Download Physics 150 Early quantum physics and photon

Physics 150 Early quantum physics and photon Chapter 27 Quan<za<on Towards the end of the 19th century, some physicists were seriously thinking that the “end” was in sight: •  Newton’s Laws were sufficient to understand mechanics •  Maxwell’s Laws were sufficient to understand electromagne<sm •  The work of Joule and Clausius and Maxwell and Boltzmann and others were sufficient to understand thermodynamics It seemed like there was nothing new leO to figure out. Well, there were s<ll a few observa<ons that didn’t seem to be described very well by those theories. For example, perfect blackbodies didn’t emit light in quite the way theory said they would. Physics 150, Prof. M. Nikolic 2 Blackbody radia<on An ideal (perfect) blackbody is an object that absorbs all the radiant energy that falls on it. The radia<on emiXed by a blackbody is a con<nuous spectrum characteris<c only of the temperature of the body. EmiXed radia<on is propor<onal to the temperature of the black body P ≈ T 4
−2
The radia<on also prefers some wavelengths to others λmaxT = 2.898 ×10 m ⋅ K
Radia<on curve would con<nue to increase with increasing frequency (decreasing wavelength) è blackbody should radiate infinite energy Physics 150, Prof. M. Nikolic 3 Blackbody radia<on and Planck Experimental results were very different from the exis<ng theory. In 1900, German physicist Max Planck published a mathema<cal model of the phenomenon that he considered more a curiosity than reality. Planck made an assump<on that the energy of light could only come in discrete quan<<es that he called quanta (singular quantum). Physics 150, Prof. M. Nikolic 4 Quan<za<on of energy Planck also proved mathema<cally that these discrete quan<<es of energy , quanta, are directly propor<onal to the frequency: E0 = hf
h = 6.626×10-­‐34 J s is called Planck’s constant. Planck’s revolu<onary discovery completely changed the fundamental ideas of classical physics and started a new approach to physics – quantum physics. It also won him a Nobel prize in physics in 1918. Physics 150, Prof. M. Nikolic 5 Photoelectric effect Under certain circumstances EM radia<on incident on a metal will eject electrons from the metal. This is the photoelectric effect. Experimental observa<ons: 1.  Brighter light causes more electrons to be ejected, but not with more kine<c energy. 2.  The maximum KE of ejected electrons depends on the frequency of the incident light. 3.  The frequency of the incident light must exceed a certain threshold, otherwise no electrons are ejected. 4.  Electrons are ejected with no observed <me delay regardless of the intensity of the incident light. Physics 150, Prof. M. Nikolic 6 Photoelectric effect Exis<ng wave theory (what we learned in previous chapters): •  EM waves carry energy •  This energy is absorbed by the metal target by electrons •  When electrons accumulate enough energy to escape, they are ejected However, the wave theory was unable to completely explain the results of the photoelectric effect experiment. Based on Planck’s findings, Albert Einstein proposed that EM radia<on is quan<zed Light is both an EM wave and a parJcle. The quantum of EM radia<on is called the photon. E = hf
Physics 150, Prof. M. Nikolic h = 6.626×10-­‐34 J 7 Explaining the experimental findings The amount of energy needed to break the bond between the electron and the metal is called the work funcJon (ϕ). Each metal has its own characteris<c work func<on. AOer the electron absorbs photon some of the energy is used to eject the electron and the rest goes into the KE of the electron. K max = hf − ϕ
