12.1 Turning points in Physics - The discovery of the electron – Mark schemes

Q1.(a)     Cathode rays/electrons move from cathode toward anode

Accept move left to right.1

The paddle wheel has gained energy from cathode rays/electrons. ✔

Accept as alternatives for energy kinetic,energy/momentum/impulse ✔Ignore references to force.Ignore references to applying a magnetic field.

1

(b)     Electrons are pulled out/escape from atoms OR gas atoms are ionised ✔Condone molecules as alternative to atoms.

1

(Positive ions generated near the cathode are attracted to the cathode causing free) electrons emitted from the cathode.✔

1

Electrons are accelerated toward the anode (by the potential difference) ✔

Do not accept attraction as an alternative to acceleration.1

(c)     Reason: Idea of fewer electrons/cathode rays ✔

Effect: Paddle wheel rotates less ✔Must score the reason mark to score the effect mark.Ignore references to air resistance.

OR

Reason: Idea of electrons/cathode rays have higher energy/speed/momentum ✔

Effect: Paddle wheel rotates more ✔If no mark is awarded, one mark can be awarded for the effect of the paddle wheel rotating more where the reasoning is limited to less collisions of electrons with air molecules.

2[7]

Q2.(a)     (i)      diffraction  

1

(ii)     the electrons in the beam must have the samewavelength  

otherwise electrons of different wavelengths(or speeds/velocities/energies/momenta) woulddiffract by different amounts (for the same order) [owtte]  

2

(b)     (i)      (eV = ½ m v2 gives) either v =

or 1.6 × 10–19 × 25000 = ½ × 9.1 × 10–31 × v2  

v =  = 9.4 × 107 m s–1  

p or mv (= 9.1 × 10–31 × 9.4 × 107) = 8.5 × 10–23  

kg m s–1 (or N s)  

alternatives for first two marks

p or mv =    =

   4

(ii)     any two of the first three mark points

increase of pd increases the speed (or velocity/energy/momentum) of the electrons  

(so) the electron wavelength would be smaller  

(and) the electrons would diffract less (when they passthrough the lenses)  

and

the image would show greater resolution (or be more detailed)  max 3

[10]

Q3.

(a)     (i)      

(ii)     heats the filament [or cathode or wire] (1)to enable electrons to gain (sufficient) k.e. to leave filament[or cause thermionic emission] (1)

(4)

(b)     (i)      electron moves towards positive platecurve in field (1)and straight beyond (1)

(ii)      = 1.67 ns (1)

(iii)    y = – at2 (1)

 

combine to give (1)

                                                = 1.8 × 1011 C kg–1 (1)(max 8)

[12]

Q4.(a)     (i)      positive (1)

(ii)     QE = mg (1)

 

(iii)    two electrons (1)missing (1)

(max 5)

(b)     upwards (1)

the electrical force is increased (1)

so there is a net upward force (1)

as the weight and upthrust are the same (1)(max 2)

[7]

Q5.(a)     (i)      arrow pointing towards centre of curvature (1)

(ii)     velocity [or direction of motion] is perpendicular

to the direction of the force (1)

work done is force × distance moved in the direction of the force (1)

no work done as there is no motion in the direction of the force (1)(max 3)

(b)     25mm (1)

mυ2 = eV (1)

= Beυ (1)

(1) = 1.8 × 1011 C kg–1 (1)(6)

[9]

Q6.(a)     path curves upwards from O to P

path is tangential to curve at P and straight beyond P2

(b)     (i)      magnetic field exerts a force on a moving charge/electron (1)magnetic force has a downwards component (at all points)[or magnetic force < electric force] (1)

(ii)     magnetic force = Bev (1)

electric force  = eE (1)

Bev = eE (gives v = ) (1)5

(c)     work done (or eV) = gain of kinetic energy (or ½mv2) (1)

 (1)

 = 1.8 × 1011 C kg–1 (1)3

[10]

Q7.

