4.5 Quantum Physics

Question 1

explain what a photon is and provide two equations for it’s energy

Answer 1

EM waves travel through space as a wave, but interact with matter as photons (discrete energy quanta)
E=hf E=hc/λ

Question 2

define the electronvolt and explain how to convert between eV and J

Answer 2

the unit for energy of a photon
1eV=1.60x10^-19J

Question 3

describe a PAG to estimate the planck constant using an electrical circuit

Answer 3

• connect an LED of known λ in a circuit with a variable resistor, a voltmeter and a milliammeter
• start w/ no current flow & adjust the variable resistor until current begins to flow & the LED lights
• record the threshold voltage (V0) & λ
• repeat with LEDs of different λ
• plot a graph of V0 against 1/λ (in metres)
• gradient = hc/e
• calculate h

Question 4

explain the photoelectric effect

Answer 4

• EM radiation of a high enough f on the surface of a metal will eject electrons
• free electrons + energy = bonds broken = electron emitted
• these electrons are called photoelectrons

Question 5

describe how to demonstrate the photoelectric effect

Answer 5

gold leaf electroscope:
- a zinc plate on a negatively charged stem w/ a negatively charged golf leaf attached
- zinc plate is exposed to UV radiation
- free electrons are ejected -> plate & stem are no longer neg -> gold leaf drops

Question 6

recall the conclusions of the gold leaf electroscope experiment

Answer 6

• photoelectrons are not emitted if f is below the threshold frequency
• photoelectrons are emitted with a variety of KE (0-KEmax), ^ f = ^ KE, ^ intensity =/= ^ KE
• intensity is directly proportional to the number of photoelectrons emitted

Question 7

explain threshold frequency

Answer 7

• if hf >= Φ, electron is emitted
• if hf <= Φ, electron isn’t emitted, the metal heats up
• f=Φ/h

Question 8

explain KEmax and how to calculate it

Answer 8

KE = hf-energy lost leaving metal (unaffected by intensity)
hf=Φ+KEmax KEmax=1/2mvmax^2
KEmax=hf-Φ (=y=mx+c)

Question 9

explain wave-particle duality and de broglie wavelength

Answer 9

• light acts as both wave and particles in context
• de broglie concluded that the opposite is true for electrons
• λ=h/p

Question 10

describe/explain electron diffraction

Answer 10

• 1927, davisson & germer
• accelerated electrons in vacuum tube interact w/ spaces between C atoms in polycrystalline graphite
• diffraction patterns are produced (on phosphate screen)

Question 11

recall the conclusions from the electron diffraction experiment

Answer 11

• spread of lines ^ = when wavelength ^
• smaller voltage = slower electrons = space of rings ^
• speed v = momentum v = wavelength ^
• shorter λ = less diffraction effects