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61 views28 pagesGyuifiutfu 75 e 5 e
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© © All Rights Reserved
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Physics 30 Workbook Unit 5: Atomic Theory
UNIT 5:
ATOMIC THEORY Physics 30 Workbook Unit 5: Atomic Theory
A. QUANTUM NATURE OF EMR
AL. Photon Theory (EMR),
Up to the end of the 19" century, EMR was viewed as a wave, which was confirmed by its many
observed properties (e.g, refraction, diffraction, interference, and polarization). However, in the early
20th century, certain experiments dealing with light (as we will soon see) could only be explained by
viewing EMR as composed of tiny particles of energy (photons). Scientists came to realize that EMR
has both wave and particle properties, which is called wave-particle duality.
Energy of a Photon
According to photon theory, EMR is composed of bundles (quanta) of energy.
‘The energy of a photon (quantum of energy) is calculated as
Exvoon = hf where _f is the frequency of the EMR (in Hz)
h is Planck's constant (6.63 x 10° Js)
Further, using the universal wave equation, v=/A or Bf s
Epon = & where is the wavelength of the EMR (in m
Note: + High frequency EMR (such as UV, X-ray, gamma) has the highest energy photons
= this is why they are more likely to do damage to human cells / DNA
‘Since photon energies are so stall, they are often measured in electron-Volts (eV)
eV = 160x105
Inten:
(Brightness) of EMR ae :
Intensity is the energy of EMR. (es wm,
Classical (Wave) Theory:
Intensity = amplitude Bright EMR = High A Dim EMR = Low A
Quantum Theory: a
Intensity = Number of photons oee
Er =n Eyton i
where mis the number of photons Bright EMR = Large # Da eS
of photons of photons
P96 Physics 30 Workbook Unit 5: Atomic Theory
HOMEWORK (Photon Theory)
A. 1. Convert:
a) 273 GeV to Joules b) 9.12x10F to keV
2. Compare how classical and photon theory would describe the following kinds of EMR:
a) dim IR light b) bright UV light
3. Accertain blue light is created with a source vibration with a period of 1.44 x 10°" s.
‘What is the total energy of 500 photons (in eV)?
4, A.40.0 W monochromatic light source creates EMR with a frequency of 5.36 x 10"? Hz.
a) What is the wavelength of this EMR (in pm)?
b) Determine the maximum number of photons emitted by this light source per second?
B, 5. Compare red light (700 nm) and violet light (400 nm) in terms of speed, frequency,
and photon energy (in J). This is the range of visible light.
6. A certain type of EMR is shown by the
graph, How many photons of this kind iy)
of EMR are required to create (am)
a total energy of 3.978 x 10°!° J?
7. A120 W monochromatic bulb creates light with a wavelength of 5.60 x 10’ m.
‘a) What is the wavelength of this light in nanometres?
b) Whats the period of the source electrical vibration in the light bulb?
©) the bulb is 20.0% efficient, then determine the time required for this bulb to
emit 8,00 x 10” photons? Answer in minutes.
8. A Physics 30 conducts an experiment with a monochromatic bulb. He varies the frequency
of the EMR and measures the number of photons emitted. All other variables are held
constant. For this relationship, sketch the straightened curve, identify the equation of the
straight line, determine the significance of the slope, and provide the units for the slope.
SOLUTIONS
1. a) 273 x 10° eV = 437% 10° b) 57,000eV = 57.0 keV
2. a) Classical: small amplitude, long. Photon: Small number of low energy photons
b) Classical: high amplitude, short 2 Photon: Large number of higher energy photons
9444 x 10" Hz ; Epiorn = 2.875 €V ; Er= 1.44 x 10° eV = 1.44 keV
4. a) A=5.597x 10% m = 56.0 um
b) Assume 100% efficiency: Er= 403 ; Eptonn = 3.5537 x 10°" J ; n= 1.13 x 10” photons
5, Red: Speed of light ; f= 4.29 x 10" Hz ; Epiowon = 2.84 x 10° J or 1.78 €V
Violet: Same speed 30x10" Hz; Eptonon = 4.97 x 10°F or 3.11 eV
6. A=3.5 nm 5 Ephown = 5.6829 x 10" J ; n=70 photons
7. a) 560nm b) f=5.3571 x 10" Hz ; 87x 10° s
©) Eyton =3.5518 x 10" ; Er= 28,4143 ; AE = 142,0703 ;
8 Sketch mvs + ; Boustion: n= 2272
19.7 minutes
5 Units for slope: Hz or a
p97 Physics 30 Workbook Unit 5: Atomie Theory
A2. Photoelectric Effect (discovered by Heinrich Hertz, 1887)
~ when high frequency EMR causes electrons to be emitted from a metal surface
Device
- when high frequency EMR is shone on High fEMR
the metal cathode, photoelectrons are emitted
- they travel to the anode and form a
photocurrent that is measured by the ammeter
This is «called a photocell.
