56 marks – 100 min exam
Question 1. [9 marks]
The diagram below shows the frequency of light used to irradiate two different type of metals. The maximum KE is plotted on the y axis.
a) What does the y axis represent [1 mark]
b) Compare and contrast the observed maximum KE between both metals when they are irradiated with light of frequency of 4, 8 and 12 * 1014 Hz. [6 marks]
c) Explain what would you expect to happen if you doubled the intensity of the light for each of the metals. [2 marks]
Question 2. [10 marks]
a) Calculate the frequency of light emitted by an electron moving from energy
level 4 to 3 in a Bohr model hydrogen atom. [3 marks]
b) Using the conservation of energy, describe the behaviour of electrons within the Bohr model of an atom for a hydrogen atom. [3 marks]
c) Now describe the hydrogen atom in terms of the standard model of an atom, using a diagram to aid you. [4 marks] (Module 8 cross over question.)
Question 3. [8 marks]
a) Compare and contrast the diffraction pattern expected for an incandescent light globe and a laser. [4 marks]
b) A student has the following set up below
What two modifications could they do, to increase the distance between maxima’s [4 marks]
Question 4. [11 marks]
a) Outline three historical advances in our understanding of the speed of light [3 marks]
b) Evaluate the wave vs particle models of light, providing evidence for each. [8 marks]
Conduct a thought experiment, used by Einstein, which the theory of General Relativity [6 marks]
Question 6. [12 marks]
The diagram below represents the predictions made be classical and quantum theory for the emission spectra of a black body.
a) Describe what a black body is. [2 marks]
b) Providing a formula to support your answer, order A, B and C according to their temperature. [2 marks]
c) Explain what the UV catastrophe is and how this was resolved using quantum theory. [8 marks]
1a) The y axis represent the KE of a photo-electron released by the metal surface. [band 3/4]
b) [band 4/5]
|Frequency (Hz) (* 10^14)||Metal X||Metal Y|
|4||The threshold frequency is less than frequency of light being absorbed, therefore, electrons will be emitted||No electrons emitted, below threshold frequency|
|8||Electrons with a greater KE are now being emitted – assuming no change in intensity of light. The number of electrons is the same||The threshold frequency is less than the frequency of light being absorbed, therefore, electrons will be emitted. However, the same number of electrons as metal A are being emitted.|
|12||Electron with greater KE and the same number are now being emitted. KE is still greater than metal B.||Electrons with greater KE and same number are now being emitted. KE is still less than metal A.|
|Identify electrons of metal A has greater KE then metal B|
Reference threshold frequency as to why no electrons are being emitted initially for metal A
Identify increase in frequency increased KE not intensity of electrons being emitted.
|2 mark each point|
Doubling the intensity of light would not change the KE of the electrons being emitted, it would increase the number of electrons being emitted, since the photons exhibit a wave-particle duality.
|Identify the number of photons being emitted increases and the KE of photons doesn’t changeExplains this with respect to the wave-particle duality of a photon.||1 mark for each|
nf = 3 and ni = 4
f = 0.016 Hz
The electron is relaxing from the excited state to a lower energy level in the electron shells of the hydrogen atom. As a result, it is releasing energy in the form of a photon. Since the energy levels are quantized, from Planck and Einstein’s works building on the Bohr model, the energy, thus frequency of light can be predicted. The calculated frequency is proportional to energy, using E = hf. This electron would have initially been exited to the excited state by absorbing a fixed energy level of light.
|Identify the electron is relaxing|
This emits EM radiation of set Energy level, E = hf
This is predictable since the energy levels are quantized or conservation of energy
|1 mark for each|
- Neutron: Quarks are DUD and Proton: UUD
- 1 proton, 1 neutron – held together by strong and weak nuclear forces,
- gluons help for the exchange particle
- Electron orbits the nucleus, it is attracted by the electromagnetic force due to opposite charges of proton (+ ⅔ + ⅔ – ⅓ = + 1 charge for proton.) and electron (-1).
1 mark each point.
One thought on “Physics the nature of light – module 7 HSC exam”
Just wondering where the remaining answers are for question 3-6. Cheers