Note that a par<cle (electron) can either absorb photon or not. You can not absorb part of the photon! 8 Physics 150, Prof. M. Nikolic Explaining the experimental findings Stopping potenJal (Vs) is the magnitude of the poten<al difference needed to stop even the fastest electrons. K max = qΔV = (−e) × (−Vs) = eVs
Physics 150, Prof. M. Nikolic 9 Explaining the experimental findings 1.  Brighter light causes more electrons to be ejected, but not with more kine<c energy. Par<cle theory predicts a more intense beam of light to have more photons so more electrons should be emiXed, but since the energy of a photon does not change with beam intensity, the kine<c energy of the ejected electrons should not change. 2. The maximum KE of ejected electrons depends on the frequency of the incident light. Par<cle theory predicts the maximum kine<c energy of the ejected electrons to show a dependence on the frequency of the incident light. Each electron in the metal absorbs a whole photon: some of the energy is used to eject the electron and the rest goes into the KE of the electron. Physics 150, Prof. M. Nikolic 10 Explaining the experimental findings 3. The frequency of the incident light must exceed a certain threshold, otherwise no electrons are ejected. Absorbed photon has to have enough energy to overcome the work func<on. K max = hf0 − ϕ = 0
ϕ
f0 =
h
4. Electrons are ejected with no observed <me delay regardless of the intensity of the incident light. Par<cle theory predicts a low intensity light beam will just have a low number of photons, but as long as f > f0 an electron that absorbs a whole photon will be ejected; no <me delay should be observed. Physics 150, Prof. M. Nikolic 11 Conclusions Some proper<es of light were explained by •  ParJcle theory •  Photoelectric effect •  Compton scaXering •  Pair produc<on •  Wave theory •  Interference paXerns (Young’s double slit experiment) Light is both an EM wave and a parJcle. Physics 150, Prof. M. Nikolic 12 Exercise: Photon energy A 200 W infrared laser emits photons with a wavelength of 2.0×10-­‐6 m while a 200 W ultraviolet laser emits photons with a wavelength of 7.0×10-­‐8 m. (a) Which has greater energy, a single infrared photon or a single ultraviolet photon? What is given: λIR = 2.0×10-­‐6 m λUV = 7.0×10-­‐8 m hc
E = hf =
λ
Shorter wavelength means greater energy. The UV photon has the greater energy è its wavelength is smaller. (b) What is the energy of a single infrared photon and the energy of a single ultraviolet photon? hc
EIR =
λIR
hc
EUV =
λUV
Physics 150, Prof. M. Nikolic 6.626 ×10 −34 J/f ⋅ 3×108 m/s
−20
EIR =
=
9.9
×10
J=0.62 eV
−6
2 ×10 m
EUV
6.626 ×10 −34 J/f ⋅ 3×108 m/s
=
= 2.8 ×10 −18 J=18 eV
−8
7 ×10 m
13 Exercise: Photoelectric effect Two different monochroma<c light sources, one yellow (580 nm) and one violet (425 nm), are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of 6.20×1014 Hz. (a) Are both sources able to eject photoelectrons from the metal? What is given: λy = 580×10-­‐9 m λv = 425×10-­‐9 m f0 = 6.2×1014 Hz First, convert wavelengths to frequencies: c= fλ
f=
c
λ
Frequency of the yellow light c
3×108 m/s
14
fy =
=
=
5.17
×10
Hz
−8
λy 580 ×10 m
fy < f0 è not able to eject electrons from the metal Frequency of the violet light c
3×108 m/s
14
fv = =
=
7.06
×10
Hz fy > f0 able to eject electrons from the metal −8
λv 425 ×10 m
Physics 150, Prof. M. Nikolic 14 Exercise: Photoelectric effect Two different monochroma<c light sources, one yellow (580 nm) and one violet (425 nm), are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of 6.20×1014 Hz. (b) How much energy is required to eject an electron from the metal? What is given: λy = 580×10-­‐9 m λv = 425×10-­‐9 m f0 = 6.2×1014 Hz K max = hf0 − ϕ = 0
ϕ = hf0
ϕ = ( 6.626 ×10 −34 Js) ( 6.20 ×1014 Hz) = 4.11×10 −19 J = 2.56 eV
