C13. Quantum

Question 1

What is wave-particle duality?

Answer 1

Where light behaves with the properties of both a wave and a particle. It interacts with matter as a particle yet propagates through space as a wave.

Question 2

How do electromagnetic waves carry energy?

Answer 2

In discrete packets ‘quanta’ called photons - named as such since they carry a quantised energy value, equal to E=hf

Question 3

Define photon.

Answer 3

A massless quanta of electromagnetic energy.

Question 4

What equation is used to quantise the energy of a photon?

Answer 4

E=hf

Question 5

Why is energy not transferred continuously by an EM source but instead in discrete amounts?

Answer 5

EM waves consist of a stream of photons each containing individual packets/quanta of energy.

Question 6

What does h represent in e=hf?

Answer 6

h represents Planck’s constant, 6.63*10^-34 Js

Question 7

How do we use the equation e=hf to show wave-particle duality?

Answer 7

The equation links the properties of both a wave and a particle to the energy of photons.

Question 8

Define the electronvolt.

Answer 8

1eV is the energy transferred when a single electron with elementary charge passes through a potential difference of 1V.

Question 9

How to we convert from Joules to eV?

Answer 9

Divide the energy transferred in Joules by elementary charge.

Question 10

What equation can we use to show the gain in kinetic energy due to an electron accelerating through a p.d.?

Answer 10

eV=1/2mv^2 (note this looks like the kinetic energy formula, with the result in Joules converted to eV).

Question 11

How can we rearrange e=hf with the wave equation?

Answer 11

We can say that also e=hc/(lambda)

Question 12

What evidence do we have for light acting as a particle?

Answer 12

The photoelectric effect.

Question 13

What evidence do we have for light acting as a wave?

Answer 13

Young’s double slit experiment.

Question 14

Define the photoelectric effect.

Answer 14

The emission of electrons when a metallic surface (containing many free electrons) absorbs electromagnetic radiation.

Question 15

Give a name for electrons released as a result of the photoelectric effect.

Answer 15

Photoelectrons.

Question 16

How did Heinrich Hertz discover photoelectric emission?

Answer 16

When UV light was shone onto a zinc plate electrode, electrons were emitted, causing a spark be produced more easily.

Question 17

How does the gold leaf electroscope experiment work?

Answer 17

1. A top plate is charged and a negative charge spreads through the gold leaf.
2. This causes the gold leaf to rise and stay in place.
3. When a source of UV is pointed at the zinc plat electrons are emitted.
4. The leaf now falls slowly as the negative charge is depleted as electrons are released.

Question 18

Why is the emission of photoelectrons instantaneous?

Answer 18

Only one photon interacts with one electron to reach a sufficient energy level for the electron to release - the energy does not accumulate but is rather provided in a one-to-one interaction.

Question 19

Why are photoelectrons only released if incident EM radiation exceeds threshold frequency, regardless of intensity.

Answer 19

A single photon transfers its quanta of energy to a single electron in a one-to-one interaction - the number of photons (intensity) does not matter.

Question 20

Why does intensity only affect the number of electrons emitted?

Answer 20

Photons and electrons interact on a one-to-one basis; one photon transfers all of its energy to one electron. Increasing intensity will only increase the rate of this energy transfer and therefore the number of electrons released at a given moment.

Question 21

What is the work function of a metal?

Answer 21

The minimum energy required for an electron to be freed by a metal due to an incident photon.

Question 22

What is the photoelectric equation?

Answer 22

Ek(max)=hf-hf0
- where hf0 is the work function

Question 23

Why is the photoelectric equation a restatement of the conservation of energy?

Answer 23

Any energy transferred to a given electron due to an incident photon exceeding its threshold frequency is transferred fully to its kinetic store, since energy can never be created nor destroyed.

Question 24

How does frequency affect the kinetic energy of an electron?

Answer 24

Once the threshold frequency is exceeded, as frequency increases, kinetic energy increases.

Question 25

How is intensity related to rate of photoelectron emission?

Answer 25

Intensity is proportional to rate of emission.

Question 26

What is the effect of intensity of the *kinetic energy* of released photoelectrons?

Answer 26

This is a trick question - they are completely unrelated since photoelectric emission is based upon one-to-one interactions.

Question 27

What is the de Broglie equation?

Answer 27

lambda = h/mv

Question 28

What idd de Broglie hypothesise?

Answer 28

All particles exhibit wave-particle duality. They have a characteristic wavelength, the de Broglie wavelength.

Question 29

What phenomenon shows that electrons act as waves?

Answer 29

They can be diffracted when passing through an aperture of similar size to their de Broglie wavelength, similarly to a wave.

Question 30

How can we practically demonstrate electron diffraction?

Answer 30

Use an electron gun to fire electrons at polycrystalline graphite. They pass through the small gaps in the graphite's structure, diffracting as they do so. They then go on to form an interference pattern on a fluorescent screen made of phosphor.

Question 31

How does the potential tail difference across an electron guns' electrodes relate to velocity?

Answer 31

eV=1/2mv^2=eK, therefore the kinetic energy and therefore speed is proportional to the p.d. across the electrodes.

Question 32

Why is graphite useful for demonstrating electron diffraction?

Answer 32

Graphite has a crystalline structure. The gaps between atoms act as slits through which electrons can pass to diffract.

Question 33

How does the diffraction pattern appear when electrons are diffracted through graphite?

Answer 33

The pattern appears as a series of concentric rings. A larger accelerating voltage results in a larger diameter between the rings.

Question 34

How is the wavelength of a particle affected by velocity?

Answer 34

lambda = h/mv therefore the two are inversely proportional

Question 35

How do we use LEDs to estimate the Planck constant?

Answer 35

- When a sufficient p.d. acts across an LED such that it just begins to glow, the energy lost by one electron equals that gained by a single photon - We use this to determine the threshold voltage by varying p.d. using a potential divider - Each electron loses the same energy eV as is gained by the released photon, hf - Plot threshold p.d. against 1/wavelength and multiply by a constant e/c