PHYC10004
SCHOOL OF PHYSICS
SESSION 1
NOVEMBER 20, 2020
EXAMINATION QUESTION PAPER
1
Write your ID at the top of each page of your solutions
STUDENT ID
SUBJECT NAME:
PHYSICS 2 PHYSICAL SCIENCE AND TECHNOLOGY
SUBJECT CODE:
UPLOADING TIME:
PHYC10004
3 hours (180 MINS) NOT including reading or uploading time.
Estimate 1 hour (60 MINS) Canvas Quiz, 2 hours (120 MINS) Long Answer
UP TO 30 MINS (15 MINS FOR CANVAS QUIZ, 15 MINS FOR LONG ANSWER)
30 MINS (LONG ANSWER)
TOTAL MARKS:
150 marks (50 marks Quiz, 100 marks Long Answer)
NUMBER OF QUESTIONS:
29 QUIZ QUESTIONS AND 8 GRADESCOPE QUESTIONS
TOTAL EXAM DURATION:
READING TIME:
AUTHORISED MATERIALS
1.
2.
3.
4.
5.
Writing materials, stationery, rulers, and other implements
Calculator
Textbooks, lecture notes
Resources on the Learning Management System (LMS) for PHYC10004
Mobile phone with camera or other apparatus for scanning your solutions
INSTRUCTIONS TO STUDENTS
1.
The conduct of the exam is governed by the rules of the University of Melbourne
2.
Apart from the LMS and the requirements of scanning and uploading your solutions, access to the
internet and communication with others, except the examiners, is not permitted during the exam
3.
Write your solutions on blank plain or ruled sheets of paper
4.
Start each question on a new page
5.
Write your student id number at the top of each page of your solutions
6.
Write your start time and finish time on the first page of your solutions and write your signature on the
next line to certify this is a true record consistent with the honour code
7.
At the conclusion of the writing time submit your solutions as a single PDF using the appropriate app
8.
To ensure your submission is not late, you must access the exam at the designated time
9.
Note that ALL answers must be submitted for all parts of the exam within the 240 minute window.
Answers submitted after this time will not be accepted. Leave sufficient time to upload all of your solutions.
10.
Evidence of technical difficulties should be recorded and reported as soon as possible. This will be
considered for late submissions.
11.
Any questions or technical difficulties can be addressed to;
I.
ph-firstyear-exam@unimelb.edu.au
II.
Exam Support Link on the LMS
III.
Lecturers Chris Chantler and Roger Rassool
Question 1 – Gauss
25 Marks
Consider Gauss’s Law for electric fields.
a) Briefly (about 1 page) explain what the symbols in the equation for Gauss’ Law mean,
especially how the integral is defined [see Formula sheets]. Also explain what charge
q is referring to on the right-hand side.
b) A point positive charge, +Q, is placed at the centre of a spherical cavity of radius R
formed in a large block of conducting material (assume the conducting material goes
from r=R to r= infinity). Use Gauss’ Law to find the magnitude and direction of the
electric field at a distance
i.
ii.
iii.
r = R/2 (inside the cavity)
r = R (at the edge between the cavity and the conducting material)
r > R from the centre of the cavity (in the conductor).
c) Two long concentric conducting cylinders of radius a and b, respectively, each carry
equal and opposite charges, with a linear charge density λ C m-1. The cylinders are of
negligible thickness.
Assuming that b > a, use Gauss's law
i. to show that E vanishes for r < a and r > b
ii.
to derive E for a < r < b,
iii.
to describe the direction of the electric field in the space between the
cylinders
[7 + (3 + 3 + 3) + (3 + 4 + 2)] = 25 marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 1 of 12
Question 2 Rupert Bear and the Castle Trap
25 marks
Rupert Bear is one of the great UK institutions and one of the first popular text comics, with
the first comic in 1920. Rupert is a bear, living with his parents in an idyllic English town of
Nutwood, but his adventures take him all over the world (just like physics without COVID!).
