Chapter 11 Quantum Physics

Question 1

The Photon

Answer 1

• Photons are fundamental particles which make up all forms of electromagnetic radiation
• A photon is a massless “packet” or a “quantum” of electromagnetic energy
• Energy is not transferred continuously, but as discrete packets of energy
• each photon carries a specific amount of energy, and transfers this energy all in one go

Question 2

Calculating Photon Energy

Answer 2

• The energy of a photon can be calculated using the formula:

E = hf

• Using the wave equation, energy can also be equal to:
• Where:
  
  • E = energy of the photon (J)
  • h = Planck’s constant (J s)
  • c = the speed of light (m s^-1)
  • f = frequency in Hertz (Hz)
  • λ = wavelength (m)

Question 3

This equation tells us what:

Answer 3

• The higher the frequency of EM radiation, the higher the energy of the photon
• The energy of a photon is inversely proportional to the wavelength
• A long-wavelength photon of light has a lower energy than a shorter-wavelength photon

Question 4

Photon Momentum

Answer 4

• a photon travelling in a vacuum has momentum, despite it having no mass
• The momentum (p) of a photon is related to its energy (E) by the equation and Where c is the speed of light:

Question 5

The Electronvolt is derived from?

Answer 5

• the definition of potential difference:
• When an electron travels through a potential difference, energy is transferred between two points in a circuit, or electric field

Question 6

electronvolt is defined as:

Answer 6

The energy gained by an electron travelling through a potential difference of one volt

1 eV = 1.6 × 10^-19 J

Question 7

(relation to kinetic energy) When a charged particle is accelerated through a potential difference, it

Answer 7

• gains kinetic energy
• If an electron accelerates from rest, an electronvolt is equal to the kinetic energy gained:

eV = ½ mv²

• Rearranging the equation gives the speed of the electron:

Question 8

• To convert between eV and J:

Answer 8

• eV → J: multiply by 1.6 × 10^-19
• J → eV: divide by 1.6 × 10^-19

Question 9

The photoelectric effect is the

Answer 9

• phenomena in which electrons are emitted from the surface of a metal upon the absorption of electromagnetic radiation
• Electrons removed from a metal known as photoelectrons

Question 10

The photoelectric effect provides important evidence that light is

Answer 10

quantised, or carried in discrete packets

• This is shown by the fact each electron can absorb only a single photon
• This means only the frequencies of light above a threshold frequency will emit a photoelectron

Question 11

The photoelectric effect can be observed on a

Answer 11

• gold leaf electroscope
• A plate of metal, usually zinc, is attached to a gold leaf, which initially has a negative charge, causing it to be repelled by a central negatively charged rod
  
  • This causes negative charge, or electrons, to build up on the zinc plate
• UV light is shone onto the metal plate, leading to the emission of photoelectrons
• This causes the extra electrons on the central rod and gold leaf to be removed, the gold leaf begins to fall back towards the central rod
  
  • they become less negatively charged, and hence repel less

Question 12

Some notable observations of photoelectric effect can be observed on a gold leaf electroscope

Answer 12

• Placing the UV light source closer to the metal plate causes the gold leaf to fall more quickly
• Using a higher frequency light source does not change the how quickly the gold leaf falls
• Using a filament light source causes no change in the gold leaf’s position
• Using a positively charged plate also causes no change in the gold leaf’s position

Question 13

• The threshold frequency is defined as:

Answer 13

The minimum frequency of incident electromagnetic radiation required to remove a photoelectron from the surface of a metal

Question 14

The threshold wavelength, related to

Answer 14

• threshold frequency by the wave equation, is defined as:

The longest wavelength of incident electromagnetic radiation that would remove a photoelectron from the surface of a metal

• Threshold frequency and wavelength are properties of a material, and vary from metal to metal

Question 15

The Photoelectric Equation

Answer 15

E = hf = Φ + ½mv²_max

• Symbols:
  
  • h = Planck’s constant (J s)
  • f = the frequency of the incident radiation (Hz)
  • Φ = the work function of the material (J)
  • ½mv²_max= the maximum kinetic energy of the photoelectrons (J)

Question 16

• Since energy is always conserved, the energy of an incident photon is equal to:

Answer 16

The threshold energy + the kinetic energy of the photoelectron

• The energy within a photon is equal to hf
• This energy is transferred to the electron to release it from a material (the work function) and gives the emitted photoelectron the remaining amount as kinetic energy

Question 17

(E = hf = Φ + ½mv²_max )This equation demonstrates

Answer 17

• If the incident photons do not have a high enough frequency (f) and energy to overcome the work function (Φ)
• >no electrons will be emitted
• When hf₀ = Φ, where f₀ = threshold frequency, photoelectric emission only just occurs
• E_kmax depends only on the frequency of the incident photon, and not the intensity of the radiation
• The majority of photoelectrons will have kinetic energies less than E_kmax