It is a unique source of electrical energy, ‘Cathode ‘Anode
and it is the basic idea behind a solar panel.
Classical Prediction
Electrons will be emitted only when they absorb enough energy.
‘The higher the intensity of the incident EMR, the faster the energy is delivered to the electrons
So, bright light would always emit electrons faster than dim light
This would be true for any frequency of EMR.
Actual Results (defied classical theory)
* Only high frequency (short wavelength) EMR can emit photoelectrons
~ the frequency must be greater than a minimum frequency, called the threshold frequency
Ifthe incident frequency is too low (f fo),
~ electrons are emitted
~ no matter how dim the EMR is
f Frequency
© The brighter the EMR, the greater the photocurrent produced
ing f> f
a Photo
current
f>f
bright light = high I
High |
ddim light = low I
Bright Intensity
p98 Physics 30 Workbook Unit 5: Atomic Theory
A3. Einstein's Explanation of Photoelectric Effect (verified wave-particle duality of EMR)
© Why must the incident EMR have a high frequency for electrons to be emitted?
- the surface electrons of the metal require a certain amount of energy to be emitted
This energy required to dig the electron out of the metal is called the work function (W)
- only one photon can be absorbed by an electron in the metal surface
= based on the equation Eyioo= Hf, the incident photon will only have enough energy to emit
electrons when the frequency is high enough
If ffa: The photon energy is greater then the work function
The photon is fully absorbed by the surface electron in the metal
~ part of the photon's energy is used to dig the surface electron out of the metal
~ the rest of the photon's energy is in the form of kinetic energy
Conservation of Energy: Incident
photon
En = Ew E=Wf
= hela
Eqvown = W+ Ek max
where Epioon = hf = ®
Note: The greater the frequency of the incident EMR,
- the greater the energy of the incident photon
- the greater the kinetic energy (speed) of photoelectron
‘© Why does bright EMR create more photocurrent?
- the more intense the light,
=> the greater the number of photons in the EMR.
= each photon is absorbed by an electron in the metal
=> thus, more electrons will be emitted
=> more photocurrent will be created
p99 Physics 30 Workbook Unit 5: Atomic Theory
HOMEWORK (Photoelectric Effect 1)
A. 1. The work function for cesium is 1.90 eV. What is the maximum wavelength that can emit
electrons from the metal surface (in nm)? What kind of EMR is this (approximately) ?
2. The minimum photon energy to emit electrons from a metal surface is 5.28 x 10°"? J.
a) What is the minimum frequency to emit electrons from this metal?
b) If'incident EMR with a frequency of 9.00 x 10'* Hz is shone on this metal,
‘then what is the maximum speed of the emitted photoelectrons?
3. Incident photons with 7.40 eV of energy are shone on a surface and the electrons are emitted
with a maximum speed of 1.12 x 10° m/s. What is the work function of the metal (in J) ?
Both A and B
4, The minimum frequency of light to emit electrons from a metal is yellow light.
Describe the resulting photocurrent (no current, low current, or high current) for the
following incident EMR:
a) dim vs bright indigo light b) dim vs bright orange light
5. Describe and explain the effect on a simple photocell when the incident EMR experiences the
following changes: a) increased frequency b) increased intensity
B. 6. When incident photons are shone on a metal surface, electrons are emitted with a maximum
Kinetic energy of 2.60 eV. If the minimum frequency to emit electrons from this metal
is 8.94 x 10 Hz, then what is the wavelength of the incident EMR (in nm)? Type of EMR?
7. ‘The maximum wavelength to emit photons from a metal surface is 428 nm. When a certain
type of incident EMR is shone on the metal, the electrons are emitted with a maximum
‘speed of 8.59 x 10° m/s, What is the frequency of the incident EMR?
8. When EMR js shone on a metal surface, it creates a photocurrent of 58.0 mA. What is the
‘minimum number of photons hitting the metal surface in 7,00 seconds? Assumptions?
9. White light is shone on a metal surface with a work function of 1.94 eV.
What is the greatest possible speed of the emitted electrons from the metal?
SOLUTIONS
1. fo = 4.5894 x 10" Hz 54 x 107 m = 654 nm, This is orange / yellow light.
2, a) W=5.28x 10S ; f= 7.96 x 10 Hz) Em = 6.87 10 F 5 Vnar = 3.88 x 10° mis
3. Ephoton = f= 1.184 x 10° I; Ek max = 5.7138 x 109 F ; W=6.13 x 10 T
4. a) f> fi, 80 electrons are emitted. Dim indigo = low current, Bright indigo = high current
b) f fi).