(a)     (use of v = gives) v = = 1.11 × 10–4 m s–1 (1)1

(b)      πr3 ρg = 6 πvr (1)

r = (1)

= (1) (= 9.7 × 10–7m)

            (allow C.E for v from (a))

3

(c)     qE = mg (1)

m = πr3 ρ = π (9.7 × 10-7)3 × 970 = 3.7 × 10-15 kg (1)

= 6.37 × 10–19 C (1)

            (allow C.E. for value of mass m)3

[7]

Q8.(a)     (i)      (vertically) upwards  (1)

(ii)     mg = qE,   (1)

  (1)      (= 2.0 × 10–5 C kg–1)3

(b)     initial downwards acceleration due to weight (or gravity)  (1)viscous force/drag/friction (or resistance) due to air                                 increases with increase in speed (1)speed increases until drag become equal to (and opposite to) weight(no resultant force) hence no acceleration  (1)

max 3[6]

Q9.(a)     (i)      filament heated by an electric current

          [or metal heated by nearby hot wire filament] (1) (conduction) electrons in the metal gain sufficient             kinetic energy to leave the metal / cathode / filament (1)

(ii)     temperature of filament depends on the current           or low current so small heating effect] (1) kinetic energy of electrons depends on temperature of filament (1) electrons must do work (or overcome work function) to leave metal (1) electrons have insufficient (kinetic) energy to leave           metal / cathode / filament (or overcome work function)                                 if the current is too low (1)

4

(b)     (i)      Ek (= eV = 1.6 × 10-19 × 4200) = 6.7 × 10-16 (J) (1)

(ii)     (use of Ek = ½mv2 gives) ½mv2 = 6.7 × 10-16 (J) (1)                                (allow C.E. for value of Ek)

(1)

  = 3.8 × 107 m s-1 (1)4

[8]

Q10.(i)      electrons [or ions] present (1)

electrons/ions accelerated by electric field[or electrons and ions collide] (1)excitation/ionisation of gas atoms/ions/molecules/particles occur (1)photons emitted on return to lower energy or ground state (1)

(ii)      electrons/ions do not gain enough kinetic energy(to produce ionisation) (1)because too many atoms/ions/molecules/particles present (1)

max 4QWC 1

[4]

Q11.(a)     k.e. (= work done = qV [or 1.6 × 10–19 × 2200]) = 3.5 × 10–16 J (1)

mυ2 = 3.5 × 10–16 J

hence υ = 2.8 × 107 m s–1 (1)(2)

(b)     (i)      all the k.e. goes to one photon (1) hf = k.e. [or 3.5 × 10–16 J] (1)

λ = (1)

= 5.7 × 10–10 m (1)

(ii)     λ = (1)

= = 2.6 × 10–11 m (1)(7)

[9]

Q12.(a)     (i)      field strength is the same everywhere / at any position

B1

field strength is force per coulomb

or force experienced by a charged particle

(not just force)B1

(2)

(ii)     lines parallel in central region

- must be shown throughout region between plates and start and end

on platesB1

curved at edgesB1

(allow if only curved at one end)

arrows correctB1

(3)

(b)     (i)      field strength E = V / dC1

95 000 V m–1 or NC–1

A1(2)

(ii)     weight = EqC1

3.16 (3.2) × 10–19 C                e.c.f. for EA1

(2)

(iii)    charge is negativeM1

force must be upwards to oppose gravity or oppose weightA1

charge must be attracted to positive (top)(upper) plateA1

or force is opposite to direction of electric field(3)

(c)     force = q2 / 4πε0r2                e.c.f. for qC1

correct substitution of dataC1

(9 – 10) × 10–20 NA1

compares answer with weightB1

N.B. no s.f. penalty here

use of charge that has not been calculated in (b)(ii)

gets 2 max of first 3 marks(4)

[16]

Q13.(a)     (i)      weight [or force of gravity] pulls droplet down (1)

no electric force to counteract weight s (1)viscous force increases with speed(1)weight = viscous force at terminal speed(1)

(ii)     viscous force = 6πη rυ (1)

weight = πr3 ρg (1)

πr3 ρg = 6πη rυ to give desired equation showing working (1)max 6

(b)     (i)      r2 = (1) (= 6.7 × 10–13 m2)

r = 8.2 × 10–7 (m) (1)

m = π × (8.2 × 10–7)3 × 950 (1) (= 2.2 × 10–15` kg)

(ii)      = mg [or Q = ] (1)

Q = (1)

Q = 4.8 × 10–19 (1)6

(c)     charge on oil droplet always a multiple of a basic amount (1)basic amount = 1.6 × 10–19 C (1)which is the charge of the electron (1)

max 2[14]

Q14.

