Quantum Phenomena

Question 1

Define the photoelectric effect

Answer 1

Electrons are emitted from the surface of a metal when electromagnetic radiation above a certain frequency is directed at the metal

Question 2

Give 3 properties of the photoelectric effect which cannot be explained with the wave theory of light

Answer 2

1) Photoelectric emission doesn’t occur unless the incident electromagnetic radiation is above a threshold frequency
2) The number of electrons emitted per second is proportional to the intensity of the incident radiation (provided the frequency is greater than the threshold frequency)
3) Photoelectric emission occurs without delay as soon as the incident radiation is directed at the surface (provided it is greater than the threshold frequency)

Question 3

Give the 2 incorrect predictions made by the wave theory of light for the photoelectric effect

Answer 3

• Emission should take place with waves of any frequency

- Emission would no take no longer using low intensity waves than using high intensity waves

Question 4

What was Einstein’s explanation for photoelectricity?

Answer 4

He assumed that light is composed of wavepackets (photons) each of energy equal to hf

• When light is incident on a metal surface, a surface electron absorbs a single photon and gains energy equal to hf
• This electron can leave the metal surface if the energy gained from the single photon exceed the work function, Φ

Question 5

Give the equation for the energy of a photon

Answer 5

E = hf

Question 6

Give the equation for the energy of a photon in terms of its wavelength

Answer 6

E = hc / λ

Question 7

Define the work function

Answer 7

The minimum energy needed by an electron to escape from the metal surface when the metal is at 0 potential

Question 8

Give the equation for the maximum kinetic energy an emitted electron can have

Answer 8

Eᵏ(max) = hf - Φ

Question 9

Give the equation for the threshold frequency

Answer 9

f(min) = Φ / h

Question 10

Describe the energy of each vibrating atom

Answer 10

It is quantised (only certain levels of energy are allowed)

Question 11

Describe a vacuum photocell and its usual set-up

Answer 11

A vacuum photocell is a glass tube that contains a metal plate (a photocathode) and a smaller metal electrode (the anode).
When light of a greater frequency than the threshold frequency is directed at the photocathode, electrons are emitted from the cathode to the anode.
An attached microammeter measures the photoelectric current, which is proportional to the number of electrons transferred from the cathode to anode per second

Question 12

Give the equation for current for a photoelectric current

Answer 12

Number of photoelectrons transferred per second = I / e

where e is the charge of the electron

Question 13

Describe the relationship between the intensity of incident light and the maximum kinetic energy of an emitted photoelectron

Answer 13

The intensity of incident light DOES NOT affect the maximum kinetic energy of a photoelectron. This is because the energy gained by a photoelectron is due to the absorption of one photon only

Question 14

Explain why the number of photoelectrons emitted per second is proportional to the intensity of the incident light

Answer 14

• The photoelectric current ∝ Intensity of incident light
• Light intensity = energy per second (carried by light)
• Energy per second ∝ no. of photons per second on cathode
• Because 1 surface electron absorbs 1 photon to escape the metal surface:
• No. of photoelectrons emitted per second ∝ intensity of incident light

Question 15

What would you use to measure the maximum kinetic energy of the photoelectrons emitted for a given frequency of light?

Answer 15

A photocell

Question 16

For a graph of Eᵏ(max) against frequency, what do the following components represent:

i) gradient
ii) y-axis intercept
iii) x-axis intercept

Answer 16

i) h
ii) c = - Φ
iii) Threshold frequency f(min) = Φ / h

Question 17

Define ionisation and give examples

Answer 17

Any process of creating ions
Examples:
- Alpha, beta and gamma radiation create ions when the pass through substances and collide with the atoms of the substance
- Electrons passing through a fluorescent tube create ions when they collide with the atoms of the gas or vapour in the tube

Question 18

Describe how you would measure the ionisation energy of a gas atom

Answer 18

A heated filament emits electrons at increasing speed which are attracted to an anode.
The potential difference between the anode and the filament is increased so as to increase the speed if the emitted electrons.
An attached ammeter only records a very small current due to the electrons reaching the anode. No ionisation takes place until the emitted electrons reach a certain speed.
At sufficient speed, the electrons can ionise gas particles, knocking more electrons out of the gas atom. Ionisation near the anode causes a much greater current

Question 19

Give the equation used to calculate the ionisation energy of a gas atom by the potential difference required to ionise it

Answer 19

Ionisation energy of gas atom = eV

where V is the pd across the ionising tube, and e is the charge of the electron

Question 20

Define the electron volt

Answer 20

The electron volt is a unit of energy equal to the work done when an electron is moved through a pd of 1V
Work done = qV
The work done when an electron moves through a potential difference of 1V is 1.6x10⁻¹⁹ J = 1 eV

