1 eV
• (b) The stopping potential x e = Max Kinetic energy of the photon
• => eVs = Kmax = hf - W = (3.1 - 2.9) eV
• Hence, Vs = 0.2 V
• a retarding potential of 0.2 V will stop all photoelectrons
• (c) hf = Kmax + W = 3 eV + 2.9 eV = 5.9 eV. Hence the frequency of
the photon is
f= 5.9 x (1.6 x 10-19 J) / 6.63 x 10-34 Js
= 1.42 x1015 Hz
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Which of the following statement(s) is (are) true?
I-The energy of the quantum of light is proportional to the
frequency of the wave model of light
II-In photoelectricity, the photoelectrons has as much
energy as the quantum of light which causes it to be
ejected
III-In photoelectricity, no time delay in the emission of
photoelectrons would be expected in the quantum theory
A. II, III
B. I, III
C. I, II, III
D. I ONLY
E. Non of the above
Ans: B
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Light made of two colors (two frequencies) shines on a
metal surface whose photoelectric threshold
frequency is 6.2 x 1014 Hz. The two frequencies are 5 x
1014 Hz (orange) 7 x 1014 Hz (violet).
(1) Find the energies of the orange and violet photons.
(2) Find the amount of energy a photon needs to knock
electrons out of this surface.
(3) Do either the orange photons or the violet photons
have this much energy?
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1. The energy for the orange frequency is:
E = hf = (6.6 x 10-34) x (5 x 1014) = 3.3 x 10-19 J.
The energy for the violet light is:
E = hf = (6.6 x 10-34) x (7 x 1014) = 4.6 x 10-19 J.
2. The threshold energy is
w = hfo = (6.6 x 10-34) x (6.2 x 1014) = 4.1 x 10-19 J.
3. The blue photons have sufficient energy to knock
electrons out, but the orange photons do not.
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Compton Effect
,Compton scattering occurs when the incident x-ray photon is deflected
from its original path by an interaction with an electron. The electron is
ejected from its orbital position and the x-ray photon loses energy
because of the interaction but continues to travel through the material
along an altered path. Energy and momentum are conserved in
thisprocess. The energy shift depends on the angle of scattering and
not on the nature of the scattering medium. Since the scattered x-ray
photon has less energy, it has a longer wavelength and less penetrating
than the incident photon.
hf   hf  Eel
Electron will acquire max Kinetic energy
when the scattered photon is at 1800
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example
In the Compton scattering the experiment the incident x ray have a
frequency of 1020 Hz at certain angle the outgoing x ray have frequency
of 8x10 19 Hz find the energy of the recoiling electron in electron volts
Eel  hf  hf  h( f  f )
 (4.135 1015 ev.s)(1020 Hz  8 1019 )
82.700eV
Example.
X-rays of wavelength 0.140 nm are scattered from a very thin
slice of carbon. What will be the wavelengths of X-rays scattered
at (a) 0°, (b) 90°, and (c) 180°?
Use the Compton scattering equation.
a.0.140 nm (no change)
b. 0.142 nm
c. 0.145 nm (the maximum)
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26-3
X-rays
• X-rays are produced when high-speed
electrons are suddenly slowed down
– Can be caused by the electron striking a metal
target
Current passing through a filament
produces copious numbers of electrons by
thermionic emission. If one focuses these
electrons by a cathode structure into a
beam and accelerates them by potential
differences of thousands of volts until they
collide with a metal plate (a tungsten
target), they produce x rays by
bremsstrahlung as they stop in the anode
material.
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As a result, we would expect a photon of much larger energy, and
hence much higher frequency, to be emitted when the atom returns
to its normal state after the displacement of an inner electron.
This is, in fact, the case, and it is the displacement of the inner
electrons that gives rise to the emission of x-rays. In addition to the
x-ray line spectrum there is a background of continuous x-ray
radiation from the target of an x-ray tube. This is due to the sudden
deceleration of those “cathode rays”.
The remarkable feature of the continuous spectrum is
that while it extends Indefinitely towards the long
wavelength end, it is cut off very sharply at the short
wavelength end. The quantum theory furnishes a
simple explanation of the short-wave limit of the
continuous x-ray spectrum (it is called Bremsstrahlung
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X-rays have a very small wavelength, no larger than 10-8 to 10-9.
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Bremsstrahlung Radiation Production
• The projectile electron interacts with the nuclear force field of the target tungsten
atom
• The electron (-) is attracted to the nucleus (+)
• The electron DOES NOT interact with the orbital shell electrons of the atom
• As the electron gets close to the nucleus, it slows down
(brems = braking) and changes direction
• The loss of kinetic energy (from slowing down) appears in the
form of an x-ray
• The closer the electron gets to the nucleus the more it slows
down, changes direction, and the greater the energy of the
resultant x-ray
• The energy of the x-ray can be anywhere from almost 0 (zero)
to the level of the kVp
1. Electron scattered in different direction with less energy
2.-ray
•
Energy dependent upon location of electron to nucleus and degree of
deceleration Energy dependent upon target atom and electron
T.Norah Ali Al
moneef the term characteristic)
binding energies of that atom
(hence,
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The X-Ray Spectrum (Changes in Voltage)
The x-ray spectrum has two distinct components
1) BremsstrahlungThe continuous spectrum is from electrons
decelerating rapidly in the target and transferring their energy to
single photons, Bremsstrahlung. Continuous broad spectrum
dependent on the applied voltage. E  eV
2) The sharp, intense lines (The
V  peak voltage across the X  ray tube
characteristic lines) are a result of
electrons ejecting orbital electrons
from the innermost shells. When

electrons from outer shells fall down
to the level of the inner ejected
electron, they emit a photon with an
energy that is characteristic to the
atomic transition. Characteristic spikes
max
p
p
in the graph are dependent on the target
material
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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
The maximum x-ray energy, and minimum wavelength results
when the electron loses all its energy in a single collision, such
that
hc
eV = hfmax = hc/lmin or therefore
lmin 
• This results in the continuous spectrum produced
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An electron is accelerated through 50,000 volts
What is the minimum wavelength photon it can produce when striking a target?
lmin
hc 6.6 1034  3 108


 .0248nm
19
eV 50000 1.6 10
 Compute the energy of the 1.5 [nm] X-ray photon.
E = hc/l = (6.6x10-34 [J s])(3x108 [m/s]) / (1.5x10-9 [m])
= 1.3x10-16 [J]
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example
The green line in the atomic spectrum of
thallium has a wavelength of 535 nm. Calculate
the energy of a photon of this light?
Planck's Constant h = 6.626 x 10-34 J • s
h•c
E =
λ
6.626 x 10-34 J  s x 3.00 x 108 m / s
E 
10-9 m
535 nm
nm
E  3.716 x 10-19 J
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example
18 kV accelerating voltage is applied across an X-ray tube.
Calculate:
a.The velocity of the fastest electron striking the target,
b.The minimum wavelength in the continuous spectrum of Xrays produced.
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NMR
10 um - 10 mm
700 to 104 nm
400 to 700 nm
10 to 400 nm
10-1 to 10 nm
10-4 to 10 -1 nm
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