(a)     force due to electric field = (1)

force due to electric field = weight or = mg (1)

for stationary droplet (1)

q = = 4.8 (+ 0.1) × 10–19 C(1)

(b)     1.6 × 10–19 C is the charge of the electron [or quantum of charge] (1)[max 4]

Q15.(a)     (i)      current heats the wire (1)

electrons (in filament) gain (sufficient) k.e. (to leave the filament) (1)

(ii)     electrons would collide with gas atoms / molecules (1)3

(b)     (i)      k.e. = (eV = 1.6 ×10–19 × 3600) = 5.8 ×10–16 (J) (1)

(ii)       mυ2 = eV (1)

υ = (1) = 3.6 × 107 m s–1 (1)4

[7]

Q16.(a)     (i)      electrons are negatively charged so beam is attracted to positive plate

[or repelled by negative plate or electron experiences force towards positive plate](1)

(ii)     beam does not spread out (1)if speeds varied, faster electrons would bedeflected less than slower electrons (1)

3

(b)     (i)      to give conduction electrons sufficientk.e. to leave metal [or to cause thermionicemission or electrons have insufficientke. in a cold filament to leave filament] (1)

(ii)      mυ2 = eVA [or υ = ] (1)2

(c)     (i)      into the plane of the diagram (1)perpendicular to the diagram [or the electric field] (1)

(ii)     Beυ = (1)

(iii)    combine the two equations to give (1)

(1)

1.75 × 1011 Ckg–1 (1)max 5

[10]

Q17.(a)     (i)      positive (1)

(ii)     electric force directed upwards = weight (1)

[or = mg]2

(b)     (i)      v =  = 8.7 × 10–5 m s–1 (1)

(ii)     weight [or mg] = 4/3 πr3 ρg (1)(since speed constant) viscous force = 6 πηrv (1)

4/3r3 g = 6πηrv to give desired equation (1)

(iii)     rearrange equation to give r =  (1)

= 8.7 × 10–7 m (1)    (8.65 × 10–7 m)

(allow C.E. for value of v from (i), but not 3rd mark)m (= 4/3 πr3 ρ) = 4/3 π (8.65 × 10–7)3 × 960 (1)     (= 2.6 × 10–15 kg)

(iv)     = mg (1)

Q =  (1)

    = 4.8 × 10–19  C (1)    (4.78 × 10–19 C)10

[12]

Q18.(a)     (i)      metal wire emits electrons when heated (1)

conduction electrons in metal gain kinetic energy whenwire is heated (1)

(ii)     electrons from wire would be absorbed/scattered/stopped bygas atomsor collide with gas atoms and lose kinetic energy or speed (1)

(iii)     electrons carry negative charge so anode needs tobe positive (to attract them) (1)

4

(b)     (i)      Ek (or ½mv2) (= work done or eV) = 1.6 × 10–19 × 2500 (1)= 4.0 × 10–16 J (1)

(ii)      =  (1)

= 3.0 × 107 m s–1 (1)

(allow C.E. for value of Ek from (i))4

[8]

Q19.(a)     (i)      emission of (conduction) electrons from a

heated metal (surface) or filament/cathode (1)work done on electron = eV (1)

(ii)     gain of kinetic energy (or ½ mv2) = eV; rearrange togive required equation (1)                                                                       3

(b)     (i)      work done = force × distance moved in direction of force (1)force (due to magnetic field) is at right angles to the direction ofmotion/velocity[or no movement in the direction of the magnetic force

 no work done] (1)electrons do not collide with atoms (1)

          any two (1)(1)

[alternative for 1st and 2nd marks(magnetic) force has no component along direction of motion (1)no acceleration along direction of motion (1)or acceleration perpendicular to velocity]

 (1)

 (1)

 (1)

(iii)     (rearranging the equation gives)  (1)

Ckg–1 (1)7

[10]

Q20.

(a)     force due to electric field is vertically upwards and proportional(or related to) plate pd (1)at V = Vc, force due to field is equal and opposite to the weight ofthe droplet (1)no resultant force (or forces balance) at Vc (droplet remains stationary) (1)

3

(b)     (i)      electric force (or qV/d) = weight (or mg) (1)

q  =  (1)

= 6.4 × 10–19C (1)

(ii)     for pd > 5700 (V), droplet moves upwards (1)due to increased electric force (1)droplet reaches terminal velocity (1)

max 5[8]

Q21.(a)     (i)      The number of electrons (per second) in the beam will increase (1)

because the filament will become hotter and will emit moreelectrons (per 2 second) (1)

2

(ii)     the speed (or kinetic energy) of the electrons will increase (1)

because the electrons (from the filament) are attracted towardsthe anode with a greater acceleration (or force) (1)

