Unit 2.7)Photons

Question 1

What are photons?

Answer 1

Photons are particles of light.
They are sometimes referred to as a quantum of energy of EM radiation. ‘A quantum’ in this context just means a set(finite) amount.

Question 2

True or false? The energy of a photon is proportional to the wavelength of the light?

Answer 2

False.
It’s proportional to the frequency: E=hf.
Energy is inversely proportional to the wavelength.

Question 3

What is ‘h’ in the equation E=hf? Give units.

Answer 3

h is the Planck constant, measured in Js.

Question 4

What quantity can be measured in electron volts (eV)?

Answer 4

Energy.

Question 5

Describe an experiment which can be used to estimate the value of the Planck constant?

Answer 5

An LED will only allow current to pass after a minimum voltage has been put across it.
At this voltage all the electrons will have the same energy as a photon emitted by the LED(which you will know the frequency of).
Finding the threshold voltage by seeing when current flows in the circuit can then be used to find h from: h=E/f.

Question 6

Describe how to improve the accuracy of the estimate of this experiment?

Answer 6

To improve the accuracy of this estimate, the experiment can be repeated with a variety of different coloured LEDs, which each emit different wavelengths of light. The values of wavelength and threshold p.d. for each can be recorded, and a graph drawn of V against 1/λ. The gradient of this graph will be equal to hc/e. As the speed of light and the electron charge are known constants, we can calculate the value of h from this.

Question 7

What is the photoelectric effect?

Answer 7

The photoelectric effect is a phenomenon where shining light with enough energy onto a metal releases electrons (and can cause a current to flow).
The electrons emitted are called photoelectrons.

Question 8

Which features of the photoelectric effect can’t be explained if light is a wave?

Answer 8

If light was a wave, the energy of the electrons released would increase with increasing intensity of the light - but this isn’t the case. Instead the energy of the electrons depends on frequency(no electrons are released below a certain threshold value, no matter how intense the light is).

Question 9

How many photons does each photoelectron absorb prior to emission?

Answer 9

1.
If it doesn’t contain enough energy the electron will re-emit the energy rather then being released.

Question 10

How does the photon model of light explain the threshold frequency seen in the photoelectric effect?

Answer 10

Each electron absorbs one single proton.
This single photon must have enough energy for the electron to be released, if it doesn’t the energy is re-emitted.
The electron can’t build up energy as it could if light was a wave.

Question 11

What is the name given to the minimum amount of energy an electron requires to leave the surface of a metal?

Answer 11

The work function.

Question 12

Write a word equation relating the energy of an incident photon to the work function and the kinetic energy of released electrons?

Answer 12

Photon energy = work function + kinetic energy.

Question 13

True or False? The rate of emission of photoelectrons is proportional to intensity (provided the light is above threshold frequency)?

Answer 13

True.
Higher intensity means more photons, this means more electrons absorb energy and be released.

Question 14

Does the maximum kinetic energy of a released electron depend on the intensity of light hitting the surface?

Answer 14

No.
Energy transferred is due to a one-to-one interaction, and so depends on frequency, not intensity.

Question 15

What experimental evidence appears to show particles behaving as waves?

Answer 15

Electron diffraction.
Electrons when passed through the spaces between atoms in graphite(like a tiny diffraction grating).
This wouldn’t happen if electrons were behaving as particles only.

Question 16

Which equation relates the wave and particle properties of electrons?

Answer 16

The de Brogile equation: λ=h/p.
Where λ = wavelength (wave-property), h= Planck’s constant, and p = momentum(particle-property).

Question 17

What are Photons?

Answer 17

Photons are particles of light.
They are sometimes referred to as a quantum of energy of EM radiation. ‘A quantum’ in this context just means a set(finite) amount.

Question 18

True or False? The energy of a photon is proportional to the wavelength of the light.

Answer 18

False.
It’s proportional to the frequency: E=hf.
Energy is inversely proportional to the wavelength.

Question 19

What is ‘h’ in the equation E=hf? Give units?

Answer 19

h is the Planck constant, measured in Js.

Question 20

What quantity can be measured in electron volts (eV)?

Answer 20

Energy.

Question 21

Describe an experiment which can be used to estimate the value of the Planck constant?

Answer 21

An LED will only allow current to pass after a minimum voltage has been put across it.
At this voltage all the electrons will have the same energy as a photon emitted by the LED(Which you will know the frequency off).
Finding the Threshold voltage by seeing when current flows in the circuit can than be used to find h from: h =E/f.

Question 22

What is a vacuum photocell?

Answer 22

A vacuum photocell is a device which can be used to measure the maximum kinetic energy of electrons.

Question 23

What are vapour lamps?

Answer 23

Tubes containing a gas. A high voltage source is used to provide electrons with energy which then transfer some of their energy to the gas atoms. This excites electrons within the atoms causing them to rise to higher energy states. They then fall back down to the lower energy state(from which they came) releasing a photon. Since there are multiple energy states within atoms, photons of different energies are released. Therefore you would see a combination of colours from the lamp.

Question 24

What is the ionisation energy?

Answer 24

The ionisation energy is the energy an electron requires to escape an atom, leaving it as an ion. This means it is raised up all of the energy states until the state with zero energy. The negative sign for the energies means the electron is bound to the atom. When the energy becomes positive, the electron escapes.
To find the ionisation energy, you need to work out the energy difference between the ground state and the zero energy state.

Question 25

How can the photo electric effect be demonstrated?