Question 18

Graphical Representation of Work Function

Answer 18

• The photoelectric equation can be rearranged into the straight line equation:
• y = mx + c
• Comparing this to the photoelectric equation:
• E_kmax = hf - Φ
• A graph of maximum kinetic energy E_kmax against frequency f can be obtained

Question 19

The key elements of the graph:

Answer 19

• The work function Φ is the y-intercept
• The threshold frequency f₀ is the x-intercept
• The gradient is equal to Planck’s constant h
• There are no electrons emitted below the threshold frequency f₀

Question 20

• The work function Φ, or threshold energy, of a material is defined as:

Answer 20

The minimum energy required to release a photoelectron from the surface of a material

Question 21

an electron can only escape the surface of the metal if

Answer 21

it absorbs a photon which has an energy equal to Φ or higher because the electrons in a metal as trapped inside an ‘energy well’ where the energy between the surface and the top of the well is equal to the work function Φ

A single electron absorbs one photon

Question 22

• Different metals have different threshold frequencies, and hence different work functions
• Using the well analogy:

Answer 22

• A more tightly bound electron requires more energy to reach the top of the well
• A less tightly bound electron requires less energy to reach the top of the well

Question 23

Alkali metals have threshold frequencies in the

Answer 23

• such as sodium and potassium, have threshold frequencies in the visible light region
  
  • This is because the attractive forces between the surface electrons and positive metal ions are relatively weak

Question 23

Alkali metals have threshold frequencies in the

Answer 23

• such as sodium and potassium, have threshold frequencies in the visible light region
  
  • This is because the attractive forces between the surface electrons and positive metal ions are relatively weak

Question 24

Transition metals have threshold frequencies in the

Answer 24

such as manganese and iron, have threshold frequencies in the ultraviolet region

• This is because the attractive forces between the surface electrons and positive metal ions are much stronger

Question 25

• Laws of Photoelectric Emission
• Observation:
  • Placing the UV light source closer to the metal plate causes the gold leaf to fall more quickly

Answer 25

• Explanation:
  
  • Placing the UV source closer to the plate increases the intensity incident on the surface of the metal
  • Increasing the intensity, or brightness, of the incident radiation increases the number of photoelectrons emitted per second
  • Therefore, the gold leaf loses negative charge more rapidly

Question 26

Laws of Photoelectric Emission

• Observation:
  
  • Using a higher frequency light source does not change how quickly the gold leaf falls

Answer 26

• Explanation:
  
  • The maximum kinetic energy of the emitted electrons increases with the frequency of the incident radiation
  • In the case of the photoelectric effect, energy and frequency are independent of the intensity of the radiation
  • So, the intensity of the incident radiation affects how quickly the gold leaf falls, not the frequency

Question 27

Laws of Photoelectric Emission

• Observation:
  
  • Using a filament light source causes no change in the gold leaf’s position

Answer 27

• Explanation:
  
  • If the incident frequency is below a certain threshold frequency, no electrons are emitted, no matter the intensity of the radiation
  • A filament light source has a frequency below the threshold frequency of the metal, so, no photoelectrons are released

Question 28

Laws of Photoelectric Emission

• Observation:
  
  • Using a positively charged plate causes no change in the gold leaf’s position

Answer 28

• Explanation:
  
  • If the plate is positively charged, that means there is an excess of positive charge on the surface of the metal plate
  • Electrons are negatively charged, so they will not be emitted unless they are on the surface of the metal
  • Any electrons emitted will be attracted back by positive charges on the surface of the metal

Question 29

Laws of Photoelectric Emission

• Observation:
  
  • Emission of photoelectrons happens as soon as the radiation is incident on the surface of the metal

Answer 29

• Explanation:
  
  • A single photon interacts with a single electron
  • If the energy of the photon is equal to the work function of the metal, photoelectrons will be released instantaneously

Question 30

The maximum kinetic energy of the photoelectrons is

Answer 30

• independent of the intensity of the incident radiation
• This is because each electron can only absorb one photon

Question 31

The maximum kinetic energy of the photoelectrons is

Answer 31

• independent of the intensity of the incident radiation
• This is because each electron can only absorb one photon

Question 32

Kinetic energy is only dependent on the

Answer 32

• frequency of the incident radiation
• Intensity is a measure of the number of photons incident on the surface of the metal
• So, increasing the number of electrons striking the metal will not increase the kinetic energy of the electrons, it will increase the number of photoelectrons emitted

Question 33

Photoelectric Current is

Answer 33

the number of photoelectrons emitted per second

Question 34

Photoelectric current is proportional

Answer 34

• to the intensity of the radiation incident on the surface of the metal
• This is because intensity is proportional to the number of photons striking the metal per second

Question 35

Photoelectric current is proportional

Answer 35

• to the intensity of the radiation incident on the surface of the metal
• This is because intensity is proportional to the number of photons striking the metal per second

Question 36

Kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity and independent of frequency

Answer 36