93/08(a) Semester 2, 2010
THE UNIVERSITY OF SYDNEY
PHYS1004 PHYSICS 1 (ENVIRONMENTAL & LIFE SCIENCES)
Solutions
NOVEMBER 2010
Time allowed: THREE Hours
MARKS FOR QUESTIONS ARE AS INDICATED TOTAL: 90 marks
INSTRUCTIONS • All questions are to be answered. • Use a separate answer book for each section. • All answers should include explanations in terms of physical principles. DATA
Density of water ρ = 1.00 × 103 kg.m−3
Density of air ρ = 1.20 kg.m−3 Atmospheric pressure 1 atm = 1.01 × 105 Pa Magnitude of local gravitational field g = 9.80 m.s-2
Avogadro constant NA = 6.022 × 1023 mol-1
Permittivity of free space ε0 = 8.854 × 10−12 F.m−1
Coulomb’s law constant K = 8.99 × 109 N.m2.C-2
Permeability of free space µ0 = 4π × 10−7 T.m.A−1
Elementary charge e = 1.602 × 10−19 C
Speed of light in vacuum c = 2.998 × 108 m.s−1
Planck constant h = 6.626 × 10−34 J.s
Rest mass of an electron me = 9.110 × 10−31 kg
Rest mass of a neutron mn = 1.675 × 10−27 kg
Rest mass of a proton mp = 1.673 × 10−27 kg
Rest mass of a hydrogen atom mH = 1.674 × 10−27 kg
Boltzmann constant k = 1.381 × 10−23 J.K-1
Stefan-Boltzmann constant σ = 5.671 × 10-8
W. m-2
K-4
Atomic mass unit u = 1.661 × 10−27 kg
Energy equivalent of 1 atomic mass unit E = 931.49 MeV
ENV_Q01
Semester 2, Y2010 Question 1
Mineral sands consist of some grains that are conducting and others that are non-conducting. These two components can be separated using the following process.
• The sand grains are charged positively by passing them through a stream of positive ions.
• The charged sand, which includes conducting and non-conducting sand grains, is then dropped on top of an earthed, rotating, metal disk.
• As the disk turns the conducting and non-conducting sand grains are separated.
(a) Briefly explain what happens when a conducting sand grain comes in contact with the metal disk.
(b) Briefly explain what happens when a non-conducting sand grain comes in contact with the metal disk.
(c) Explain how the differing behaviour in parts (a) and (b) can be used to separate the conducting and non-conducting sand grains.
(5 marks)
Solution
All the positively charged particles will induce negative charges on the metal disk. (1 mark) (a) The charge on the conducting sand particles is neutralised by the flow of electrons from the metal disk and they soon fall off the disk as it turns. (1 mark) (b) The charge on the non-conducting sand particles cannot be neutralised, and the force of attraction between the positive charge on the sand and the induced negative charge on the roller holds the particle to the roller. (2 marks) (c) Thus, the conducting sand particles come off the disk first and land in a pile, while the non-conducting sand particles have to be scraped off the disk. Thus refinement is achieved. (1 mark)
ENV_Q02
Semester 2, Y2010 Question 2 You have constructed a circuit and to test it you need to measure the potential difference between locations in the circuit. You would use a voltmeter (or a voltage setting on a multimeter) to measure this. You may also need to measure the current flowing at different locations in the circuit. You would use an ammeter (or a current setting on a multimeter) to measure this.
(a) What is the property of an ideal voltmeter that ensures it does not affect the measurement that you wish to make?
(b) What is the property of an ideal ammeter that ensures it does not affect the measurement that you wish to make?
Suppose that your circuit needs a voltmeter that goes from 0.0 V to a full-scale reading of 5.0 V . But all you have is an ammeter that goes from 0 to a full-scale reading of 500μA. You can use this meter to measure voltages by putting it in a measuring circuit as shown in the figure.