Physics 150, Prof. M. Nikolic 15 Exercise: Photoelectric effect Two different monochroma<c light sources, one yellow (580 nm) and one violet (425 nm), are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of 6.20×1014 Hz. (c) What is the kine<c energy of the electron aOer using the violet source of light? What is given: λy = 580×10-­‐9 m λv = 425×10-­‐9 m f0 = 6.2×1014 Hz K = hfv − ϕ
K = 6.26 ×10 −34 Js ⋅ 7.06 ×1014 Hz − 4.11×10 −19 J
K = 3.16 ×10 −12 J = 0.2 eV
(d) What is the stopping poten<al? K = eVs
Physics 150, Prof. M. Nikolic Stopping poten<al is Vs = 0.2 V 16 X-­‐ray produc<on Reverse process from photoelectric effect •  Electrons move through a large poten<al difference – large kine<c energy (eΔV = K) •  High energy electrons impact a target and cause emission of EM radia<on (X rays) •  Electrons loose energy and slow down – bremsstrahlung (German for breaking radia<on) Physics 150, Prof. M. Nikolic 17 X-­‐ray produc<on Electrons can emit many photons as they slow down. Cutoff frequency – frequency of a single photon that takes all kine<c energy from the electron hfmax = K
X-­‐ray spectrum •  Con<nuous – bremsstrahlung •  Characteris<c x-­‐rays – characteris<cs of the target material Physics 150, Prof. M. Nikolic 18 Exercise: X-­‐rays What is the cutoff frequency for an x-­‐ray tube opera<ng at 46 kV? Cutoff frequency occurs when a singe photon takes all energy from electron. hfmax = K
We need to find kine<c energy of the electron. Electron speeds up in poten<al difference: K
fmax =
h
eΔV = K
K = 1e ⋅ 46kV = 46keV
h = 6.26x10-­‐34 J s = 4.136x10-­‐15 eV s 46 ×10 3 eV
19
fmax =
=
1.1×10
Hz
−15
4.136 ×10 eV s
What about photon’s wavelength? c = fmax λmin
Physics 150, Prof. M. Nikolic λmin =
c
fmax
19 Compton scaXering Another proof of par<cle nature of light. American physicist Arthur Compton observed that x-­‐ray photons sent into a metal would be deflected and have longer wavelengths aOerwards. •  Could not be explained by classical mechanics and wave nature of light Physics 150, Prof. M. Nikolic 20 Compton scaXering But if light was a par<cle, it could go under the elas<c collision with the electron (like billiard balls) and give part of its energy to electron Energy is conserved Ei = E f
E = Ke + E '
Photon energy before collision hc
hc
= Ke +
λ
λ'
Photon energy aOer collision Electron’s kine<c energy Electron acquires maximum kine<c energy when photon backscaXers (θ=1800). Physics 150, Prof. M. Nikolic 21 Compton scaXering Momentum is conserved ! !
pi = p f
x : p = p'cosθ + pe cos ϕ
y : 0 = p'sin θ − pe sin ϕ
Manipula<ng the previous expressions gives us the Compton shiP: h
λ '− λ =
(1− cosθ )
mec
h
= 2.426 pm
mec
Physics 150, Prof. M. Nikolic The Compton wavelength 22 Exercise: Compton scaXering A photon is incident on an electron at rest. The scaXered photon has a wavelength of 2.81 pm and moves at an angle of 29.5° with respect to the direc<on of the incident photon. (a) What is the wavelength of the incident photon? What is given: λ’ = 2.81 pm θ = 29.50 λ '− λ =
h
(1− cosθ )
mec
and h
= 2.426 pm
mec
h
λ = λ '−
(1− cosθ )
mec
λ = 2.81 pm − ( 2.426 pm ) (1− cos29.5°)
= 2.81 pm − 0.314 pm = 2.5 pm
Physics 150, Prof. M. Nikolic 23 Exercise: Compton scaXering A photon is incident on an electron at rest. The scaXered photon has a wavelength of 2.81 pm and moves at an angle of 29.5° with respect to the direc<on of the incident photon. (b) What is the final kine<c energy of the electron? What is given: λ’ = 2.81 pm θ = 29.50 And we found λ = 2.5 pm hc
hc
= Ke +
λ
λ'
K e = ( 6.626 ×10
−34
Ke =
hc hc
−
λ λ'
$
'
1
1
J/f ) ⋅ (3×10 m/s) &
−
)
−12
−12
% 2.5 ×10 m 2.81×10 m (
8
K e = 8.77 ×10 −15 J = 55 keV
