5 nature of light test 10_11_answers

8/2/2019 5 Nature of light test 10_11_answers

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Advanced Level - ASUnit 2: Physics at Work: Topic 5 - Nature of light I Quantum PhysicsTime: 1 hour 30minutes

Scientific calculatorRuler

Advice Read each question carefully before you start to answer it. Keep an eye on the time. Try to answer every question. Always check your answers.

Instructions Use black ink or ball-point pen (blacklblue). Fill in the boxes at the top of this page with your names and the date. Answer all questions (there are 17 questions and 1 challenge / bonus question). Answer the questions in the spaces provided - there may be more space than you need. In calculations you should show all the steps in your working, giving your answer at each stage.Information The total mark for this paper is 82 (+ 15bonus marks). The marks for each question are shown in brackets - use this as a guide as to how much time to spend

on each question. Questions labelled with an asterisk (*) before the number of the question, are ones where the qualityof your written communication will be assessed - you should take particular care with spelling,

punctuation and grammar, as well as the clarity of expression, on these questions. The question with an asterisk (*) after the number of the question is a challenge question. It is more

searching and deeper and it is intended to challenge your knowledge and understanding of physics.Itis a bonus question and you should attempt it only if you have time. The list of data, formulae and relationships is provided separately. You may use a scientific calculator

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2/19

The diagram shows some of the electron energy levels for the hydrogen atom with fourpossible transitions.Energy I eV

O - " - " - - " -- , -- - - .- - " - .. " . - - - - '- . - -- - '-'--r---"""-~~"--'---~c ID-0.85 ,----"--""---.---,----,-------"'-.,'----,-,----"----,--1.51

A B

-13.6 -,".------------"---""'""-,-----------,.-"--,---.,,--.---

The transition that would result in the emission of the longest wavelength is0 A0 B0 C~ D

. . 2 . The behaviour of light can be described in terms of waves or particles. The particlenature of light can be demonstrated byo A light being diffracted as it passes through a narrow slit.D B the speed of light reducing when it is refracted by glass.~ C light causing electrons to be emitted from a metal surface.D D light being polarised.

. . . 1 3


3/19

.3 In the 1930s, experiments were performed where beams of X-rays or beams of highenergy electrons were directed through a crystal as shown in the diagram.

Beam of X-rays Diffraction patternor electrons

Which of the following statements about the experiments is correct?D A They show that the X-rays are behaving like particles.~ B They show that the electrons have a wavelength similar to the size of the atoms.o C They show that electrons are charged.o D They show that electrons can have all of the properties of electromagnetic

radiation.

The di~gram repr~sents t~e experimental arrangement used to produce interference fringes inYoung s double slit expenment.

x l. Imonochromatic Y Ilight source singleslit doubleslit screenThe spacing of the fringes on the screen will increase ifA the width of the single slit is increasedB the distance XY between the two slits is increased@ a light source of lower frequency is usedD the distance between the single and double slits is decreased

.. .1 4


4/19

-4-

5 A parallel beam of monochromatic light is directed normally at a plane transmission gratingwhich has 600 lines per millimetre. A third order diffracted beam is observed at an angle of54to the zero order diffracted beam.

. 5~ 3'" order

incident 7-__ - - j I f - V S_ 4~+_---__..:;;..-----zero orderbeam ~ ~ diffracted beam

grat3rd order

Which line, A to D, in the table gives the wavelength of the light and the angle of diffractionof the first order beam?

wavelength / angle of diffractionnrn of first order 450 16

B 450 18C 520 16D 520 18

t 6 ' An electron initially at rest is accelerated through a potential difference. It is then brought to restin a collision, and all of its kinetic energy is converted into a single photon of electromagneticradiation. Which one of the following quantities is not required to find a value for the wavelengthof the photon?@ The mass of the electronB The charge on the electronC The velocity of electromagnetic waves

D The value of the potential difference

... /5


5/19

-5-

: r - Electrons with charge e are accelerated by a voltage V in an electron diffraction tube and exhibitwave properties when aimed at a thin film of graphite. The wavelength associated with theelectrons is 4.Which row of the table gives correct formulae for the kinetic energy and momentum of theelectrons?

kinetic energy momentum

A V 4e hB V h}. ,e

C eV 4h@ eV h1"

(Total for Questions 1 - 7 = 7 marks)

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6/19

-6-iB ) The diagram shows, not to scale, an experimental arrangement for studying the

transmission of light by a double slit.