(c) What value of R must you use so that the meter will read 500μA when the potential
difference VΔ is 5.0 V ? Assume that the ammeter is ideal. (5 marks)
Solution (a) An ideal voltmeter has infinite resistance. (1 mark) (b) An ideal ammeter has zero resistance. (1 mark) (c) When the potential difference VΔ is equal to 5.0 V then the current flowing through the ammeter and hence also the resistance R is 500μA. Therefore
6
4
5.0 Ω500 10
=1.0 10 Ω
VV I R RI −
ΔΔ = ⇒ = =
××
(1 mark for Ohm’s law; 1 mark for correct answer; 1 mark for units)
ENV_Q03
Semester 2, Y2010 Question 3
Air flows horizontally past the wing of a small airplane so that the speed is 170.0m.s− over the slightly curved top surface and 160.0m.s− past the flat bottom surface. The wing has an area of
220.0 m on both its top and bottom surfaces. Assume that the flow of air past the wing is laminar. Assume also that the air does not compress during the flow and that the density of air is
31.20 kg.m− .
(a) Draw a diagram showing the streamlines of the flow of air above and beneath the wing.
(b) Calculate the pressure difference between the top and bottom surfaces of the wing.
(c) Calculate the net force on the wing and its direction.
(5 marks)
Solution (a)
The essential features of this diagram from Knight Field and Jones (edition 2) is that the streamlines are closer together above the wing and less close together below the wing. (1 mark) (b) Bernoulli’s equation applied to the top and bottom surfaces of the wing is:
2 21 12 2t air t b air bp v p vρ ρ+ = +
where tp and bp are the pressures at the top and the bottom surfaces of the wing; and tv and bv are the flow velocities at the top and the bottom surface of the wing. This can be written in terms of the pressure difference between the top and bottom surfaces as
2 21 1
2 2b t air t air bp p v vρ ρ− = −
Substituting values we get:
( )2 2(0.5)(1.20) 70 60 780 Pab tp p− = − =
(2 marks) (c) Pressure and force F are related by
FpA
=
where A is the area of the wing surface. (½ mark) Substituting values we get 4(780)(20) 1.56 10 NF p A= = = × which is directed upwards. (1 mark for answer and ½ mark for direction)
ENV_Q04=TEC_Q04
Semester 2, Y2010 Question 4
A piece of iron is glued to the top of a block of wood. The block is placed in a tank of water with the iron on top (see diagram) and is found to float.
(a) Does the water level in the tank, rise, fall, or stay the same when the block is placed into the tank? Briefly justify your answer.
The block is now turned over so that the iron is submerged beneath the wooden block.
(b) Does the block still float or does it sink to the bottom? Briefly justify your answer.
(c) Does the water level in the tank, rise, fall, or stay the same as compared with the situation when the block is floating with the iron on top? Briefly justify your answer.
The blocks were removed from the tank and separated. When they were returned to the tank, the piece of iron sank to the bottom of the tank and the wood floated.
(d) Does the water level in the tank, rise, fall, or stay the same as compared with the situation before the iron is attached to the underneath of the block of wood. Briefly justify your answer.
(5 marks)
Solution
(a) The level in the tank rises when the block is put into the water. Archimedes’s principle tells us that the block, when floating, displaces a volume of water having the same weight as the weight of the block. It is this volume of displaced water that causes the rise in the level of water in the tank. This can be written as (not needed for the marks): ( )
( )1
1 /wood iron water
wood iron water
m m g gV
V m m
ρ
ρ
+ =
⇒ = +
where 1V is the volume of water displaced, waterρ is the density of water, woodm and ironm are the masses of the wood and iron respectively. (½ mark answer; 1 mark justification) (b) The block still floats. Archimedes’s principle does not depend on the orientation of the object. (½ mark answer; ½ mark justification) (c) The level stays the same. The same volume of water is displaced and this causes the same rise in the water level in the tank. (½ mark answer; ½ mark justification) (d) The water level in the tank will go down. The volume of water displaced by the floating block of wood remains the same ( )/wood waterm ρ . But the volume of water displaced by the more dense
iron is now just the volume of the iron block itself ( )/iron ironm ρ . Previously this volume was the volume of the equivalent amount of water having the same weight as the iron block( )/iron waterm ρ . This can be written as (not needed for the marks)
2 / /wood water iron ironV m mρ ρ= + We have
2 1V V< because iron waterρ ρ> (½ mark answer; 1 mark justification)
ENV_Q05
Semester 2, Y2010 Question 5 A collimated beam of electrons, accelerated through a potential difference of 600 V, passes through a thin aluminium foil onto a photographic plate.