6. Ex mac = 4.16 x 10" J ; W= 5.9272 x 10"? ; A= 1.97 x 10m = 197 nm (UV)
7. W= 4.6472 x10 J Ey moe = 3.361 x 10S; f= 1.21 x 10" Hz,
8. g=0.406C ; n=2.54x 10"* photons (Each photon emits 1 e- ; all e- reach the other plate)
9. Shorter wavelengths have more photon energies. So, choose 400 nm.
W=3.104x 10S; f= 7.50. 10! Hz 5 Ex mar= 1.8685 x 10°F 5 vnax = 6.40 x 10° m/s
p.100 Physics 30 Workbook
A ikan's Verification of Einstein
- Millikan verified Einstein's Photoelectric
equation using the device shown,
‘The back voltage acts to slow down
‘the photoelectrons (decrease their Ei).
As the back voltage increases, the
deeper electrons do not have enough
Ex to reach the other plate (but the
surface e- still do). So, current decreases.
If the back voltage increases until the
current just becomes zero (called Vip),
even the surface electrons have lost all
their Zi by the time they reach the other plate,
Equatior
Vop is sufficient (barely) to reduce the surface
photoelectron’s max kinetic energy to zero.
AE, =-AE
So,
Exmax = 4 Vesey
where
Unit 5: Atomic Theory
Back voltage
(increased until = 0)
Voack
Ex max is the max kinetic energy of the surface electrons (in J)
is the magnitude of the charge of the photoelectron (in C)
Vooo is the minimum back voltage required to stop the photocurrent (in V)
Vo gave Millikan a way to measure the Ek max of the surface photoelectrons
Millikan's Experiment
Millikan discovered that there was a linear relationship
between E¢ mar and f
p.101
This verified Einstein's photoelectric
equation:
yroomxt b
Eima = h f - W
or
Wf=W+ Ex mee
Eplown = W+ Ekmas Physics 30 Workbook Unit 5: Atomic Theory
HOMEWORK (Photoelectric Effect II)
A. 1. The minimum potential difference to stop photocurrent is 3.60 V. What is the maximum
speed of the photoelectrons as they leave the cathode?
2. ‘The minimum energy to remove electrons from a metal surface is 2.08 eV. If the incident
EMR shone on the metal has a frequency of 1.67 x 10'° Hz, then what is the minimum
back voltage to prevent photocurrent?
Both A and B
3. A Physics 30 student investigated
the relationship between.
incident frequency and the
resulting Et mar of the oy
photoelectrons. The results
are shown inthe graph,
Using only the line of best-fit, determine:
a) the threshold frequency
b) Planck’s constant
©) the work function
Frequency (« 10" Hz)
4, When a certain type of EMR is shone on a metal,
the cut-off (stopping) potential difference
for a metal is 2.75 V. Ifthe minimum frequency to emit photoelectrons from this metal
is 4.35 x 10'* Hz, then what is the wavelength of the incident EMR (in nm).
5. The maximum wavelength to emit photoelectrons from a metal is 565 nm. Ifthe incident
photons have 6.70 eV of energy, then what is the minimum back voltage to stop the
photocurrent?
SOLUTIONS
1. Ex mac = 5.76 X10 F 5 Var = 1.12 x 108 m/s
2, W=3.328 x 10S 5 By mar = 7.7441 x 10°F 5 Voy = 4.84 V
3. a) fo=xint = 5.8 x 10! He,
b) b= slope = 6.7 x 10 Js
0) W=hfy3.9x 10
4. Ex mac= 4.40% 10° J 5 W= 2.8841 x 10S 5 4=2.73 x 107m = 273 nm
5. W= 3.502% 10 J; E=hf= 1.072 x 10 5 Ex mex = 7.1996 x 10S 5 Vorap = 4.50V
p.102 Physics 30 Workbook Unit 5: Atomic Theory
AS. Compton Effect (1923)
Photon Momentum
- according to classical theory, photons are pure energy and have no mass
- thus, they should have no momentum (p= mv = 0)
- but, Compton suggested that photons could have equivalent mass
‘That is, they can behave as if they do have mass and collide with other particles.
~ he calculated the momentum of a photon in the following way:
Since m and v=c ,itfollows that p=mv = (Z)o -£
e ©
E=pe | where Eis the photon energy (in Joules)
pis the photon momentum (in kg-m/s)
ccis the speed of light in air / vacuum
Based on this equation, high energy (ie. high frequency, short wavelength) photons have the
‘most momentum
eg, X-rays and gamma rays
‘Compton Effect
- provided evidence for photon momentum (irrefutable evidence that EMR behaves as a particle)
Experiment:
- directed high frequency (short 2) X-rays through thin carbon foil
emitted
UU
short 2 x-rays
C foil
longer A x-rays
deflected
Results;
‘© an electron was emitted (photoelectric effect)
‘+ most X-rays were deflected and they emerged at a longer This is called the
(ie. ata lower frequency, lower energy) Compton Effect
p.103 Physics 30 Workbos
ok
Unit 5: Atomic Theory
‘The results of the Compton experiment defied classical theory:
- there should be no deflection
- frequency (and thus wavelength, if the medium does not change) should never change once
the wave leaves the source
Compton's explan:
ation
He showed that the x-ray photon was having an oblique collision with an electron in orbit around
‘carbon atom. In this way, the photon was acting like a particle. This is illustrated below:
e _ emitted electron
Incident x-ray
photon
OVW &
a
© deflected x-ray
photon
‘Compton applied the laws of conservation of energy and momentum to this collision.