It is a precursor of the Belgian and French TinTin [from 1929] and the French Asterix comics
[from 1959] and there are still annuals every year. In this adventure Rupert meets up with
his friend Bingo, the brainy pup, to help rescue the kidnapped Squire and foil some crooks
with a metal-tracer machine. Without electricity and magnetism, they could not have been
successful.
a) We needed to rewrite Ampere’s Law to include a time-dependent term.
Explain the revised equation, what displacement current means, how it relates to a
time-varying electric flux. Inside a material, what values should permeability and
permittivity have?
b) Faraday’s Law showed another time-dependent effect. Explain what it is and all
elements of the equation.
c) Time-dependent magnetic flux can arise from several alternate experimental
components changing with time. Explain three.
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 2 of 12
d) Bingo finds the trowel
Rupert’s father has lost his trowel in the garden, so Bingo
and Rupert find it using the metal-tracer.
Explain how the metal detector can detect the metal of a
trowel buried just under the surface. From lectures, and
demonstrations, discuss the characteristic of the trowel
which the metal detector can detect, knowing that the
metal detector has a big magnetic field which is either
static or moving.
e) Rupert and the metal detector
The metal detector can detect iron (in a trowel). Is it likely to be more sensitive to silver and
gold if the object is of the same size. Why?
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
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f) The French Trouve made a metal detector in 1874, and Alexander Graham Bell (the
telephone guy) made one in 1881. They were used in Admiral Byrd’s second Antarctic
expedition in 1933 and could detect objects 8 feet underground. They were used by
the Allies to detect mines in the second world war. Many prospectors look for gold,
jewellery or archaeological object like ancient coins or bullets with Bingo’s metaltracer. Sometimes these lie well below the surface. How can the machine be sensitive
enough to detect these?
g) By analogy with Bingo’s metal-tracer, explain the demonstration where a bar magnet
was falling through an aluminium pipe. What happened, and why?
[6 + 6 + 3 + 3 + 3 + 2 + 2 = 25 marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 4 of 12
Question 3 Blackbody Radiation
8 marks
A glowing candle, giving off a warm “yellowish” glow, can be considered a blackbody
radiator of T=800 K. In comparison the sun which can also be modelled as a black body
radiator of temperature of T=5500 K produces a much “whiter” spectrum of light.
a) With the aid of a sketched graph, compare the blackbody emission spectrum of the sun
and a candle. Label your axis and highlight the key points of difference.
b) Assuming the glowing radiating part of the candle has an area of approximately 10-3 m²,
estimate the power output of the candle. Outline any assumptions you have made in
your calculation.
The ideal spectrum of visible sunlight is shown in the diagram below. Upon closer
examination of the solar spectrum as measured at the surface of the earth reveals very fine
structure and specific dark bands.
c) Briefly explain one likely cause of these variations in your answer include a specific
statement about the actual process which is taking place to cause these differences.
[ 3 + 3 + 2 = 8 Marks]
Question 4 Heisenberg Uncertainty Principle
10 marks
a) Explain what is meant by the Heisenberg Uncertainty Principle (∆p∆x > h/4π)
In the Bohr model, an electron in the ground state of a hydrogen atom is considered to have
a kinetic energy of 13.56 eV.
b) Show that the momentum of this electron is approximately 2 x10-24 kg.m.s-1.
c) Use the Uncertainty Principle to estimate the diameter of the hydrogen atom. Make sure
you explain the reasoning behind your estimate.
[ 3 + 3 + 4 = 10 Marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
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Question 5 Bohr Model
7 marks
Data: Z=1 for H, Hydrogen, Z=2 for He, Helium and Z=3 for
Li, Lithium.
According to the Bohr model, the energies of a hydrogen
like atom or an ion which has one electron orbiting a
positively charged nucleus are given by
𝑍𝑍 2 𝑚𝑚𝑒𝑒 4 1
𝐸𝐸𝑛𝑛 = − 2 2 . 2
8𝜀𝜀𝑜𝑜 ℎ 𝑛𝑛
where ‘n’ is the principal quantum number.