(a) Draw a neat sketch showing the pattern observed on the photographic plate after several thousand electrons have passed through the foil.
(b) Briefly explain why the observed pattern occurs.
(c) What would happen to the pattern if the total exposure were the same but the intensity of the beam were reduced to the stage where on average there was only one electron at a time passing through the foil? What does this reveal about the nature of electrons?
(d) What would happen to the pattern if the accelerating voltage were increased to 2400 V ? Briefly explain your answer
(5 marks)
Solution (a)
(b) Electrons have both particle and wave character. Here electron waves diffract and interfere as a result of interacting with the foil producing bright circles where constructive interference occurs and circular nulls where destructive interference occurs. (1 mark) (c) Points are: * The pattern would be the same. (1 mark) * The pattern indicates the probability that individual electrons will strike particular points
on the photographic plate. It is not due to interactions between electrons! * It reveals that individual electrons exhibit wave-like properties under certain
circumstances. * It reveals the particle nature of electrons as each electron strikes a single point. (½ mark to a maximum 1 mark) (d) The radii of the circles in the pattern will decrease. (½ mark) This is because the electron’s kinetic energy will quadruple, their momentum double, and their de Broglie wavelength will decrease by a factor of two. (½ mark) Hence the radii of the circles will decrease by a factor of two, since
/diff dθ λ≺ where d is the effective (average) crystal size. (bonus ½ mark)
ENV_Q06
Semester 2, Y2010 Question 6 The isotope 99
42 Mo of Molybdenum is produced in a nuclear reactor. It has too many neutrons to be stable and decays into a metastable state of Technetium-99, 99
43Te.m This then decays into 9943Te and a gamma ray with energy of 140.4 keV .
(a) Write down the nuclear equations for these two decays.
(b) An 80 kg technician absorbs 20% of the gamma rays from a 510 Bq sample of 9943Tem
for 5 minutes. Calculate the absorbed dose in grays ( 11Gy 1J.kg−= ).
(c) Would the same absorbed dose of alpha particles cause more, less, or the same biological damage to the technician than the gamma rays in part (b)?
(5 marks)
Solution (a) 99 9942 4399 99
43 43
Mo Te + e
Te Te +
me
m
ν
γ
−→ +
→
(1 mark for each) (b)
5 7Number of decays activity* time 10 Bq*300s 3 10= = = × (½ mark)
7 3 19
7
Energy deposited fraction *number of decays*energy per decay(0.2)(3 10 )(140.4 10 )(1.6 10 )1.3 10 J
−
−
=
= × × ×
= ×
(1 mark)
7
9
Absorbed Dose Energy / mass1.3 10 J / 80 kg1.7 10 Gy
−
−
=
= ×
= ×
(½ mark) (c) Greater because alphas cause more ionization. (1 mark)
ENV_Q07=TEC_Q07
Semester 2, Y2010 Question 7
A large blue-black cumulonimbus cloud towers above its base which is flat and 2.0 km 2.0 km× square, 500 m above the earth. Just before a lightning strike the electric field at the ground is measured at 6 11.2 10 V.m−× vertically upwards.
