Quantum

Question 1

What is wave-particle duality?

Answer 1

• All matter (including light, EM waves) 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.

Question 3

Define photon.

Answer 3

• A massless quanta of electromagnetic energy
• Carries energy proportional to the frequency of EM radiation.

Question 4

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

Answer 4

E=hf=hc/lambda

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.
- Energy is a quantity associate with particles.
- Frequency is something we typically associate with waves.

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

Question 11

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

Answer 11

• The photoelectric effect
• Light is comprised of photons which interact with electrons at a metal’s surface one-to-one transferring their quanta of energy

Question 12

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

Answer 12

• Young’s double slit experiment.-
• Light can undergo diffraction to form an interference pattern

Question 13

Define the photoelectric effect.

Answer 13

• The emission of electrons when an incident photon interacts with a metallic surface

Question 14

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

Answer 14

Photoelectrons

Question 15

How did Heinrich Hertz discover photoelectric emission?

Answer 15

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

Question 16

How does the gold leaf electroscope experiment work?

Answer 16

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 plate electrons are emitted.
4. The leaf now falls slowly as the negative charge is depleted

Question 17

Why is the emission of photoelectrons instantaneous?

Answer 17

• Only one photon interacts with one electron to deliver sufficient energy to exceed the metal’s work function and free an electron

Question 18

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

Answer 18

• A single photon transfers its sum of energy to a single electron in a one-to-one interaction
• Thus the number of photons/second (intensity) does not matter

Question 19

Why does intensity only affect the number of electrons emitted?

Answer 19

• Photons and electrons interact on a one-to-one basis
• If intensity increases then more photons per second reach the metal
• So more electrons per second are released
• Energy is not accumulated continuously

Question 20

What is the work function of a metal?

Answer 20

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

Question 21

What is the photoelectric equation?

Answer 21

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

Question 22

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

Answer 22

• 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 23

How does frequency affect the kinetic energy of an electron?

Answer 23

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

Question 24

How is intensity related to rate of photoelectron emission?

Answer 24

Intensity is proportional to rate of emission.

Question 25

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

Answer 25

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

Question 26

What is the de Broglie equation?

Answer 26

lambda = h/mv

Question 27

What idd de Broglie hypothesise?

Answer 27

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

Question 28

What phenomenon shows that electrons act as waves?

Answer 28

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

Question 29

How can we practically demonstrate electron diffraction?

Answer 29

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

Question 30

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

Answer 30

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

Question 31

Why is graphite useful for demonstrating electron diffraction?

Answer 31

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

Question 32

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

Answer 32

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

Question 33

How is the wavelength of a particle affected by velocity?

Answer 33

lambda = h/mv therefore the two are inversely proportional

Question 34

How do we use LEDs to estimate the Planck constant?

Answer 34

- 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

Question 35

Why are quantum effects difficult to observe practically, in the world around us? Think of scale.

Answer 35

- We know that lambda = h/mv - So for objects of relatively large mass moving at fairly low velocities like objects or humans, their frequency remains very small - This is more difficult for us to observe.