Slits Light Q

1 sensor ~ o faserP

6.0m

Monochromatic light from a laser falls normally on two narrow, closely spaced parallelslits. The intensity of light transmitted is studied by moving a small light sensor alongthe line PQ, at a perpendicular distance of 6.0mfrom the slits. The graph shows howthe light intensity varies with distance y from the mid-point O.

Intensity/arbitrary units

-6.0 -4.0 o 2.0 4.02.0Distance y/cm

6.0

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-1-Explain with the aid of a diagram why the two light waves from the slits produce a Leaveminimum intensity at X. blank

. "//l

0 J ) / V = - - - - - - -VVV. . .A . x . . . ~ I.A~ ~ f ! : t . t r . ? 4 . s .l tt? f .( .~ f r .: .~ . .~~f~ .. . .. J . ! f . .fk'!l.~.~ ..{o. ~dt,~ ~f.?rf . . .~ /f1 :!!.\t.~ t;.c .vr .;~ .. . . M . ~ / . . . . f , ! ; I ~ ~ ( 4 . R .. . ~ . . . t y , J . . , . . . . . ~ . ~ . .u....~~.~t~..' .~f. I4if~'t ! .{ .. 5 . : .

(2)Point 0 is equidistant from the slits. State, in terms of the wavelength A , the pathdifference between the waves arriving at Y.

Path difference = ~ 4 _ .2-What is the phase difference, in radians, between the waves arriving at point Y?. . . .r = 1 f . . " . : . &2. ' . .=~.~ g .

~ . S.,,-Phase difference = rad.(2)The spacing of the slits in the experiment was 0.20 rnrn. Use this, together withinformation from the diagrams, to calculate the wavelength of the light........;;.~ ~ ... J . . ~ ~ . .~ . / . g ( " . . 4 ok O2. IJ ~Itt ..

................. '. ~ . .;?- L : i , k t ! : f i . ; ; . . J c C.:ux.to-31 y _ : .({:tJ4. . . . . . . . . . . . . . . . . . . . . ' : : : . f : . o . . q . ~ r .< J .~ ~ ~ ( ~ 6 . .o _ g . 4 _ ~ . . . .. .. . .

Wavelength = t f .p f ! .d ':~ i ..~ ! ! . .(3)One of the two slits is now covered up and the experiment is repeated. Add a line tothe graph opposite to show how you would expect the light intensity to vary with thedistance y.{] lI;t( . . I~C ~J. 'Y ~ ~ . ~ f A .I ,e./: ~'Ir.~~ w/{f k (2)~fW;L ~"" i4fe4ff~( 1 4 f I ' U Iu 'tk. ~~~ Y'(~I~kS)

t( e n . .L,-"t:a'ye.-:!>f dl h.

~ :ZA ffJ"'! f? > ~ Z~f.e4f"~( I i t -k .hf l } c< ' i1b1" t .J_ ]


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9) (a) A helium-neon laser produces monochromatic light of wavelength 632.8 nm which fallsnormally on a diffraction grating. A first order maximum is produced at an angle of 18.50measured from the normal to the grating.Calculate

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(i) the number of lines per metre on the grating,$J ;If.H . ~? 'k < L9.H . ,[:::H } l& hH m~ ~ m mm H H m H H ..mm

..../1.:;./ : : ; # ) . [ . : ; ; . . ~~~~ . " f t ~. .1 . ~ ."1.."&/t? ~ 5' -f~ .~ t (Mk- : :T .Z : 1tlf..~.#I.~~=. /. ff~;'. -6 "'.s::Q(.k>':~.1It

(ii) the highest order which is observable .