Conservation of energy
Eony
= Ex) + Ee
Conservat
Br =r
Baoay = Be + Bay
jomentum
This would be a 2-D
vector analysis
Compton combined these two laws and derived a formula to predict the change in wavelength of
the x-ray photon:
a + (~coso)
me
where
Ade
"— A. is the change in wavelength of the x-ray photon (in m)
‘mis the mass of the particle (usually an electron) hit by the photon (in kg)
‘his Planck’s constant (use 6.63 x 10™ J's)
cris the speed of light
@ is the angle of deflection of the x-ray photon (w.r. its initial direction)
Quantum nature of EMR is conclusive
Because the Compton effect could not (in any way!) be explained by classical theory, most,
scientists became convinced of the particle nature of EMR. Physies 30 Workbook Unit 5: Atomic Theory
HOMEWORK (Compion Effect) For the questions below, ignore relativistic effects
Both A and B
1. A5.80 pm photon deflects off
a stationary electron, as shown, BS) an iowa
Determine the wavelength of the
deflected x-ray photon.
‘Answer in pm.
‘A. 2. A photon has 450 MeV of energy.
a) What is its wavelength? b) What is its momentum?
3. What wavelength of photon would have the same momentum as an alpha particle travelling
at 5.20 x 107 m/s?
4, Sketch the vector triangle that illustrates conservation of momentum.
An x-ray photon, moving towards the South, collides with a stationary electron.
After the collision, the electron is travelling East.
5. A 2.80 pm x-ray photon deflects off a proton (initially at rest). If the proton has a final kinetic
energy of 4.33 x 10 J, then determine the wavelength (in pm) of the deflected x-ray photon.
6. 4.00 x 10’ m (i.e. 4.00 fin) x-ray photon is moving East and collides with a
stationary proton. After the collision, the x-ray photon is travelling due South. Determine:
) the frequency of the x-ray photon after the collisi
') the kinetic energy of the electron after the collision
©) the proton’s angle of deflection and final momentum
B. 7. A photon has a momentum of 5.07 x 10 kg-m/s.
a) What is its frequency? b) How much energy does it have (in keV)?
8. How fast must a proton be travelling to have the same momentum as a 700 keV photon?
9. When an x-ray photon deflected off an electron, its wavelength increased by 1.70 pm.
Determine the x-ray photon’s angle of deflection (w.r:. its initial direction).
10. Sketch the vector triangle that illustrates conservation of momentum:
A photon, travelling West, collides with a stationary electron. After the collision,
the electron is moving at 20° N of W and the photon is moving at 50° S of W.
11. A 2.10 pm x-ray photon, moving East, collides with a stationary electron. After the collision,
the x-ray photon is moving at 75.0° N of E. Determine:
2) the wavelength (in pm) of the x-ray photon after the collision
) the kinetic energy of the electron after the collision
©) the electron’s angle of deflection
p.105 Physics 30 Workbook Unit 5: Atomic Theory
12, Compare and contrast the Compton effect with the photoelectric effect.
SOLUTIONS
if = 110° ; AA=3.26pm ; 2=9.06 pm
2. B=72x10"y a) A=2.76x10% m b) p=2.40x 10" kg-m/s
3. pa=3.458x 10? kgm/s ; A=1.92x 10m Be
4. See diagram, .
Prony
5. Esray= 7104 x 103 ; Cons of Energy: py = Ek py +E pay
So, E hry = 2.775 x 105 ; 2'=7.17 pm
6 a) O=90° p,.AA=1.3234 fm ; A= 5.32 fm ; f=5.64* 107 Hz
b) Conservation of E: E,= 1.24 10 Prom
¢) Cons ofp: Prny= 1.65810 kg-mi/s 5 pry = 1.245 x 10° kg-m/s =
Solving the triangle, p%, = 2.0710’ ke-m/s ; 0 = 36.9° Pree
7 a) 2.29% 10" Hz b) 1.521 x 10° J = 9.50625 x 10° eV = 951 keV
8. E=112x10°F 5 pproon = 3.7333 x 10 kemm/s ; v= 2.24 x 10° m/s
9. 726°
10. See diagrem. Pe
20°
11, a) AA=1.798pm ; 4=3.90 pm
b) Conservation of Energy: £,=4.37 x 10"
157% 107 kg-m/s 3 Pray = 1.701 x 107 kgsmis
a Prony = Piggy COSTS? + Bi, Bl, = 2.717 «10 ke-m/s
O= ply Sin75°+ Pi, BL, = 1.643 x 107 kg-m/s
Solving the triangle, @ = 31.2°
12. Photoelectric Effect - the incident photon is fully absorbed by the electron, the electron is emitted,
and the electron keeps the rest of the energy as kinetic energy
‘Compton Effect - the incident x-ray photon is only partially absorbed, emitting the electron and
giving the electron kinetic energy. The rest of the photon energy is reemitted as an x-ray
photon with less energy (lower f, longer 2).