(a) Assume that the base of the cloud and an equal area of the Earth ( 2.0 km 2.0 km× ) directly below it form the electrodes of a parallel plate capacitor. Calculate the effective capacitance of this system.
(b) Calculate the potential difference between the surface of the earth and the base of the cloud.
(c) Calculate the amount of charge on the base of the cloud and determine its sign.
(d) What is likely to be the sign of any charge on the raindrops that fall from the cloud?
(e) A lightning flash occurs lasting for 1.0 ms and transfers all the charge on the cloud to the earth in a single bolt of lightning. Calculate the average current that flows in the lightning discharge.
(f) What is the magnitude of the magnetic field during the current flow at a horizontal distance of 5.0 m from the site of the lightning strike?
(10 marks)
Solution (a)
0 ACdε
=
We have ( )23 6 22.0 10 4.0 10 mA = × = × and 500 m.d = Thus
( )( )12 68
8.85 10 4.0 107.08 10 F 71nF.
500C
−−
× ×= = × =
(2 marks) (b)
( )( )6 81.2 10 500 6.0 10 V 600 MV.
V E d=
= × = × =
(2 mark) (c)
( )( )8 87.08 10 6.0 10 42.5C 43CQ CV −= = × × = = . (2 mark) The electric field at the earth is pointing upwards; therefore the earth is positive and the charge on the cloud is negative because electric field lines are drawn from positive charges to negative charges. (1 mark) (d) Since the base of the cloud is negative and the charge is held on water droplets in the cloud, then it follows that the raindrops formed in the cloud that fall to earth will be negatively charged. (1 mark) (e) The current in the lightning strike is
43
42.5 4.25 10 A 43kA1.0 10
QIt −= = = × =
Δ ×
(1 mark) (f)
0
2IBr
μπ
=
Using 5.0 mr =
( )( )( )
7 43
4 10 4.25 101.7 10 T
2 5.0B
π
π
−−
× ×= = ×
(1 mark) (do not cascade penalties from earlier errors)
ENV_Q08
Semester 2, Y2010 Question 8
A single circular loop of wire of radius 0.010 mR = carries a current of 10.0 AI = in the anti-clockwise direction about the centre of the loop at C (see above figure). A smaller loop of radius
0.0050 mr = encircles the centre point C and is initially in the same plane as the large loop. The small loop is open at points A and B where a potential difference can be measured.
(a) What is the strength of the magnetic field at the centre point C due to the current flowing in the large loop?
(b) If the large loop consisted of 20N = turns of wire with current I in each loop, what would be the strength of the magnetic field at point C?
(c) If the current remains constant in the large loop and the small loop remains stationary. Would there be a potential difference between points A and B? Briefly justify your answer.
The current in the large loop of wire is now reduced smoothly from 10.0 AI = to 5.0 AI = in a time of 31.0 10 st −= × . Assume that the magnetic field is uniform over the area of the small coil.
(d) What will be the value of the potential difference measured between contacts A and B on the small coil as the current falls?
With a current 10.0 AI = flowing in the single large loop, the small loop is rotated through an angle of 90 in a time 31.0 10 st −= × . After rotation, the small loop is edge-on as seen in the diagram above. Assume that the magnetic field is uniform over the area of the small coil.
(e) What will be the value of the potential difference measured between contacts A and B on the small coil as the loop rotates?
10 marks
Solution (a)
( )7 40 104 10 6.28 10 T2 2 0.01
IBR
μ π − −⎛ ⎞= = × = ×⎜ ⎟×⎝ ⎠
(2 marks) (b)
5 20 20 6.28 10 =1.26 10 T2N IBR
μ − −= = × × ×
(2 marks) (c) No. The magnetic field at point C (and across the area of the small loop) is constant and the area of the small loop perpendicular to the field remains constant. By Faraday’s law there is no emf induced around the small loop and no potential difference between points A and B. (2 marks) (d) The area of the small loop is
2 3 2 5 2(5 10 ) 7.85 10 mA rπ π − −= = × = × . Magnetic flux through small loop
( )( )4 5 8Area= 6.28 10 7.85 10 4.93 10 WbBϕ − − −= × × × = ×
If the current is halved in 31.0 10 s−× then the flux is halved, and 82.46 10 Wb,ϕ −Δ = × and 31.0 10 st −Δ = ×
Faraday’s Law gives: 8
53
2.46 10emf 2.46 10 V1.0 10t
ϕ −−
−
Δ ×= = = ×Δ ×
(2 marks) (e) This time the flux goes from the initial flux to zero in time 31.0 10 st −Δ = × .