(or gain more kinetic energy in crossing a greater pd)2

(b)     (i)      (magnetic) force on each electron in the beam is perpendicularto velocity (1)

no work is done on each electron by (magnetic) force so ke(or speed) is constant (1)

magnitude of (magnetic) force is constant because speedis constant (1)

(magnetic) force is always perpendicular to velocity sois centripetal (1)

max 3

(ii)     rearranging r =  gives  (1)

 = 1.81 × 1011 (1) C kg–1 (1)

for correct answer to 2 sf (1)4

(iii)     specific charge for the electron ≈ 2000 × specific charge of H+ (1)(accept = and accept any value between 1800 and 2000)

which was the largest known specific charge before the specificcharge of the electron was determined/measured (1)

(or which could be due to a much greater charge or a much smallermass of the electron)

2[13]

Q22.(a)     (i)      unit A: supplies current/power/energy to the filament or heats the

filament (1)0 – 50 V (1)

(ii)     unit B: to make the anode positive w.r.t. the filament, so thatelectrons are attracted/accelerated to the anode (1)> 250 V (1)

max 3

(b)     (i)      beam current or intensity is reduced (1)(because) fewer electrons are emitted (per sec) from the filament (1)[or no beam as no electrons emitted if voltage of Areduced enough (1)(only)]

(ii)     electrons travel faster [or more kinetic energy] (1)(because the force of) attraction to the anode is greater (1)

4[7]

Q23.(a)     Electrons collide with atoms.✔

Electron in an atom is excited into a higher energy level.✔

Emits a photon when the electron relaxes / moves to lower energy level.✔3

(b)     Substitutes in ✔

Manipulates and gives answer to 2 or more sfs.✔2

(c)     Deduces e/m = v/Br.✔

Substitutes data (condone power of 10 errors).✔

 

1.7 × 1011 (C kg−1).✔Substitution may come before manipulation.

3

(d)     Electron velocity decreases when they collide.✔

v is proportional to r

OR r = vm/Be and m, B and e are constant.✔

r (gradually) decreases

or path with be an inwards spiral.✔3

[11]

Q24.(a)     (i)      either

(at terminal speed (v)) the viscous force on the droplet =its weight (or mg or the force of gravity on it)

or

viscous force = 6π rv (where r is the radius of the droplet and  is its viscosity) and weight (= mg) = 4πr3ρg/3  

4πr3ρg/3 = 6π rv  

(which gives r = (9 v/2ρg)½)2

(ii)     r (can be calculated as above then) used in theformula m = 4 πr3ρ/3 to find the droplet mess, m [owtte]  

alternatively; (from 6π rv = mg) (as all values are knownuse) m = 6π rv/g  

1

(b)     (i)      electric force (or QV/d) = the droplet weight (or mg)  

Q =  C  

2 sf answer  3

(ii)     any two from

the charge on each droplet is a whole number × 1.6 × 10–19 C(or × charge of the electron)  

the least amount of charge (or the quantum of charge) is thecharge of the electron  

the quantum of charge is 1.6 × 10–19 C [owtte]  max 2

[8]

Q25.(a)     (i)      (at terminal velocity v ), weight of droplet ( or mg) = viscous drag (or 6 r v)

Backward working 3 marks max;viscous force ( = 6 r v ) = 6 × 1.8 × 10−5 × 1.0× 10−6 × 1.1

× 10−4 = 3.7 × 10−14 N  

mass (m) of droplet = (4 r3 / 3 ) × ρ , (where r is the droplet radius ) weight = mg =

 (allow 3.7)

(therefore) (4 r3 / 3 ) × ρ g = 6 r v ( or rearranged)  

(hence) r (= (9 ɳ v / 2 ρ g)1/2

 (therefore) viscous force = weight as required for constant velocity  

note; some evidence of calculation needed to give final markAllow final answer for r in the range 1 to 1.05 × 10−6 to any number of sig figs

4

(ii)      Allow ecf for r from a(i) in a correct calculation that gives m in the range 3.6 to 4.0 × 10−15 kg

( or correct calculation of 6 r v / g)1

(iii)    electric force ( or QV / d) = droplet weight ( or mg)  Allow ecf m (or r) from a(ii) (or a(i)).

 Accept values in 1st mark line

[or Q (= viscous force × d / VUse of e instead of Q or q = 2 marks max

= 6 × 1.8 × 10−5× 1.0 × 10−6× 1.1 × 10−4 × 6.0 × 10−4 / 680  ]For the 2nd mark, allow use of viscous force calculation. Use of viscous force method does not get 1st mark.