106 Physics 30 Workbook Unit 5: Atomic Theory
B. MODELS OF THE ATOM
BL. Barly Models of the Atom
rd Ball Model (1810)
all matter is composed of atoms
= atoms are the smallest particles (quanta) of matter @ atom
~ there exist no subatomic particles
Dal
Advantages: Explained chemical reactions very well
‘* law of conservation of matter (the number of atoms remains constant)
‘the nature of elements (each element has a unique atom)
‘the nature of compounds (the law of multiple proportions: H30 , CO:)
Disadvantages: It could not explain:
‘* why molecules form (what holds them together?)
‘the repeated properties of elements on the Periodic Table
Cathode Rays (Crookes, 1879) Higa
In a cathode ray tube (CRT), two parallel plates were placed
within a glass tube and attached to a high voltage source. 1 | | | |
‘When the gas pressure was reduced inside the tube, coloured = +
rays were observed moving from the negative plate (cathode) ss
to the positive plate (anode). | >|
‘Thus, these rays were called cathode rays.
Cathode ‘Anode
Were cathode rays particles or waves (EMR)?
Scientists showed conclusively that cathode rays were discrete negative particles, and not EMR.
How?
Cathode rays deflected in magnetic fields and electric fields as a negative charge would.
IMR does not deflect in either field
ee ee
cathode cathode _
rays rays is
ona)
Since this obeys LHR #3, cathode Only negative particles would attract
rays are negative particles to the positive plate and repel from
the negative plate
p.107 Physics 30 Workbook Unit 5: Atomic Theory
‘Thomson's Raisin Bun Model (1897) Also called the "Plum Pudding model"
The discrete negative particles inside cathode rays were called electrons by J.J. Thomson. When
electrons were emitted by the cathode, there was no change in mass or deterioration of the cathode
metal. This was clear evidence that electrons were particles smaller than atoms, which defied Dalton's
‘model. Thus, a new model was needed to account for the existence of electrons inside the atom,
‘Thomson suggested that the electrons were distributed evenly in a positive fluid, much like raisins
ina bun,
Electrons Positive Fluid
~ embedded in the
positive fluid
- distributed evenly
= very tiny mass
- uniform density
However, the raisin bun model needed to be tested. This was done by Ernest Rutherford,
Rutherford's Scattering Experiment (1911 - 1913)
Rutherford tested the Thomson model of
the atom by firing alpha particles through
very thin gold foil
- most of the mass and volume
~ equal but opposite charge as
electrons (if neutral atom)
5K
Prediction: as
Thomson's model predicted that the gold
atoms were of uniform density.
Asa result, there should be no deflection,
The alpha particles should go straight lead
through the foil (close to 0°)
‘lbha paroles
Actual Resul:
most (99.99%)of the alpha particles went through with little deflection (< 10°)
however, | in 10,000 were deflected by more than 10°
* very rarely, an alpha particle even deflected 180°
‘Thomson's model had no explanation for the significant deflections (> 10°).
p.108
— fuorescent zine
suiphide screen
~ongle of detection Physics 30 Workbook Unit 5: Atomic Theory
Rutherford’s Planetary Model (1913)
Rutherford postulated that most of the atom’s mass is contained in a tiny nucleus, and this nucleus
had a very large positive charge. To account for the volume of the atom, Rutherford suggested that the
electrons were in orbit around the nucleus.
Empty Space
a ~ takes up 99.99% of the
Electrons + atomic volume
~ in orbit around
the positive nucleus
‘Nucleus
- tiny, dense, positively-charged
~ contains most of the atomic mass
Rutherford’s planetary model well explained the results of the scattering experiment:
© Most a-particles went through with no deflection
Why? The gold atoms are mostly empty space. The a-particles move through with no
interaction with the gold atoms.
«A few scattered at significant angles (~10°)
Why? a-particles that get close to the nucleus are
deflected by electric repulsion (like charges)
The closer to the nucleus, the greater the deflection,
‘This was verified using Coulomb’s law.
Note: If the a-particle travelled “head-on”,
it would reflect straight backward,
Major Problem with the Planetary Model
ee
~ if the electron is in orbit around the nucleus, then it
is constantly accelerating (centripetally)
- an accelerating charge should emit EMR (Maxwell) ®
- thus, the electron should constantly be losing energy
- it would spiral in to the nucleus in 10° s!
