Section 1: Origins of QM

Question 1

Two equivalent formulations of quantum mechanics

Answer 1

1. wave mechanics
2. matrix mechanics

Question 2

blackbody

Answer 2

perfect absorber and emitter of radiation

totally absorbs all ration that falls upon it

is in thermal equilibrium so has temp T

Question 3

a blackbody is in thermal equilibrium so has

Answer 3

a temperature T

Question 4

emitted radiation depends only upon

Answer 4

the radiator’s temperature

follows stefan-boltzman law R=sigma T^4

Question 5

typical classical model for a perfect blackbody

Answer 5

is a black cavity with a small hole: all the light entering is reflected multiple times across the black walls and is (almost completely) absorbed.

Since the cavity is in thermal equilibrium, the emitted radiation depends only on its temperature, so the cavity emits like a black body.

Question 6

total emittance obtained by

Answer 6

integrating over all wavelengths

Question 7

wavelength (or frequency) monochromatic energy density

Answer 7

u(λ,T) (or u(v,T))

Question 8

since u(λ,T)=4pi/c R(λ,T) - from a measurement of the emission spectrum one could

Answer 8

determine the energy density

Question 9

samuel langley and his bolometer

Answer 9

Ohhhh…
Langley invented the bolometer,
A very fine sort of thermometer.
It can measure the heat,
Of a polar bear’s feet
At the distance of half a kilometer.

Question 10

From general thermodynamical arguments, in 1893 Wien showed that the black-body energy density had to take the form

Answer 10

u(λ,T)=dE/dλ - λ^-5 f(λT)

or in terms of f:
u(v,T) = dE/dv = v^3g(v/T)

Question 11

f(λT) and g(v/T) are

Answer 11

universal functions that depend solely upon λT and v/T respectively and cannot be derived thermodynamically.

Have to be found empirically

Question 12

u(v,T) = dE/dv = v^3g(v/T) is called

Answer 12

Wien’s Law (NOT wien’s displacement law!!)

Question 13

wien’s law is a good model for

Answer 13

the observed spectrum at short wavelength (high freq)

hc&raquo_space; kbTλ

not accurate at high wavelengths

Question 14

explicit form of wien’s approximation for the spectral emittance

Answer 14

from relation between spectral emittance and energy density

planck later introduced fundamental constants to the approximation

Question 15

problem’s with Wien’s approximation

Answer 15

This model works well at small wavelengths but it is not very good at high wavelengths - low frequencies.

Thermodynamical reasoning is not sufficient to derive an accurate model.

Question 16

Wien’s displacement law

Answer 16

λmax = b/T

Question 17

Wien’s displacement law derivation

Answer 17

1. start from energy density in wien’s law - peak by imposing du/dv=0
2. define x=v/T

3.only depends on x, not v or T individually so do not need o solve explicitly

only get proportionality, cannot solve for constant

Question 18

The Rayleigh-Jeans law is a good approximation to the observed spectrum at

Answer 18

long wavelengths - low freq

hc«kbTλ

Question 19

big problem with the Rayleigh-Jeans law

Answer 19

UV CATASTROPHE

the radiance keeps increasing indefinitely at short wavelengths (higher frequencies). If it were true, the power emitted at short wavelengths would be infinite!

Question 20

the UV catastrophe indicates the

Answer 20

failure of classical physics to explain the behaviour of thermal radiation

Question 21

Planck was able to derive a spectrum that fits the observed data of the blackbody spectral emission by assuming

Answer 21

that energy comes in discrete quanta of energy

E=hv

Question 22

graph to compare planck, wien and RJ

Answer 22

plot spectral radiance against wavelength

wien follows standard curve
planck agrees on far side, but increased radiance slightly

RJ off to the right and displaying UV catstrophe

Question 23

According to the classical theory, the average energy per mode can be obtained starting from the

Answer 23

maxwell boltzmann distribution and using the classical equipartition of energy

Question 24

demonstrating the result of planck’s model

Answer 24

1. start with LHS, sub in x=e^-hv/kbT

2, use geometric series to reach RHS

Question 25

In classical physics the occupation of each mode was equally possible, but since modes are quantised and each quanta has energy E=hv, exciting higher modes is

Answer 25

less probable as it requires more energy

Question 26

the probability that a mode will be occupied is given by

Answer 26

the Bose-Einstein distribution function

Question 27

The classical theory stated that each mode needed equal energy of kbT to be excited, in accordance with the equipartition of energy. However, the average energy per “mode” (or “quantum”) is given by

Answer 27

its energy hv times the probability that this will be occupied

Question 28

The energy density is then given by the

Answer 28

number of modes, per unit volume, per unit frequency, times the average energy per mode

Question 29

understanding the planck distribution - for small hv/kbT

Answer 29

the discrete steps are small enough that the distribution is pseudo-continuous: recover classical Rayleigh–Jeans form

Question 30

understanding the planck distribution - at large hv/kbT

Answer 30

the first non-zero mode is highly Boltzmann-suppressed: finite energy in the cavity solves the ultraviolet catastrophe!

Question 31

photoelectric effect - what is expected classically

Answer 31

if we treat light as a wave, we know that its energy depends on the intensity, and therefore on the amplitude of the EM wave.