Physics 150, Prof. M. Nikolic 24 Atomic spectra A hot, solid object will emit a con<nuous spectrum. A hot gas will show an emission or line spectrum. Electrically excited hydrogen gas Physics 150, Prof. M. Nikolic Emission spectrum of hydrogen (seen through a diffrac<on gra<ng) 25 Atomic spectra In 1885, Swiss mathema<cian Johann Balmer developed a strictly empirical formula to determine wavelengths of these 4 lines of hydrogen: "1 1%
1
= R $$ 2 − 2 ''
λ
# n f ni &
R = 1.097×107 m-­‐1 is the Rydberg constant ni = 2 nf = 3, 4, 5, … Later on, it was found that this equa<on gives wavelengths of all hydrogen lines (ni can be any number). Physics 150, Prof. M. Nikolic 26 Discovery of atomic nucleus Ernest Rutherford (1907) Atom – small, massive and posi<vely charged core (nucleus) – large shell filled with light electrons Physics 150, Prof. M. Nikolic Same charges repel and opposite charges aXract each other. 27 Rutherford’s model of atom Atom – small, massive and posi<vely charged core (nucleus) – large shell filled with orbi<ng light electrons A few serious faults to this theory: 1.  Electrons are nega<ve and protons are posi<ve è why don’t electrons spiral onto nucleus? 2.  According to Maxwell’s theory, accelera<ng charges radiate EM waves è electrons should loose energy and spiral onto nucleus. 3.  Why electrons are radia<ng EM waves only at specific wavelengths (Balmer’s discovery)? Physics 150, Prof. M. Nikolic 28 The Bohr model of the hydrogen atom In 1913, the Danish physicist Niels Bohr postulated the idea of atomic structure based on Rutherford’s experiments: 1.  The electron can exist without radiaJng energy only in certain circular orbits called staJonary states. The electron orbits have quan<zed radii, energy, and angular momentum. 2.  Newtonian physics applies to an electron in a sta<onary state. 3.  The electron can transi<on between sta<onary states through the emission or absorp<on of a single photon. The energy of the photon is equal to the difference between the energies of the two sta<onary states: ΔE = hf
Physics 150, Prof. M. Nikolic 29 The Bohr model of the hydrogen atom 4. The sta<onary states have quan<zed angular momentum in the amount h
Ln = n
= n!
2π
n is an integer and defines the sta<onary state at which you can find the electron The allowed radii are of Bohr orbits n 2! 2
2
rn =
=
n
a0
2
me ke
Where a0 is called Bohr radius Physics 150, Prof. M. Nikolic !2
a0 =
= 52.9 pm
2
me ke
30 Energy levels of hydrogen atom The state when n=1 is called the ground state – smallest radius and lowest energy: me k 2 e 4
−18
E1 = −
=
−2.18
×10
J= −13.6 eV
2
2!
States when n>1 are called the excited states. The energy levels of excited states are given by me k 2e 4 E1
En = − 2 2 = 2
2n !
n
Physics 150, Prof. M. Nikolic n = 1, 2, 3,.. n is called the principal quantum number 31 Energy levels of hydrogen atom The energy of a photon emiXed by an electron during a transi<on is "1 1%
hc
E=
= Ei − E f = −E1 $$ 2 − 2 ''
λ
# n f ni &
1
E1 " 1 1 %
= − $$ 2 − 2 ''
λ
hc # n f ni &
Balmer equa<on R=−
Physics 150, Prof. M. Nikolic E1
= 1.097 ×10 7 m −1
hc
32 Exercise: Hydrogen atom If an electron drops from the n = 4 level to the n = 1 level, what is the energy of both states and what is the energy of light emiXed? "1 1%
hc
E=
= Ei − E f = −E1 $$ 2 − 2 ''
λ
# n f ni &
Here is very important to define ini<al and final states: ni = 4 nf = 1 1
1
=
−13.6eV
= −0.85eV
2
2
ni
4
1
1
Final state, n=1: E f = (−13.6eV ) 2 = −13.6eV 2 = −13.6eV
nf
1
Ini<al state, n=4: Ei = (−13.6eV )
Energy of the emiXed photon: "1 1 %
E = E4 − E1 = −E1 $ 2 − 2 ' = −0.85eV − (−13.6eV ) = 12.75eV