Emits EMR
ofall
frequencies
p.109 Physics 30 Workbook Unit 5: Atomic Theory
2, Properties of the Electron (Charge and mass)
1 Charge-to-Mass Ratio of the Electr jomson, 1897)
Thomson used the modified CRT below to investigate the properties of the electron:
f electric field plates
Lyjifi|4 a IN
large potential difference cols fo produce magnetic field
Step 1: Thomson turned on both fields so that the cathode rays went through undeflected
‘Newton's 1 Law: (balanced forces)
Fé CSS
FFl-l ~ fe
qvB. = q Al e—>- t te
vB.= |B) sor xB,
SS
Using this approach, Thomson was able to determine the speed of the electrons
Step 2: Thomson then tured off the electric field,
so that there was only a magnetic field Fn
- the electron would then be deflected into x Ee re
uniform circular motion =
x x\ xB
1
1
x xr x
:
x -7x x
Fn
p.110 Physics 30 Workbook Unit 5: Atomic Theory
Step 3: Thomson then combined the two equations to find the charge-to-mass ratio:
aa
1.76 x10"! C/ kg.
@
m
2. Charge of the Electron (Millikan, 1913)
- Recall, Millikan suspended oil droplets
between charged parallel plates
‘Newton's 1" Law: (balanced forces)
So,
~he discovered two things:
© the smallest possible charge was 1.6 x 10~'® C, which he called the elementary charge
all other oil-drop charges were multiples of the elementary charge
ie.
qzne
where eis the elementary charge (1.6 x 10°C)
nis the number of electrons in excess or deficit
- Millikan believed that the elementary charge was the charge of an electron (magnitude)
Charge of an electron = 16x 10°C (magnitude)
3. Mass of the Electron
- combining the results of Thomson and Millikan, the mass of the electron could be determined
4 = 1.76 « 10" Ckg
m
a = STE = 91 107g
176x10" Ge 1.7610" i
pil Physics 30 Workbook
HOMEWORK (Charge-to-mass Ratio)
Atomic Theory
A. 1. A positively-charged particle travels undeflected
through mutually perpendicular electric and x x x x B
‘magnetic fields, as shown. If the charge is
travelling at 6.80 x 10° m/s and the magnetic field
strength is 770 mT, then find the magnitude and
direction of the electric field between the plates. _———s
2. Analpha particle is entering a perpendicular magnetic
field at a speed of 4.0 x 10° m/s. If the magnetic . . a bg
field is 290 mT acting out ofthe page, then determine og? ———__>
a) the radius ofits circular path (in em)
+) whether the path is clockwise or counterclockwise see
3. A charged particle enters a 75.0 mT magnetic field and goes into circular motion with a
diameter of 14.0 cm. If it is moving at a speed of 8.10 x 10° mis, then find
its charge-to-mass ratio.
4. A proton is accelerated from rest through a potential difference of 2.7 kV.
What is its final speed?
An oil droplet, with a weight of 2.80 x 10°" N, =~; -=—_—_
is suspended between two parallel plates 60cm 1
a) What is the charge of the oil drop? cE
Is it positive or negative? *q TT 700V
b) Find the number of electrons in 6.0 =f
excess or deficit. ———
6. For the following, sketch the straight-line graph, provide the equation of the line,
and determine the physical significance of the slope (including its units):
Millikan’s oil-drop experiment: Potential difference as a function of charge
for suspended oil-drop
B. 7. A negatively-charged particle travels undeflected
through mutually perpendicular electric and —_! + plate
‘magnetic fields. The charge is travelling at a speed
of 1.80 x 10” m/s. If the voltage between the plates ee eeeeeeeeeeeeeeeeeeeceeee
is 500 V and they are 8.00 cm apart, then what
is the magnitude and direction of the magnetic field? ——— - plate
p.ll2 Physics 30 Workbook Unit 5: Atomic Theory
8. A particle with a charge-to-mass ratio of 1.33 x 10° C/kg enters a 70.0 mT perpendicular
‘magnetic field and goes into circular motion with a radius of 16.0 mm. What was its speed?
9. An electron first travels undeflected through a region with mutually perpendicular electric
and magnetic fields (|E| = 7.00 x 10 Vim and B, =2.37 mT). Then, it enters a
perpendicular magnetic field (2 = 5.00 mT) and goes into circular motion.
What is the diameter of the circle?
10. A charged particle is accelerated from rest through a potential difference of 950 V and
it attains a final speed of 1.80 x 10” m/s. What is the particle’s charge-to-mass ratio?
(Note: Can you show that the units will become C/ke?)
11. An alpha particle is accelerated from rest through a potential difference V. When it then enters
a perpendicular 760 mT magnetic field, it goes into circular motion with a diameter
of 6.00 cm. What is the potential difference ¥? Answer in kV,
12. An oil drop has 7 electrons in excess and it has a mass of 5.40 x 10°" kg.
Ifit is suspended between two parallel plates that are 11.0 cm apart, then what is the
required potential difference between the plates?