An energy diagram of the quantum states for the H atom is shown in the figure above.
a) Write down the value of the of the term:
𝑍𝑍 2 𝑚𝑚𝑒𝑒 4
8𝜀𝜀𝑜𝑜2 ℎ2
1
. 𝑛𝑛2 (Include units in your answer)
When H atoms are excited in a gas discharge tube, they may exist briefly in a state with n ≥ 2.
As these atoms then de-excite to a lower quantum state they will emit radiation. Recall from
lectures that the Balmer series, which is emitted when electrons in an excited state drop to
the n = 2 state contains four lines that are in the visible region of the spectrum.
b) What is the lowest energy of a photon of visible light emitted in the Balmer series of
hydrogen.
Atoms of helium have Z = 2. If one of the electrons in this atom is removed, it is called a
singly ionised He ion, namely He+.
c) What is the binding energy of the He+? Explain how you arrived at your answer.
[ 2 + 2 + 3 = 7 Marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 6 of 12
Question 6 Ionisation
8 marks
The figure below shows the binding energy i.e. the energy required to remove the least
strongly bound electron of the light elements up to F, (Fluorine, Z=9).
a) Briefly explain the general trends indicated by the straight lines on this graph Include
in your answer
i. The reasons for the large difference in ionization energies of He and Li
ii. The reason that the ionization energies gradually increase from Li to F.
b) Estimate the ionization energy of the next element in the series Ne (Neon, Z =10).
c) Provide an approximate value for the ionization energy of Na (Sodium Z=11) and
explain your reasoning.
[ (2 + 2) + 2 + 2 = 8 Marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 7 of 12
Question 7 X-rays
11 marks
The spectrum of x-rays produced when an intense beam of energic electrons is directed
onto a Molybdenum (Mo) target is shown below. Note that there are two distinct peaks,
one at 63 pm, the other at 72 pm.
a) Briefly explain how the scattering of these electrons can produce the features of the
spectrum. Include in your explanation a reference to the formation of the distinct
peaks and the sharp cut-off at around 35 pm.
b) Calculate the energy of the incident electrons.
c) Assuming that the two peaks are the results of K transitions in Mo, estimate the
separation in the energies of the L and M shell in Mo.
d) Consider now the case where the energy of incident electrons is now doubled. On a
clearly labelled diagram with axes similar to that above, sketch the new spectrum of
x-rays you would expect these electrons to produce.
[ 3 + 1 + 3 + 4 = 11 Marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 8 of 12
Question 8 Characteristic X-rays
6 marks
Moseley observed that characteristics x-rays from an element were unique. By carefully
studying a large number of elements, he showed a linear relationship between the square
root of the frequency of emission and an element’s position in the periodic table.
Ultimately, Moseley showed that the energy of the 𝐾𝐾𝛼𝛼 x-rays produced by any element
satisfy the empirical relation:
𝐸𝐸𝐾𝐾𝛼𝛼 = ℎ𝑓𝑓 = 13.56𝑒𝑒𝑒𝑒(𝑍𝑍 − 1)2 �
~ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. × 𝑍𝑍 2
1
1
− 2�
2
1
2
With reference to the Bohr model, explain how one may interpret the terms in this equation.
For full marks, make sure you include a rationale for the “constant × (Z-1)2” term and the
quantities in the square brackets.