Question 32

Simulated experiment 1: changing the material and the wavelength

Answer 32

Nothing happens until the wavelength reaches a certain threshold where then electrons start being emitted If we change the material, the critical frequency changes

Question 33

the more we increase the frequency...

Answer 33

the faster the electrons move - this means they have more kinetic energy

Question 34

Simulated experiment 2: changing the intensity and the voltage - null voltage

Answer 34

the current increases with increasing intensity, and in fact we can observe that more electrons are released and all of them reach the collector at higher intensities.

Question 35

Simulated experiment 2: changing the intensity and the voltage - positive voltage

Answer 35

now the electrons are moving faster because of the potential, and the system reaches a saturation current

Question 36

Simulated experiment 2: changing the intensity and the voltage - negative voltage

Answer 36

increasingly higher proportions of electrons start travelling towards the collector but come back, as now the potential is opposed to the initial motion of the electrons.

Question 37

Since the intensity changes the number of electrons emitted and therefore the photocurrent, at a fixed negative voltage the photocurrent will

Answer 37

decrease with decreasing intensity

Question 38

regardless of light intensity, we can eventually reduce the photocurrent to 0 with

Answer 38

a negative stopping voltage V0

Question 39

The stopping voltage decreases linearly with

Answer 39

the light frequency, until it reaches 0 there is no photocurrent below a threshold frequency

Question 40

Electrons are emitted when

Answer 40

a work of at least the binding energy is paid to free them

Question 41

minimum work that needs to be done to emit electrons is

Answer 41

Eb=hv-W

Question 42

minimum threshold frequency necessary to eject electrons

Answer 42

vth = W/h

Question 43

the additional energy results in

Answer 43

kinetic energy which determines a maximum possible velocity

Question 44

The intensity determines the

Answer 44

number of photons present in the light sent to the metallic plate

Question 45

higher intensity means that

Answer 45

more electron can be extracted (if freq above vth)

Question 46

bohr model postulates

Answer 46

e- move in quantised circular orbits depending on newton and coulombs laws must emit/absrob quanta of energy to move orbits atomic angular momentum quantised

Question 47

Why did the quantum and classical results agree in atoms with large mass number?

Answer 47

Because in large systems classical physics is a good approximation of quantum physics

Question 48

Bohr’s correspondence principle

Answer 48

Predictions of quantum theory must correspond to the predictions of classical physics in the region of sizes where classical theory is “known to hold”. The classical limit should be recovered from the quantum results, for n --> infinity

Question 49

problems with Bohr's model

Answer 49

failed to predict the observed intensities of spectral lines limiting for multi-electron atoms

Question 50

What happens if electrons interact with light at high frequencies such as X-ray?

Answer 50

reasonable to consider the electrons free, since the binding energy of the atom is very small compared to the X-ray energy, so the interaction between the high-frequency radiation and the electron would result in scattering.

Question 51

compton scattering - classical expectation

Answer 51

when the EM wave is scattered off atoms, the wavelength of scattered radiation should be the same as the wavelength of the incident radiation

Question 52

Compton found that with increasing scattering angle, a

Answer 52

second peak was appearing at longer wavelength compared to the one of the incident light

Question 53

compton shift

Answer 53

the change in wavelength between the peaks of the incoming beam and the scattered beam

Question 54

p=

Answer 54

h bar k =hv/c =h/λ

Question 55

k=

Answer 55

2pi/λ

Question 56

de broglie wavelength

Answer 56

λdb = h/p

Question 57

de broglie wavenumber

Answer 57

k=2pi/λdb

Question 58

how de broglie gives a qualitative explanation of Bohr

Answer 58

electrons would interfere, leaving areas of destructive interference where “electrons are forbidden”, while only standing waves would be allowed to “fit” in the Bohr’s circular orbits.

Question 59

de broglie explaining bohr - the standing waves allowed to fit would be given by

Answer 59

a discrete set of wavelengths that interfer constructively if an integral number of wavelengths fits exactly into the circumference of the orbit nλ=2pir (r= radius of the circular orbit)

Question 60

kinetic energy of the electron

Answer 60

p^2/2m

Question 61

why experimental de broglie is difficult

Answer 61

ideally would use experiments already used in optics ie diffraction but order of nm - impossible to design a diffraction grating for such small distances

Question 62

classical physics is

Answer 62

deterministic can determine what will happen next the process of observing (measuring) a classical system does not affect the final configuration of the system.

Question 63

quantum physics is

Answer 63

probabilistic

Question 64

electron orbitals are characterised by

Answer 64

quantum numbers

Question 65

principle quantum number

Answer 65

n indicates the energy level and relative size of the orbital positive integer values: 1,2,3,...

Question 66

orbital angular momentum quantum number

Answer 66

l defines the shape of the orbital takes values from 0 to n-1 each value corresponds to a specific type of orbital (s,p,d,f)

Question 67

magnetic quantum number

Answer 67

ml describes the orientation of the orbital in space takes integer values from -l to +l

Question 68

spin quantum number

Answer 68

s for an electron =1/2

Question 69

sin magnetic quantum number

Answer 69

ms specifies the electron's spin direction which can either be +1/2 or -1/2