. . .. .J. .r(~ ~" ..:;;:~d ~ ../ . . ! t ~ ~ ~ .fk{~~~~4'.~ . 1 . ~ . . 1 f . s ( : 9 . h . ~ . .:f. .m~HItH~H-(:~ f f : ~ ~ " : : .. .mH9.km~H~.m( i . .H4r . ) ;AH( { .~~~ j " (6)~ h~4J. - = . 3 !7 te / t & 4 l l . rf l 5 V ' I # . r ~~.I( U ~ " '3 r l

(b) When the grating is used with a different monochromatic source, the first order maximumis observed at an angle of 17.20.Calculate the wavelength of this second source.

sri It . t J : : - tt ). ........... ........... . 11 .} .J -'.

10) (a) State the approximate wavelength limits of the visible spectrum.. .. .. . .. .. .. .. .. .. .. .. . .. .. .. .. .. .. .. .. . .. .. .. .. . ' 1 : : . q t ; !, !: I. . ~ ~ . .?.t ; ? . t ? . ~ . ~ .(b) (i) Describe the spectrum of sunlight. [2 ] ; t ~ ~ 6 2 : : f ~ : : ~ ~ f i l ; / : : ; ; ; _ ~ ~ ~ : . y .~...........................t v . ~ . y . . .(ii) The Sun is known to contain helium gas. Explain how this fact is deducedfrom spectral data. [~]. 1 i . . ~ .~


9/19

- < 7 -I) The following passage describes some important aspects of the photoelectric effect.

Insert the missing words.In the photoelectric effect, a single Ib~ interacts with a single electronat the surface of a Ir!.~~.{ In this interaction ~~(J;Y isconserved. This was summarised by Albert Einstein in the following equation

hf = < D + ' i2mv 2h 11 2' h . k i f h dec' 6 r 6 1 ,ere r smv IS t e maximum inetic energy 0 t e .

and < D is the ..? t ? . 6 Y l . . . f t ! . ! . q : ~ . .(Total for Question II _ = 5 marks)

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10/19

-/()-I~ Ultraviolet radiation incident on a zinc plate releases electrons from the zinc's surface.

The energy of each incident photon is 5.4 eV. Zinc has a work function of 4.3 eV.(a) (i) State the name given to this effect.

(1)1~1?6.fr,~(A;(y'~C:~,

(ii) State the speed of the photons.(1):rl g . f. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .! / . : t ! . ..f ? J 0~.:7 < ; : = . : ~k(:? ~ .. .

(iii) What is meant by the work function of a metal?(1)

. . . . . . . . . . . . .. i . ; ( f . ' - ~ 1 . ' ! < ' . ! . 1 . / ~ . ~ ~ ~~ . ~ . ~ . f.!f ~~.~'gJ T .4 . .~ ~ , .~~.~1. ' : " ". . . . . . . . . . . . . . .! ~ ~ . ~.~ ~ ! . ! .~~!. I . .+ . 4 . ~ ~ .k ~~/A~ ~e..fk. .~ ~ . c g. .~ .(b) An electron is emitted from the surface of the zinc.

(i) Calculate the maximum kinetic energy of the electron in joules.(3)

. . . . . . . . . . . . . .t ; I H . ) - ~ ~ . ~ = ~ = ~ f..~.t(=t.~.~.~. ..~ ..{./..~.v. t:l#.~(fk.~.;.~~= /. g Xl'o_IJ ..T.............................................................................................. " .Maximum kinetic energy = /~ ?.. .kf.~. ' . ! . . . ~ .

(ii) Calculate the maximum speed of the electron.(2)

. . . . . . . . . . . .( E " A . ~ . .F . i : . .l t . . . . ./,.:mm ..~m(/'"... ; ; ( i { , . k ; . l ~ > : : : / : ~ f J ;

........................................................................................................................................................~.~.~(? ..~.. Y . ' . .~ . ~ .. _ C2. )e(OShe/" -IMaximum speed - .

0', //1


11/19

-!(-(c) The intensity of the ultraviolet radiation is doubled.

State what happens to the maximum speed of an electron emitted from the zinc.(1)

..........~..~.1.~.1f~L . ~ T 4 . ~. ~ .. f .~1. . ! .? .~ ~.t~':~.~~.~d:t. ~.:.~~J= .

. . . . . . . . . . . . . . . . . . .4 . ~ . I . I ! . I . : ~y . : ! . 4 . . .f.~~4.~.t.~.~ 4..~ f!L!.~ &..~ ~d .(Total for Question 1 . 2 = 9 marks)

. r 2 . .