[6 Marks]
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 9 of 12
End of Exam
Formula Sheet over page
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
Page 10 of 12
Formula Sheet and Useful Data for PHYC10004 (may be detached)
Page 1 of 2
Constants
Mass of electron, me
9.109 x 10-31 kg = 0.00055 u
Mass of proton, mp
1.673 x 10-27 kg = 1.00728 u
Mass of neutron, mn
1.675 x 10-27 kg = 1.00866 u
Electron charge magnitude, e
1.6 x 10-19 C
Permeability of free space, µo
4π x 10-7 T m A-1
Permittivity of free space, εo
8.85 x 10-12 C2 N-1 m-2
Bohr radius, aB
0.529 x 10-10 m
Universal gas constant, R
8.31 J K-1 mol-1
Avogadro’s number, NA
6.02 x 1023 mol-1
Specific heat capacity of ice
2220 J K-1 kg-1
Specific heat capacity of water
4187 J K-1 kg-1
Atomic mass unit u
1.6605 x 10-27 kg = 931.49 MeV/c2
-19
1 eV = 1.6 x 10 J
Speed of light in vacuum, c
299 792 458 m s-1
Gravitational acceleration, g
9.8 m s-2
Planck’s constant, h = 6.63 x 10-34 J s = 4.14 x 10-15 eV s
Rydberg constant, R
1.097373 x 107 m-1
Boltzmann constant, k
1.38 x 10-23 J K-1
Stefan-Boltzmann constant, σ
5.67 x 10-8 W m-2 K-4
1 atm = 101325 Pa = 760 mm Hg