Q = 3.2 × 10−19C  If both methods are given and only one method gives Q = ne (where n = integer >1), ignore other method for 2nd mark and 3rd mark.For the final mark, Q must be within n e ± 0.2 × 10 −19 from a correct calculation.

3

(b)     The weight of the second droplet is greater than the maximum electric force on it  

Alternative for 1st mark;

weight = drag force + elec force ( owtte)

Scheme using V for next 5 marks;

If n =1 for the second droplet , pd to hold it = 1580 V ( = mgd / e)  

which is not possible as V max = 1000 V  

If n = 2 , it would be held at rest by a pd of 790 V ( = 1580 / 2 or 680 × 4.3 / 3.7 V)

if n > 2 , it would be held at rest by a pd of less than 790 V ( or 790 / n V)  

So n =1(e ) must be the droplet charge  Alternative schemes for last 5 marksQ scheme Using QV /d = mg for a stationary droplet gives Q = mgd / V = 2.53 − 10−19 C  which is not possible as Q = integer x e  (so) Q (=ne) < 2.53 × 10−19 C   owtte)Calculation to showQ= 1e fits above condition  Q= 2e does not fit above condition  F scheme;- Calc of mg to give 4.2 (+0.2) × 10−14 N Calc for Q = 1e of QV / d to give 2.6(+0.2) × 10−14 N Calc for Q =2e of QV / d to give 5.3 (+0.2)× 10−14N  mg> elec force for Q =1e or <2e for Q=2e So n =1(e ) must be the droplet charge  

Max 4[12]

Q26.(a)     force due to electric field acts (vertically) downwards on electrons  

vertical (component) of velocity of each electron increases  

horizontal (component of) velocity unchanged (so angle to initial directionincreases)  

3

(b)     (i)      magnetic flux density should be reversed and adjusted in strength(gradually until the beam is undeflected)  

1

(ii)     magnetic (field) force = Bev

and electric (field) force = eV / d  

(Accept Q or q as symbol for e (charge of electron)

Bev = eV/d (for no deflection) gives v = V / Bd  2

(c)     (gain of) kinetic energy of electron = work done by anode pd or ½ m v2 = e V(A)  

 

= 1.8 × 1011 C kg–1.3

[9]

Q27.(a)     (i)      There is a (constant) force acting which is (always) at right angles /

perpendicular to the path / motion / velocity / direction of travel / to the beamOr mentions a centripetal force ✓

First mark is for condition for circular motionNot speedSecond mark is for a statement relating to the origin of the force

Force is at right angles to the magnetic field and the electron motionOrdirection given by left hand rule ✓

Any mention of attraction to the plates is talk out (TO)2

(ii)     States Bev = and evidence of correct intermediate stage showing

manipulation of the formulaor

Quotes r = from formula sheet and change of subject to v =Ber / m seen

 Accept delete marks

or rewrite as Be =

or rearrangement as = 1

(iii)    States  Bev =

or F = Bev F = (or F=Ee and E = in any form) Allow use of e or Q

and

states v = ✓ 

No mark for just quoting final equation. There must be evidence of useful starting equations

1

(b)     Equates the formulae for v and shows equated to  Must include ‘e / m =’ not just ‘specific charge =’Note there is no ecf. Candidates who use an incorrect equation in (a) (iii) will lose this mark unless they restart from first principlesCondone Q / m

1

(c)     Using band marking 

Marks awarded for this answer will be determined by the Quality of Written Communication (QWC) as well as the standard of the scientific response.

Level 1 (1−2 marks)

Answer is largely incomplete. It may contain valid points which are not clearly linked to an argument structure.Unstructured answer.Errors in the use of technical terms, spelling, punctuation and grammar or lack of fluency.

Level 2 (3−4 marks)

Answer has some omissions but is generally supported by some of the relevant points below:- the argument shows some attempt at structure- the ideas are expressed with reasonable clarity but with a few errors in the use of technical terms, spelling, punctuation and grammar.

Level 3 (5−6 marks)

Answer is full and detailed and is supported by an appropriate range of relevant points such as those given below:- argument is well structured with minimum repetition or irrelevant points- accurate and clear expression of ideas with only minor errors in the use of technical terms, spelling and punctuation and grammar.