12/19

-12-The diagram shows monochromatic radiation falling on a photocell connected to acircuit.

IncidentmonochromaticradiationAnodeVariable d.c.supply-6V to +6 V

VDigitalvoltmeter

Photocathode

The incident radiation has a wavelength of 215 nm. The metal surface of thephotocathode has a work function of 2.26 eV.Calculate the energy in eV of a photon of the incident radiation ........i.I4..:-::..j . . . : ; ; ; . .1 . . ( , . , . . 1 . .: :: ::. . .~ . , Kl")~.k 3Xl" ~.P"'~ ';' r :j7 21.s-r~fNk/' K(o-f...!?'e{/-I'................................................................................................................................................

=S,?-8eV. ! " ,

............................................................................................................................................._ s : :tte:VEnergy - eV( 4 )What is the maximum kinetic energy in eV of the emitted electrons?..K .~ . : ; . .~.{~.. 1 . .; . .-?:. . . t : f (~. l ! .~.~.~~f.


13/19

-/3-If the wavelength and intensity of the incident radiation is kept constant, a graph of thecurrent I through the photocell against applied p.d. V is as shown.

----t~~~---------------------VN.5Mark a letter S on. the graph to show the stopping-potential.

The photocathode is .replaced with one whose metal surface. has a greater workfunction. On the graph above, sketch how I would vary with V given that thewavelength and intensity of the incident radiation remain unchanged.

(Total 9 marks)(3)

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14/19

-If-

(3)

(1)

(2)

. 1 4 ) The graph shows how the maximum kinetic energy E of photoelectrons emitted from thesurface of aluminium varies with the frequency / of the incident radiation.

;y1~-~'H-!~-'!~~-~H-'H~~~'+'H~#~~'H~"~~H++~-~!-H~'~-~'~~-+-Ho J ~ 1 t B d ~ ~ ~ t ~ l ~ , ~ 5 1 ~ f D , ~ ~ i E 0 1 1 Bo 2 4 6 8 10 12 14 16 18 20

jll014Hz(a) Explain why no photoelectrons are emitted below a frequency of 10x 1014Hz.

(1)'Ie f ? / ' o / n . (Zh a J - t P (-.!ft) .. r #OJ;" ~II~~ - I i "If!.w . ~ . .iAC-f.o., !f ? f l u ~"ii./ Ul1 A ..(b) Calculate the work function of aluminium in electron volts.

/~ tiA ~ ~ ~ ~ /~ .ty ':/;'tg~ ..~ ~h f-::h ! .. .k l.~A /: ;/. f..~ < ? ~ ~ . .~ 4 . J ; ~ ~ .~ W - : = t.,k(O-3("::[eJc/(Jk(O(',e~ _ ./ -e 1 /7. . ..1~.6j;(o~/!17(zv..;;.i;:;e.==..- .. .Work function =.... 1/eV

(c) State the quantity represented by the gradient of the graph.. . . . . . . . . .( 4 . 4 . ~ 1 . . . 7 . .


15/19

I~The energy level diagram shows the ground state and two excited states E] and E2 of aneon atom.

-----E2-----E

----- ground state

In a helium neon laser, collisions occur between helium atoms and neon atoms. Thisresults in the helium neon atoms being excited from the ground state to level E2. Theythen emit photons and move to level E].(a) What is meant by 'energy level'?

(1).....)q ~ 1 ~ ~ .'(c.r .e.& a.:~~"ift 'k #,e; f ~t; ~ .~ .(4 .1 Iy~ ~.~ ..~l.4 ..f .{ ~ ~ ~ ) . .4.f.4l~.e..{ - h ? 4 t L ~ . . . .

(b) What is a photon?( 1 )

....A ~r.~r.e .&/~cftz .: flJ .~~ /(f~.~ . ..J . : . . . q . ~ . .'!i t f : ' ! ~ ~ & 7 ( ~ f ~ d . ~ ~ 8 g.~ 4 : t . ~ t~.!. ' . ' .r(.6 ~ . ~ 4g /to /J7 )~ .(c) Write a formula in terms of E1 and E2for the energy of an emitted photon.