Question 21

Describe excitation by collision

Answer 21

Where a gas atom absorbs all the kinetic energy of a colliding electron without being ionised. This causes the electrons in the gas atoms to become excited and thus promoted to higher energy levels.
This happens at certain energies which are characteristic of the gas

Question 22

Define excitation energies

Answer 22

The energy values at which an atom absorbs energy

Question 23

Describe an experiment used to calculate the excitation energies of a gas

Answer 23

By increasing the pd across a gas filled tube containing an electron emitting filament and an anode and measuring the pd when the anode current falls

Question 24

Define ground state

Answer 24

The lowest energy state an atom can be in

Question 25

What happens to an atom in the ground state that absorbs energy

Answer 25

One of its electrons move to a shell at higher energy, so the atom is now in an excited state

Question 26

Describe the energy levels when the ionisation level is considered as the 0 reference level for energy

Answer 26

The values for the energy levels show the energy levels relative to the ionisation level, (rather than showing the energy needed to excite them to that level if the ground state was considered as the 0 reference level for energy)

Question 27

Explain the cause of de-excitation

Answer 27

Excited electrons are more unstable than electrons at a ground state, so emit a photon of light to become more stable and reach the ground state

Question 28

Describe the properties of the emitted photon when an electron is de-excited to a lower energy level

Answer 28

The change in energy levels is equal to the photon emitted: | hf = E₁ - E₂

Question 29

Explain the excitation of an electron by photon absorption

Answer 29

An electron in an atom can absorb a photon and move to an outer shell where a vacancy exists - but only if the energy of the photon is exactly equal to the gain in the electron's energy

Question 30

Explain why a fluorescent object may emit visible or white light, even though it absorbs UV

Answer 30

Because an atom in an excited state can be de-excited directly or indirectly to the ground state (regardless of how the excitation took place). The atom can therefore absorbs photons of certain energies, and then emit photons of the same or lesser energies.

Question 31

Explain how a tube containing mercury gas at a low pressure can emit visible light if the emission of mercury is mostly UV

Answer 31

The tube is coated in a fluorescent coating. - Ionisation and excitation of the mercury atoms occurs as they collide with each other and with electrons in the tube - The mercury atoms emit ultraviolet photons, as well as visible photons and photons of much less energy when they de-excite - The UV photons are absorbed by the atoms of the fluorescent coating, causing excitation of the atoms - The coating atoms de-excite and emit visible photons

Question 32

Give the average wavelengths of deep violet and red light

Answer 32

deep violet ≈ 400nm | red ≈ 650nm

Question 33

Describe an emission spectrum

Answer 33

Coloured lines on a black background where the position of the line indicates the wavelengths of light emitted by an atom

Question 34

How can elements be identified by their emission spectra?

Answer 34

The wavelength of the lines of the emission spectra of an element are characteristic of the atoms of that element. No other element produces the same pattern if light wavelengths because the energy levels of each type of atom are unique to that atom

Question 35

When is the particle nature of light observed?

Answer 35

The photoelectric effect

Question 36

When is the wave nature of light observed?

Answer 36

The diffraction of light

Question 37

What did de Broglie hypothesise?

Answer 37

- Matter particles have a dual wave-particle nature - The wave-like behavior of a matter particle is characterised by a wavelength, its de Broglie wavelength, which is related to the momentum, p, of the particle by the means of the equation: λ = h / p

Question 38

Give the de Broglie wavelength of a particle in terms of its velocity

Answer 38

λ = h / mv

Question 39

Describe an experiment that shows the dual wave-particle nature of an electron

Answer 39

- A narrow beam of electrons in a vacuum tube is directed at a thin metal foil. The rows of atoms in the foil cause the electrons in the beam to be diffracted (just as a beam of light is diffracted when it passes through a slit - The electrons in the beam pass through the metal foil and are diffracted in certain directions only - The electrons form a pattern of rings on a fluorescent screen at the end of the tube

Question 40

Describe how a beam of electrons can be produced

Answer 40

- The beam is produced by attracting electrons from a heated filament wire to a positively charged metal plate, which has a small hole in its centre - Electrons that pass through the hole form the beam - The speed of the electrons can be increased by increasing the potential difference between the filament and the metal plate

Question 41

State the effect on the de Broglie wavelength when electrons are faster, and state the effect this has on the diffraction pattern seen on a fluorescent screen

Answer 41

As the speed of the electrons increases, the de Broglie wavelength decreases, since λ = h / mv Since the de Broglie wavelength decreases, the diffraction rings become smaller because there is less diffraction