Latent heat of fusion for water
333 kJ kg-1
Density of water
1000.0 kg/m3
Formulae
Electromagnetism
𝑞𝑞1 𝑞𝑞2
𝑭𝑭 =
𝑟𝑟̂
4𝜋𝜋𝜀𝜀𝑜𝑜 𝑟𝑟 2
𝑊𝑊 = −∆𝑈𝑈
� 𝑬𝑬. 𝑑𝑑𝑨𝑨 =
𝑞𝑞𝑒𝑒𝑒𝑒𝑒𝑒
𝜀𝜀𝑜𝑜
𝑉𝑉 = 𝑖𝑖𝑖𝑖
𝐶𝐶 =
𝑄𝑄
𝑉𝑉
𝜀𝜀 = 𝜅𝜅𝜀𝜀0
𝑭𝑭 = 𝑞𝑞𝒗𝒗 × 𝑩𝑩
𝐵𝐵 = 𝜇𝜇𝑜𝑜 𝑛𝑛𝑛𝑛
𝜇𝜇0 𝑖𝑖𝑖𝑖𝒔𝒔 × 𝒓𝒓�
4𝜋𝜋 𝑟𝑟 2
𝑑𝑑Φ𝐸𝐸
𝑖𝑖𝐷𝐷 = 𝜀𝜀𝑜𝑜
𝑑𝑑𝑑𝑑
𝑑𝑑𝑩𝑩 =
𝑭𝑭
𝑞𝑞
𝑞𝑞
𝑬𝑬 =
𝒓𝒓�
4𝜋𝜋𝜀𝜀𝑜𝑜 𝑟𝑟 2
𝑬𝑬 =
∞
𝑉𝑉(𝑃𝑃) = � 𝑬𝑬. 𝑑𝑑𝒔𝒔 =
𝑃𝑃
𝑞𝑞
4𝜋𝜋𝜀𝜀𝑜𝑜 𝑟𝑟
𝒑𝒑 = 𝑞𝑞𝒅𝒅
𝝉𝝉 = 𝒑𝒑 × 𝑬𝑬
𝑈𝑈 = −𝒑𝒑. 𝑬𝑬
𝑉𝑉𝑆𝑆 𝑁𝑁𝑆𝑆
=
𝑉𝑉𝑝𝑝 𝑁𝑁𝑝𝑝
𝐴𝐴𝜀𝜀𝑜𝑜
𝐶𝐶 =
𝑑𝑑
𝑑𝑑𝑑𝑑
𝑖𝑖 =
= 𝑒𝑒𝑒𝑒𝑒𝑒𝑣𝑣𝑑𝑑
𝑑𝑑𝑑𝑑
𝑉𝑉 =
𝑃𝑃1
∆𝑉𝑉 = − � 𝑬𝑬. 𝑑𝑑𝒔𝒔
𝑃𝑃2
𝑉𝑉 2
𝑃𝑃 = 𝑖𝑖𝑖𝑖 =
= 𝑖𝑖 2 𝑅𝑅
𝑅𝑅
1
1 2 1 𝑞𝑞 2
𝑈𝑈 = 𝑞𝑞𝑞𝑞 = 𝐶𝐶𝑉𝑉 =
2
2
2 𝐶𝐶
1
𝑈𝑈𝐸𝐸 = 2𝜅𝜅𝜀𝜀𝑜𝑜 𝐸𝐸 2
𝑭𝑭 = 𝑖𝑖𝒍𝒍 × 𝑩𝑩
Φ𝐸𝐸 = � 𝑬𝑬. 𝑑𝑑𝑨𝑨
𝜀𝜀 = � 𝑬𝑬. 𝑑𝑑𝒔𝒔 = −
𝑑𝑑Φ𝐵𝐵
𝑑𝑑𝑑𝑑
𝜇𝜇 = 𝑖𝑖 × (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙)
𝑈𝑈(𝑃𝑃)
𝑞𝑞
� 𝑩𝑩. 𝑑𝑑𝑨𝑨 = 0
Φ𝐵𝐵 = � 𝑩𝑩. 𝑑𝑑𝑨𝑨
� 𝑩𝑩. 𝑑𝑑𝒔𝒔 = 𝜇𝜇𝑜𝑜 𝑖𝑖𝑒𝑒𝑒𝑒𝑒𝑒 + 𝜇𝜇𝑜𝑜 𝑖𝑖𝐷𝐷
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
𝝉𝝉 = 𝝁𝝁 × 𝑩𝑩
𝑈𝑈 = −𝝁𝝁. 𝑩𝑩
Page 11 of 12
Formula Sheet and Useful Data for PHYC10004
Page 2 of 2
Fluids and Thermal Energy
𝑅𝑅 = 𝐴𝐴1 𝑣𝑣1 = 𝐴𝐴2 𝑣𝑣2
𝑝𝑝𝑝𝑝 = 𝑛𝑛𝑛𝑛𝑛𝑛
1
𝑝𝑝 + 𝜌𝜌𝜌𝜌ℎ + 𝜌𝜌𝑣𝑣 2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝑝𝑝 = 𝑝𝑝𝐴𝐴𝐴𝐴𝐴𝐴 + 𝜌𝜌𝜌𝜌ℎ
𝜌𝜌 =
𝑝𝑝𝑉𝑉 𝛾𝛾 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑐𝑐
𝑓𝑓
𝛾𝛾 = 𝑐𝑐𝑝𝑝
𝑐𝑐𝑣𝑣 = 2 𝑅𝑅
𝑓𝑓
𝑛𝑛𝑛𝑛𝑛𝑛
2
𝑄𝑄 = 𝑛𝑛𝐶𝐶𝑝𝑝 Δ𝑇𝑇
𝑈𝑈 =
𝑉𝑉
𝑐𝑐𝑝𝑝 = 𝑐𝑐𝑣𝑣 + 𝑅𝑅
𝑊𝑊 = − � 𝑝𝑝𝑝𝑝𝑝𝑝 = −𝑝𝑝∆𝑉𝑉
Modern Physics
𝜆𝜆𝑀𝑀𝑀𝑀𝑀𝑀 𝑇𝑇 = 2898 𝜇𝜇𝜇𝜇 𝐾𝐾
𝑃𝑃 = 𝜎𝜎𝜎𝜎𝜎𝜎𝑇𝑇 4
ℎ
𝑝𝑝
ℎ2 2
𝐸𝐸𝑛𝑛 =
𝑛𝑛
8𝑚𝑚𝐿𝐿2
𝜆𝜆 =
𝑚𝑚𝑒𝑒 4 𝑍𝑍 2
13.6𝑍𝑍 2
𝐸𝐸𝑛𝑛 = − 2 2 2 = −
𝑛𝑛2
8𝜀𝜀0 ℎ 𝑛𝑛
1
1
2 − 2�
𝑛𝑛2 𝑛𝑛1
𝐿𝐿 = �𝑙𝑙(𝑙𝑙 + 1)ℏ
𝐿𝐿𝑧𝑧 = 𝑚𝑚𝑙𝑙 ℎ
𝑓𝑓 = 𝑐𝑐𝑐𝑐 �
𝑁𝑁(𝑡𝑡) = 𝑁𝑁0 𝑒𝑒 −𝜆𝜆𝜆𝜆
𝑀𝑀
𝑉𝑉
𝑊𝑊 = −𝑛𝑛𝑛𝑛𝑛𝑛ln
𝑢𝑢𝜆𝜆 (𝑇𝑇) =
𝑎𝑎𝐵𝐵 =
𝑉𝑉𝑓𝑓
𝑉𝑉𝑖𝑖
𝑐𝑐 = 𝑓𝑓𝑓𝑓
log e 2
𝜏𝜏1⁄2
𝑐𝑐
𝑠𝑠𝜆𝜆 (𝑇𝑇) = 𝑢𝑢𝜆𝜆 (𝑇𝑇)
4
𝐸𝐸 = ℎ𝑓𝑓 =
ℎ𝑐𝑐
𝜆𝜆
ℎ
(1 − cos 𝜃𝜃)
𝑚𝑚𝑚𝑚
2𝑑𝑑 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
Δ𝑥𝑥Δ𝑝𝑝 ≥
1
𝑟𝑟𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = (1.2 × 10−15 𝑚𝑚)𝐴𝐴3
𝜆𝜆 =
𝑄𝑄 = 𝑚𝑚𝑚𝑚
𝑄𝑄 = 𝑚𝑚𝑚𝑚∆𝑇𝑇
∆𝜆𝜆 =
4𝜋𝜋𝜖𝜖𝑜𝑜 ℏ2 2
𝑛𝑛 = 𝑎𝑎𝐵𝐵 𝑛𝑛2
𝑚𝑚𝑒𝑒 2
𝑆𝑆𝑧𝑧 = 𝑚𝑚𝑠𝑠 ℎ
4√2𝜋𝜋 (𝑁𝑁/𝑉𝑉)𝑟𝑟 2
3𝑘𝑘𝑘𝑘
𝑣𝑣𝑅𝑅𝑅𝑅𝑅𝑅 = �
𝑚𝑚
8𝜋𝜋𝜋𝜋ℎ
1
� ℎ𝑐𝑐
�
5
𝜆𝜆
𝑒𝑒 𝑘𝑘𝑘𝑘𝑘𝑘 − 1
𝜖𝜖𝑜𝑜 ℎ2
= 52.92 𝑝𝑝𝑝𝑝
𝜋𝜋𝜋𝜋𝑒𝑒 2
1
∆𝑈𝑈 = 𝑄𝑄 + 𝑊𝑊
ℎ𝑓𝑓 = 𝐾𝐾𝑀𝑀𝑀𝑀𝑀𝑀 + 𝜙𝜙
𝑟𝑟𝑛𝑛 =
𝜆𝜆 =
𝛼𝛼𝑒𝑒𝑒𝑒
ℏ
2
𝜈𝜈𝑛𝑛 = 𝑐𝑐𝛼𝛼𝑒𝑒𝑒𝑒
1
𝑛𝑛
𝑒𝑒 2
1
=
≈
4𝜋𝜋𝜖𝜖𝑜𝑜 ℏ𝑐𝑐 137
𝐵𝐵. 𝐸𝐸.
= (𝐴𝐴 − 𝑍𝑍)𝑚𝑚𝑛𝑛 + 𝑍𝑍𝑚𝑚𝑝𝑝 − 𝑚𝑚𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛
𝑐𝑐 2
2020 PHYC10004 Physics 2: Physical Science and Technology Exam
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