84.93 10 Wbϕ −Δ = × and 31.0 10 st −Δ = × Faraday’s Law gives
85
3
4.93 10emf 4.93 10 V1.0 10t
ϕ −−
−
Δ ×= = = ×Δ ×
(2 marks)
ENV_Q09
Semester 2, Y2010 Question 9 The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. After the aorta, the blood flow is split up and finally circulates through smaller blood vessels known as capillaries.
(a) Calculate the average speed of the blood in the aorta if the flow rate is 15.0L.min− . Assume that the aorta has a radius of 21.0 10 m.−× ( 3 31L 10 m−= )
(b) When the rate of blood flow in the aorta is 15.0L.min− , the speed of the blood in the capillaries is 4 13.3 10 m.s− −× . If the average radius of a capillary is 64.0 10 m,−× calculate the number of capillaries in the blood circulation system.
(c) Due to fatty deposits, the inside diameter of the aorta is reduced so that it now has half of its original diameter. Calculate the percentage increase in blood pressure that must be applied by the heart in order to maintain the same blood flow rate.
(d) Can changing the viscosity of the blood assist people who have blocked arteries? Briefly explain your answer.
Solution (a)
Volume flow rate in the aorta is a a aVQ v At
Δ= =Δ
where v is the flow velocity and A is the area of the flow. For the aorta we have
( )( ) ( )
3 31
22 2
1
5.0 10 m0.265m.s
60s 1.0 10 m
0.27 m.s
aa
a a
Q V VvA A t t rπ π
−−
−
−
×Δ Δ= = = = =
Δ Δ ×
=
(2 marks) (b) Assume there are N capillaries each of radius 64.0 10 mcr
−= × . The flow velocity in the capillaries is 4 13.3 10 m.s .cv − −= × Equation of continuity gives that
( )( )( ) ( )
3
22 4 6
9
5.0 10
60 3.3 10 4.0 10
5.0 10
a a c c
a a
c c c c
v A N v A
v A VNv A t v rπ π
−
− −
=
×Δ⇒ = = =
Δ × ×
= ×
(2 marks) (c) Poiseuille’s equation is
4
8R pQ
Lπ
ηΔ
=
(1 mark) If the flow rate is maintained then we have
4 41 1 2 2
441
2 1 1 142
8 8
2 16
R p R pQL LRp p p pR
π πη ηΔ Δ
= =
⇒ = = =
Blood pressure is multiplied by a factor of 16. (2 marks) Percentage increase in blood pressure is
2 1 2
1 1
1 15 1500%p p pp p−
= − = =
(1 mark) (d) Poiseuille’s equation is
4
8R pQ
Lπ
ηΔ
=
If viscosity of the blood can also be changed then 4 4
1 1 2 2
1 24
1 2 22 1 14
2 1 1
8 8
16
R p R pQL L
Rp p pR
π πη η
η ηη η
Δ Δ= =
⇒ = =
If 2 1η η< then 2 116p p< and a lower blood pressure is required to maintain the same volume flow rate. (2 marks)
ENV_Q10
Semester 2, Y2010 Question 10 In a test of the mechanical properties of animal tendon, a sample of tendon is subjected to stretching. The sample is cylindrical in shape, with a diameter of 3.00 mm and an initial unstretched length of 50.0 mm .