13. For the following, sketch the straight-line graph, provide the equation of the line,
and determine the physical significance of the slope (including its units):
‘A charged particle being deflected into circular motion by a perpendicular
magnetic field: Radius of citcular motion as a function of speed
SOLUTIONS
1. 5.24 10° V/m ; Fy acts up, F, acts down, and so E acts downward (4)
2. a) 29cm b) Fy acts down as it enters the field, so it goes clockwise
3. 1.54% 10° Cikg
4. 7.2.x 10° mis
5. a) +4.80x10"C b) 3 electrons in defic
1 k
6 Graphars © 5 ¥ SHORE Slope is mgd ; Units forslope: V-C, Nam, J, or “27
s
7. 347x10*T ; Fracts up, Facts down, so B acts into the page (x)
8. 1.49 10° mis
9, v=2.9536 x 107 m/s ; r=3.36 om ; diameter = 6.73 cm
10. 1.71 x 10"' Cikg.
11, v= 1.0971 x 10° m/s ; AV=1.
12. 520V
13. Graph rvsv ; r=(slope)v 5 Units for slope: —"—or s
9B, te
p.ll3 Physics 30 Workbook Unit 5: Atomic Theory
‘83. Spectra and the Bohr Model
Emission and Absorption Spectra
~ there are three types of spectra:
1. Continuous Spectra (dense, heated material)
~ heated solids, liquids, and dense gases
cemit light with a continuous spectrum
Hot object prism
~ the spectrum contains all frequencies
(colours), with no gaps
- EMR is due to the collisions between atoms
2. Emission Spectra
- low pressure (rarefied) gases exposed to high voltage give off specific colours (e.g. neon signs)
— the spectra show a discrete number of bright bands
eg.
Low pressure,
excited hydrogen gas
< pox
- emitted EMR is due to the excited atoms themselves
~ each element (atom) has a unique emission spectrum (spectral signature)
~ classical theory could not explain why most frequencies are missing
3. Absorption Spectra
~ when white light passes through unexcited gas, there are dark lines (missing frequencies)
in the spectrum
Hot object Unexcited hydrogen gas
** A gas absorbs the same frequencies of EMR that it emits **
p.lld
< won
<"waxon Physics 30 Workbook Unit 5: Atomic Theory
HOMEWORK (Emission and Absorption Spectra)
1, Identify the type of spectrum associated with each of the following descriptions:
4) low-pressure neon gas is excited by current and gives off light
b) the light from the sun, after it goes through its (relatively) cold atmosphere
©) the light emitted from a hot metal
4) the entire spectrum of colours are present, but there are a few dark lines
) all frequencies of EMR are present
f) there are a few bright lines, but the rest are missing
Use the information below to answer the next 2 questions.
‘The emission spectra for 4 common elements are shown below:
650 600 550 x 450 490 | Wavelength (am)
| | I | Hydrogen (11)
ll ‘Sodium (Na)
| | Ae Helium (fe)
I | l I | Mercury ci)
2. Determine the identity of the gases in the samples below, based on the spectra:
650 6 550 500 450 400
> | WH Pee
» | TE i Ht |
o Il Wl al
SOLUTIONS
1. a) emission b) absorption) continuous d)_ absorption
€) continuous) _ emission
2. a) HeandNa —b) Hand Hg ©) H,He, and Na
polls Physies 30 Workbook Unit 5: Atomic Theory
Bohr Model of Hydrogen (1913) Combined classical and quantum theory
‘Orbital radii and energies are quantized
Electrons can only be located at certain radii
(but not in between). Ifthe electron remains at an
allowed orbit, it will not emit EMR despite its
acceleration. The lowest possible orbit (n= 1) is
lowest energy state (called unexeited or ground state),
which is the most stable orbit. All other orbits are
called excited states, and they are unstable. Outside
Each allowed orbit (called a stationary state) has atom
a required energy level, The electron must have the (free)
‘exact energy to be there. The energy levels Bohr n=0 ev
calculated are shown in the energy diagram: a ae
Note: When the electron goes beyond n = co, Taide a -151eV
it leaves (ionizes) the atom with kinetic ‘tom
energy. The minimum energy to remove ‘ n=2 340 ev.
‘an electron from the atom (Le. to get it to Go ort
n= 00) is called the ionization energy.
n=l -13.6eV
‘* Electrons are in uniform circular motion (orbits)
‘The electric force of attraction between the proton (nucleus)
and the electron is what holds the electron in orbit. That is,
the electric force is the centripetal force.
Fe = ma.
© Energy Transitions
- EMR is emitted or absorbed only when the electron jumps orbits (quantum leap)
Upper Level «
e) @ Photon
WW @ Photon absorbed
Lower Level emitted (exact energy)
(&)
Based on the law of conservation of energy, the energy gained or lost by the electron
(called the transition energy AE) must be equal to the photon energy.