# 1 4i &
When an electron absorbs an atom, ini<al state becomes the lower state and final state becomes higher state and E = Ef -­‐ Ei Physics 150, Prof. M. Nikolic 33 Problems with Bohr model •  The Bohr model correctly predicted the wavelengths of the spectral lines of hydrogen. •  The Bohr model cannot predict spectra of any element with more than one electron. •  Only hydrogenic atoms (ionized atoms that only have one electron; Li2+ for example) can have their spectra computed using the Bohr model. For a hydrogenic element with atomic number Z n 2 a0
The allowed radii are rn =
Z
The energy levels are Physics 150, Prof. M. Nikolic Z 2 E1H
En =
n2
E1H = -­‐13.6 eV 34 An<maXer MaXer è Composed of atoms with posi<ve nuclei and nega%ve electrons AnJmaTer è Composed of atoms with negaJve nuclei and posi%ve electrons (positrons) Physics 150, Prof. M. Nikolic 35 An<maXer •  Positron was discovered in 1932 •  The first ar<ficial an<-­‐atom was constructed in 1995 Every charged par<cle has an an<par<cle of the same mass and opposite charge. Even neutrons have an<-­‐neutrons. Physics 150, Prof. M. Nikolic 36 Pair produc<on A photon can interact with an atomic nucleus and change itself into an electron-­‐an<electron pair (or some par<cle-­‐an<par<cle pair.) An an<electron is called a positron. Energy and momentum are conserved E photon = Eelectron + E positron
The energy of the photon must be at least 2mec2 = 2x0.511 MeV = 1.022 MeV. If E > 2mec2 then the addi<onal energy goes into the kine<c energy of the electron-­‐
positron pair. 37 Physics 150, Prof. M. Nikolic Pair annihila<on When an<maXer meets the maXer they mutually annihilate each other conver<ng the equal amounts of maXer and an<maXer to energy (E=mc2) −
+
e + e → 2 photons
Eelectron + E positron = 2E photon
•  Gravita<onal force does not dis<nguish between maXer and an<maXer •  Both maXer and an<maXer emit the same light There could be galaxies made of an<maXer and we would not know! Physics 150, Prof. M. Nikolic 38 Exercise: Pair annihila<on A muon and an an<muon, each with mass 207 <mes greater than an electron, were at rest when they annihilated and produced two photons of equal energy. What is the wavelength of each of the photons? What do we know: Rest mass energy of an electron: mec2 = 0.511 MeV Rest mass energy of a muon is then: mpc2 = 207 x mec2 = 207 x 0.511 MeV = 105.8 Mev Emuon + Eantimuon = 2E photon
2E photon = 2 ( m p c 2 + K )
Since muons are at rest, kine<c energy K=0 E photon = m p c 2 = 105.8 MeV
Each photon has the same energy and the same wavelength: hc
E = hf =
λ
Physics 150, Prof. M. Nikolic hc
λ=
E
4.13×10
(
λ=
−15
eV s) ⋅ (3×10 −15 m/s)
106 ×10 6 eV
= 1.17 ×10 −14 m.
39 Exercise: Hydrogen atom A hydrogen atom in its excited state (n=4) emits a photon with the wavelength of 2000 nm. (a) What is the energy of the absorbed photon? What is given: ni = 4 λ = 200 nm Energy of the photon: hc
E = hf =
λ
6.626 ×10 −34 J/f ⋅ 3×108 m/s
−20
E=
=
9.9
×10
J = 0.62 eV
−9
2000 ×10 m
Physics 150, Prof. M. Nikolic 40 Exercise: Hydrogen atom A hydrogen atom in its excited state (n=2) absorbs a photon with the wavelength of 2000 nm. (b) To what energy level is the electron transi<oned? What is given: ni = 4 λ = 200 nm And we found E = 0.62 eV E = Ei − E f
ni = 4 ⇒ Ei = E2 =
E f = Ei − E
E1
42
E2 =
−13.6 eV
= −0.85 eV
16
E f = −0.85 eV − 0.62 eV = − 9.6 eV
But, we also know that: Ef =
E1
n 2f
Physics 150, Prof. M. Nikolic n 2f =
E1
Ef
n 2f =
−13.6 eV
−9.6 eV
41 Compton scaXering But if light was a par<cle, it could go under the elas<c collision with the electron (like billiard balls) and give part of its energy to electron Energy is conserved Ei = E f
E = Ke + E '
Photon energy before collision Momentum of a photon: hc
photon energy λ h
p=
=
=
c
c λ
Physics 150, Prof. M. Nikolic hc
hc
= Ke +
λ
λ'
Electron’s kine<c energy Photon energy aOer collision pc = K e + p'c
42