It is found that the sample obeys Hooke’s law up to a tensile force of 250 N , at which point the sample has increased its length to 53.0 mm . At a larger tensile force the sample shows plastic deformation, and then breaks when the tensile force is 580 N and the length is 59.0 mm .
(a) Calculate both the stress and strain values at: (i) the elastic limit; (ii) the breaking point.
(b) Draw a graph of stress versus strain for this process. Include labels and representative values on your graph.
(c) On your graph from part (b) indicate: (i) the Hookean region; (ii) the elastic limit; (iii) the breaking point.
(d) Calculate Young’s modulus for this sample (in the Hookean region) (10 marks)
Solution
(a) (i) At the elastic limit, i.e. the limit of Hooke’s Law, 250 NF = and length 53.0 mm.=
Stress = FA
( )23
250
1.50 10π −=
× × 6
2507.07 10−=
×
73.50 10 Pa= × (1 mark)
Strain 0
LLΔ
= 3.050.0
= 26.0 10−= ×
(1 mark) (ii) At the breaking point
Stress FA
= 6
5807.07 10−=
×
78.2 10 Pa= × (1 mark)
Strain 0
LLΔ
= 9.050.0
= 0.18=
(1 mark)
(b)
(general shape 1 mark) (named axes 1 mark) (c) (values on axes ½ mark, labeling of regions requested in parts (i), (ii) and (iii) ½ mark each). (d) In the linear (Hookean) region, stress and strain are related by
0
F LyA L
Δ=
Taking data from the elastic limit 0LFy
A L=
Δ
7
2
3.5 106.0 10−
×=
× [from part (a)(i)]
85.8 10 Pa= × 0.58GPa=
(2 marks)
ENV_Q11
Semester 2, Y2010 Question 11 (a) The radionuclide radium-226 (often found naturally in granite and certain soils) decays to
produce the radioactive gas radon-222, which can build up in the poorly ventilated basements of houses. The decay can be written
226 222 488 86 2Ra Rn He→ +
Neutral atoms of the three nuclides have the following masses: 226
88Ra 226.025406 u 222
86 Rn 222.017574 u 42 He 4.002603u
(i) What is this kind of decay called?
(ii) What is the maximum kinetic energy with which the 42 He nucleus can be ejected?
Express your answer in MeV.
(iii) What is the binding energy of the radium-226 nucleus?
(b) An archaeologist is trying to establish the age of an ancient tree branch. A sample of the branch is converted to 32.50 10 kg−× of pure carbon and its 14 C activity is found to be 0.45 Bq .
How old is the tree branch, given that the 14 C activity for a living tree is 255Bq per kgof carbon? The half life of 14 C is 5730 years.
(10 marks)
Solution (a)
(i) Alpha decay (1 mark)
(ii) Excess energy is due to mass lost ("mass defect").
Use atomic masses only because electrons cancel Mass lost (Ra) (Rn) (He)m m m= − −
3
226.025406u 222.017574u 4.002603u5.229 10 u−
= − −
= ×
( )( )3 27 305.229 10 1.661 10 kg 8.685 10 kg− − −= × × = ×
(1 mark)
( )( )22 30 8
13
13 19
8.685 10 kg 2.998 10
7.806 10 J7.806 10 /1.602 104.87 MeV (allow up to 5 sig fig)
E mc −
−
− −
= = × ×
= ×
= × ×=
(1 mark) The answer above is strictly the total energy rather than the kinetic energy. Calculation of the kinetic energy needs consideration of momentum. Give students who consider this more complicated situation a bonus mark. (iii) Mass defect of 226
88Ra nucleus is: ( )
27 27 27
27
( ) 88 226 88
(226.025406)(1.661 10 ) (88)(1.673 10 ) (138)(1.675 10 )2.945 10 kg.