Answer 25

You can demonstrate the photoelectric effect with an electroscope and a short wave UV-C lamp. By placing a negative charge on the electroscope, and shining the short wave UV light on top, it will discharge. Short wave UV is usually blocked by glass, but visible light is not, thus a pane of glass can be used to show that it is not just regular light that is causing the discharge.

Question 26

How a vacuum photocell can be used to
measure the maximum kinetic energy, Ek max,
of emitted electrons in eV and hence in J

Answer 26

KE = hf -BE
hf=photon energy
BE=binding energy (or work function) of the electron to the particular material.

Question 27

How does a photon picture of light lead to Einstein’s equation , Ek max = hf – φ, and how this equation relates with the graph of Ek max against frequency?

Answer 27

1. Photoelectric Effect:
   
   • When light (composed of photons) shines on a metal surface, electrons are emitted from the metal. This phenomenon is called the photoelectric effect.
   • The energy of a photon is given by E= hf, where h is Planck’s constant and f is the frequency of the incident radiation.
   • If the frequency of the incident photons is above a certain threshold frequency ((f_0)), electrons are emitted from the metal.
2. Einstein’s Equation:
   
   • The photoelectric equation relates the energy of the photon to the work function Phi(φ) and the maximum kinetic energy of the emitted electrons:
     Ek max = hf - Phi(φ)
   • Here:
     
     • Ek max represents the maximum kinetic energy of the photoelectrons.
     • Phi(φ) is the work function (the energy required to release an electron from the metal).
     • h is Planck’s constant.
3. Graphical Representation:
   
   • We can rearrange the photoelectric equation into a straight line equation: Ek max = hf - phi(φ).
   • When we plot a graph of (E_{K_{\text{max}}}) against (f), we observe the following:
     
     • The y-intercept represents the work function ((\Phi)).
     • The x-intercept corresponds to the threshold frequency ((f_0)).
     • The gradient of the line is equal to Planck’s constant ((h)).
     • No electrons are emitted below the threshold frequency ((f_0)).
4. Example Calculation:
   
   • Suppose we have a graph showing the variation of (E_{K_{\text{max}}}) with (f).
   • By identifying the threshold frequency ((f_0)) from the graph, we can calculate the work function ((\Phi)).
   • Finally, we convert the work function from joules to electronvolts (eV).

Question 28

The visible spectrum?

Answer 28

Look at an image
The visible spectrum runs
approximately from 700 nm (red end) to 400
nm (violet end) and the orders of magnitude
of the wavelengths of the other named
regions of the electromagnetic spectrum.

Question 29

Typical photon energies associated with different radiations based on their

Answer 29

Certainly! Let’s discuss the typical photon energies associated with different radiations based on their wavelengths. The energy of a photon depends on its wavelength, and this relationship is crucial in various scientific contexts.

1. Photon Energy and Wavelength:
   
   • The energy of a photon ((E)) is directly proportional to its frequency ((f)) and inversely proportional to its wavelength ((\lambda)).
   • Planck’s equation relates photon energy to wavelength:
     [E = hf = \frac{hc}{\lambda}]
     where:
     
     • (E) is the energy of a single photon.
     • (h) is Planck’s constant (approximately (6.6261 \times 10^{-34} \, \text{J} \cdot \text{s}) or (4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s})).
     • (c) is the speed of light ((299,792,458 \, \text{m/s})).
     • (\lambda) is the wavelength of the radiation.
   • The higher the frequency (shorter wavelength), the greater the photon energy, and vice versa.
2. Typical Photon Energies:
   
   • Let’s look at some representative submicroscopic energies in electronvolts (eV):
     
     • Visible light: (2.0 - 3.1 \, \text{eV})
     • Ultraviolet (UV): (3.1 - 12.4 \, \text{eV})
     • X-rays: (100 \, \text{eV} - 100 \, \text{keV})
     • Gamma rays: (> 100 \, \text{keV})
3. Example Calculation:
   
   • Suppose we have a visible light photon with an average wavelength of 580 nm (nanometers).
   • Using Planck’s equation:
     [E = \frac{hc}{\lambda} = \frac{(6.6261 \times 10^{-34} \, \text{J} \cdot \text{s}) \cdot (299,792,458 \, \text{m/s})}{580 \times 10^{-9} \, \text{m}}]
   • Calculate the energy to find the typical photon energy for visible light.

Question 30

How to produce line emission and line
absorption spectra from atoms?

Answer 30

Line emission spectrum produced by bombarding atoms with electrons or by other atoms.
when atoms absorb certain wavelengths they produce a line absorption spectrum.

Question 31

Simple atomic energy level diagrams,
together with the photon hypothesis, line
emission and line absorption spectra?

Answer 31

Search up what they look like.

Question 32

How to determine ionisation energies from
an energy level diagram?

Answer 32

Search it up.

Question 33

momentum of a particle equation?

Answer 33

p=h/λ
p=momentum of a particle
h=planks constant
λ-wavelength associated with the particle or wave.

Question 34

The calculation of radiation pressure on a
surface absorbing or reflecting photons?

Answer 34

Candidates will be expected to be able to use the
equations for photon energy and momentum.
Therefore they should be able, for example, to
calculate the momentum arriving per second at a
surface when a beam of light of a given power strikes
the surface normally. According to Newton’s 2nd and
3rd laws, this gives the force on the surface if it
doesn’t reflect. If the surface is 100% reflecting, then
the force is double because momentum is a vector
and the change in momentum of the photons is double
when their momentum is reversed. Pressure is
calculated as normal force divided by (beam) area.