0)

(d) The wavelength of an emitted photon is 6.33 x 1 O - 7m.Calculate the energy of this photon. (3)~L'(f~ A X : : c & : '~3~~::Jf~fs- t ( . !t.L~/

.-3..~('t.~J{L~? .. : r . . . .-(.:!}_- '$'(Yk(>o JEnergy - .(Total for Question 15 = 6 marks)

.. ,~,


16/19

/~ Figure 1 shows part of an energy level diagram for a hydrogen atom.Figure 1

n=4n=3 J Tn=2

\V \ Vn=1

-O.85eV-1.50eV-3.40eV-13.60eV

(a) The level, n = 1, is the ground state of the atom.State the ionisation energy of the atom in eV.

_ 13'60answer - eV(J mark)

(b) When an electron of energy 12.1eV collides with the atom, photons of three differentenergies are emitted. -/"3 ~()e.V.fI.2 le V - /f"Oe.V(i) On Figure 1 show with arrows the transitions responsible for these photons.

(3 marks)(ii) Calculate the wavelength of the photon with the smallest energy. Give your

answer to an appropriate number of significant figures.h f = !J E ~ ~ := IJ E - ) ~ = : ~~ , 4E _~= 9 , . . . ) : : : : (.{')c(a-1ti;;;; ~ 3.0,.+-("0 b..../- = (;'~/k(O ~E IS-oe V..f S. )o rO e ~ k /. '):-(-0-I),.re r:r- ( . r o e .. 11- I .l e - : : : . : - 5' ~OJ2 V~ - A c : : . - l : [OIZ V-fj.f( ) (2 V

(= !-'foe v),. rlkftl-1-answer= m

(5 marks)

.,-./Ir

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17/19

- / : : ; - -I~ (a) State what is meant by the wave-particle duality of electrons .

. .I f . A ( , I J . . .. . ! f !I 1 ! . ~ . t . . 1 i . t y. . . . .~:?.~...../ f . . . . .?p.#.(r.~., :~...~ ..~..,#.,~~~~~~~~o/..~/~t:~:~(t!:~~t(!!!~~-::.~~(1.c;{:.:.~:.~.~ .. . ! . f ? I J ? ~ f . ~ ~ : e ;. . .k/u& . t !. IU. . .. .~L..... . l ! . j I . . . .~ . . . . .tf~ ...~.~!l~....~~.(~: .

(1 mark)(b) Electrons of wavelength 1.2 x lO-lom are required to investigate the spacing between

planes of atoms in a crystal.(i) Calculate the momentum of an electron of this wavelength stating an appropriate

unit.

_ S ' 5'.k(


18/19

-/S-r - - - - - = - - - - - - - - - - - - - - - - - - - - - - - ' - - - ~ Q - - C - - W ~ - ' - - _ ,1 , 3 t - This question is about a crude theory of the singly charged helium ion, consisting ofthe nucleus and one electron, using the ideas of electron standing waves.The diagram (Fig. , ) shows an electron standing wave fitting into a spherical shape ofradius 0.5 x 10- 10m.The shaded region is where an electron is unlikely to be found.

Fig. I(a) (i) State the wavelength of the electron standing wave shown.

\ fa -(0~ = 2r: z: .2. .K 0'5' )c( 0- t..:: /.


19/19

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(b) The potential energy, E, of the electron can be obtained using the equationE = Q IQ 2 /4n8or, where r is the distance between' the electron and the nucleus.

Obtain a value for the potential energy of the electron in a helium ion (nuclear charge.+2e ) when r = 0.5 x 10-10 m. -(2_ 2 . . ~I/Cf f?_ rA1 f l f / v~" ! ,7 .5 T r . e ~ f!6.C.t2_/ s; = 8gS-xr 'O C - . : 7 - ' ' ' - . = - ' ; ,f> C / ~ = g Ql . _ = (2 Xl''')rc 10- I~).>c(_(. (,,x(O-I.7_k::) _N

C p r J . ' G r: ~l1k 2-g >)ao-f~ r !~O-.rkd ' . .x= = _ c : r . 2 . )

5 nature of light test 10_11_answers

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