p nm Ra m m− − −
−
− − −
× − × − ×
= ×
(1 mark) Binding energy is therefore
( )( )227 8 102.945 10 2.998 10 2.65 10 J.− −× × = × (1 mark) The question does not ask for conversion to eV or MeV but the answers are:
91.65 10 eV1650MeV
×
(b)
Activity per kg of sample 3 10.45 / 2.50 10 180 Bq.kg− −× = (1 mark)
From formula sheet 0/ e 180 / 255 0.706tR R λ−= = = (1 mark)
( )01/2
ln 2 ln /t R Rλτ
− = − = (1 mark)
( ) 1/20ln / 2879 2900years
ln 2t R R τ
∴ = − = ≈
(allow up to 5 sig.fig) (2 marks)
ENV_Q12
Semester 2, Y2010 Question 12 (a) An experimenter finds that more electrons are emitted from a clean metal surface when
she illuminates it with light of frequency 1f than when she illuminates it with light of a different frequency 2f . The intensity of the illumination is the incident power per unit area.
(i) If the intensities of the two light beams are the same then what can she deduce about the two frequencies and about the maximum kinetic energies of the electrons emitted? Briefly justify your answer.
(ii) What would happen if she reduced the intensity of the light beams to an extremely low level? Briefly justify your answer
(b) The surface of the Sun has a temperature of approximately 6000 K. To a good approximation, we may treat it as a blackbody.
(i) What is the wavelength of the peak intensity of the radiation produced by the Sun?
(ii) In what part of the electromagnetic spectrum does this wavelength fall?
There are wavelengths within the Sun’s spectrum where much less light is emitted than expected from a blackbody. These appear as narrow, dark lines at wavelengths some of which are related by simple formulae.
(iii) How are these lines produced?
(iv) Describe how these lines can be used to determine the energy levels of atoms and ions, and to give the chemical composition of the Sun’s atmosphere.
(10 marks)
Solution (a) (i) If the photon energy h f is greater than the work function 0E of the metal then each photon produces one photoelectron. So more electrons emitted from the surface implies that the beam of frequency 1f contains more photons 1n than the beam of frequency 2f ( 2n photons). (1 mark) Since the intensities are the same then
1 1 2 2n h f n h f= So
1 2 1 2n n f f> ⇒ < (1 mark) The energy of a photon of frequency 1f must be less than that of a photon of frequency 2f . The maximum kinetic energy ( )maxK of an electron emitted from the metal surface is given by: max 0K h f E= − where 0E is the work function of the metal. Since 2 1f f> then the maximum kinetic energy of photoelectrons emitted from the surface illuminated by frequency 1f will be larger than the maximum kinetic energy emitted by the surface illuminated by frequency 2f . (1 mark) (ii) Since the intensity is not reduced to zero some electrons would still be emitted since there are still photons that would interact with electrons in the metal. (1 mark) The number of photons is reduced but the energy of each photon remains the same. The work function of the metal remains the same. So the maximum kinetic energy of the photoelectrons remains the same. (1 mark) (b) (i)
3max
3
max
7
2.90 10 m.K
2.90 10 m6000
4.8 10 m
Tλ
λ
−
−
−
= ×
×⇒ =
= ×
(1 mark) (ii) It falls in the visible part of the spectrum.
(1 mark) (iii) The narrow dark lines that appear in the blackbody spectrum of the Sun are formed by the absorption of light by atoms and ions. The lines have specific frequencies that are determined by electron transitions in the absorbing atoms. (2 marks) (iv) The lines arise from electron transitions in the absorbing atoms and for each line the photon energy gives the difference in energy between the two electron energy levels, with
1 2.h f E E= − By looking at these energy differences we can build up an energy level diagram. (½ mark) Since each atom and ion has a different energy level diagram we can determine which atoms and ions are present in the solar atmosphere. (½ mark) Moreover, if line intensities are known then we can determine the relative fraction of each identified element or ion in the solar atmosphere (and even the temperatures) (bonus ½ mark)