- Et
AE = Epioton where AE.
pll6 Physics 30 Workbook Unit 5: Atomic Theory
Bohr’s Explanation of the Hydrogen Spectra
Light from
excited H atoms
Hydrogen Emission Spectrum
R
= Rk 656.20m)
= 6 (86.1 am)
— 5 340nm)
— V (410.1 nm
ty
Why does excited, low pressure hydrogen
gas emit only certain frequencies of EMR?
Fins, the high voltage creates current within the cathode ray tube. The electrons
collide with the hydrogen atoms and excite them to higher energy levels
Since the excited energies level are unstable, the atoms will drop to lower energy
levels. When they drop levels, they lose energy and emit photons.
Bohr was able to show that each bright line of the emission spectrum is a unique energy drop
mat: OV
0.378 eV
a 0.544 eV
im, =2: Visible 0.850 eV
Red: 32 | 8
Green: 492 -
Blue: 52 u8 1Slev
Violet 6->2 Ik
ae n=2 *) -3.40 eV
a ANLIR
ms
otc
, ae
n=l -13.6eV
sorption Spectrum
‘Bohr was also able to explain why a gas absorbs the same frequencies of EMR that it emits.
To illustrate, Bohr knew that when a hydrogen atom drops from =4 to n=2, it emits a green
photon with an energy of 2.56 eV (= 486.1 nm). Clearly, in order to raise the hydrogen atom
from n= 2 to n=4, it requires the same amount of energy (2.56 eV). The only photon capable
of providing this exact energy is the same green photon.
‘The same reasoning applies to every other wavelength emitted (and absorbed) by hydrogen.
Evaluating Bohr's Model
It could explain the emission and absorption spectra for hydrogen (and any other I-electron atoms)
It could NOT explain - the spectra for more complex atoms (with more than one electron)
- why only certain orbits were allowed
- why the orbiting electrons did not emit EMR, as predicted by Maxwell
- why some spectral lines were brighter than others
pl? Physics 30 Workbook Unit 5: Atomic Theory
HOMEWORK (Bohr's Model)
1. An electron in Bohr’s H model has a radius is 3.3856 nm, What is the orbital speed?
2. If the orbital speed of the electron in Bohr’s hydrogen model is 2.185 x 10° m/s, then.
what is the orbital radius?
For #3 - 9, use the energy diagram for the Bohr hydrogen model shown.
3. An electron in the Bohr hydrogen model drops from 7 Oev
to n=3, What is the wavelength of light emitted? 0278 eV
Which type of EMR is this (IR, visible, or UV)? 0378 eV
4, Anunexcited hydrogen atom absorbs a photon and it rises 0.544 eV
to n= 4, What is the frequency of the incident photon?
0.850 eV
5. An electron in the Bohr hydrogen model drops from n= 6
and emits a photon with a wavelength of 2.618 um.
To which lower level did it fall? nas -151eV
6. An unexcited hydrogen atom absorbs a photon with a
frequency of 3.154 x 10" Hz. To what level does it rise? n 3.40 eV
7. An electron is at m= 5 in Bohr’s hydrogen model.
What is the maximum wavelength of EMR to ionize this atom?
n -13.6eV
8. Describe and explain what happens when an unexcited hydrogen
atom receives low energy incident photons with the following energies,
@) 121eVv (b) 11.0eV © 30ev
9. Identify the transition in the Bohr hydrogen model associated with the following events.
‘Then, using the energy diagram for hydrogen, verify that the proper wavelength results.
a) emission of violet light (410.1 nm) b) absorption of red light (656.2 nm)
10. Which type of EMR is emitted for each transition?
a) 733 b) 41 @ 3-42 d) 932
SOLUTIONS
1, 2.73 x 10° m/s
2 r=5.2915x 10° m
3. Epioton = 1.232 eV ; A= 1.01 um (IR, since 2 > 700 nm)
4. Ephotn = 12.75 eV ; 08 x 10'° Hz
5. Eptoion = 0.4744 eV ; E,=—0.8522 eV ; m=4
6. Epon = 13.058 eV ; Ey=—-0.542eV ; my =5
7. Fromn=Ston=0 ; Eptown = AE. = 0.544eV ; = 2.28 pm
8. (a) The photon is fully absorbed, It has the exact energy to raise the atom ton = 3
(b) The photon is not absorbed. It does not have the exact energy required.
(c) The photon is fully absorbed and it ionizes the atom.
The electron leaves the atom with 16.4 eV of kinetic energy.
9% a) 6-2 b) 233
10. a) Infrared b) UV c) Ble d) UV
p.ll8, Physics 30 Workbook Unit 5: Atomic Theory
‘B4. Quantum Mechanics
Matter Waves (de Broglie, 1923)
Compton showed conclusively that high frequency EMR can have momentum and can thus
behave like particles. de Broglie suggested, due to a belief in symmetry, that matter could
show wavelike properties. This introduced the idea of wave-particle duality in particles.
He combined the photon formula _p -4 with the momentum formula for a particle p= mv
oe Note: Works only if v<