AMeasure the terminal speed of the falling dropletAt the terminal speed weight = viscous force (+ upthrust)

mg = 6πƞrv and m = 4πr3ρ / 3 so r2 = r could be determined as density of drop, viscosity of air and g are known (r is the only unknown)

Bm can be determined if r is knownApply pd between the plates so electric field = V / d and adjust until droplet is stationaryQV / d = mg so Q can be found

CMake a number of measurements to find QResults for Q are in multiples of 1.6 × 10−19C so Q can be found

e.g.1-2Superficial with some sensible comments about the procedure with significant errors in attempts at use of equations. May do one part of A B or C reasonably well. Relevant Equations without little explanation may be worth 13-4Should cover most of the point in two of A, B & C coherentlyA & B may be well done in an answer that is easy to followOR B and C may be well explained but there may be significant errors or omissions in the determination of rOR a bit of all A B and C with significant errors or omissions5-6Will cover the points made in A B & C with few omissions in an answer that is easy to followThe candidate will define some terms used in equations1-2Attempt to explain how to determine radius with detail of how to use dataORMakes a relevant point about some part of the procedure about the determination3-4Radius determination explained with sensible equationsExplanation of how to use data to find mass of the dropIdea of holding the drop stationary5-6Answer includes all steps to determine the charge of a droplet with correct equations showing how to use the measurementsFor highest mark the answer should include idea of interpreting results of many measurements

6[11]

Q28.(a)     (i)      electrons pulled out of (gas) atoms so (gas) atoms become (+) ions

OR ionisation by collision (also) occurs OR (+) ions (that) hit cathode causing it to release electrons ✓ conduction due to electrons and positive ions ✓

; Allow ‘electrons ionise atoms’ as compensation mark (if no marks elsewhere)

2

(ii)     ions and electrons (moving in opposite directions) collide (with each other) and recombine and emit photons ✓

Owtte

electrons excite gas atoms (by collision ) and photons are emitted when de-excitation occurs ✓

If light not photons given in 1st 2 mark points, 1 max for 1st two mark points

gas needs to be at sufficiently low pressure in order that the particles (or uncharged gas atoms / ions / electrons) in the gas are widely spaced ✓

Owtte

otherwise (+) ions and / or electrons / particles would be stopped by gas atoms OR so that ions / electrons are accelerated (or gain enough ke) to cause excitation ✓

3max

(b)     Specific charge = charge / mass (and charge(s) of ion does not depend on the type of gas) ✓

Mass of ion depends on the type of gas ✓Accept Q / m in symbols Q / m but not e / m if e / m is specifically stated as specific charge

2[7]

Q29.(a)     Experiments suggested cathode rays were negatively charged particles ✔

Particle has mass much smaller than mass of an atom / hydrogen ionORCompares Specific charge with that of hydrogen ion / atom ✔

Particles were part of the substructure of matter / atoms ✔

Particles emitted in each case were the sameORParticles emitted were the same for different gases / for photoelectrons and particles from thermionic emission ✔

MAX 2Specific charge defined =0Millikan / Rutherford deductions=0Do not allow small mass aloneAllow protonAllow two correct deductions in 1 or 2 provided that the other comment is not relevant but does not contradict,

2

(b)     (i)      electrons collide with atoms of gas ✔ (condone molecules)

Reference to collisions with nucleus = 0 for the question

atoms / electrons are excited

or atoms / electrons change to higher energy states ✔

light / photon emitted when relaxation / de-excitation occurs or as electrons move / fall back to lower energy level ✔

Do not allow•             collide with gas unless atoms mentioned later•              particles•              electrons absorbed by atomsAllow move from ground stateAllow return to ground state

3

(ii)      

Correct substitution of data in the question allowing errors in powers of 10 ✔

1.9 × 1011 ✔

C kg-1 ✔Do not allowMust be seenSubstitution of values of e and me can gain 1st and last marks only

4[9]

Q30.(a)    

  Tick (✔) if correct

Beta particle emission  Electron diffraction  Photoelectric effect  Thermionic emission ✔

1

(b)     Use of seen including correct substitution1

λ = 2.4 × 10–11 (m)1

Statement to the effect that this is similar to or less than 0.1 nm/atomic dimension/diameter of the atom (so individual atoms can be resolved).

1

Condone missing unitAllow a correct conclusion that follows from an incorrect value of λ

3

(c)     The mark scheme gives some guidance as to what statements are expected to be seen in a 1 or 2 mark (L1), 3 or 4 mark (L2) and 5 or 6 mark (L3) answer.

Guidance provided in section 3.10 of the ‘Mark Scheme Instructions’ document should be used to assist in marking this question.

 

Mark Criteria QoWC

6 At least six of the likely statements will be covered to a good standard including at least three from image formation and at least three from quality and detail.

The student presents relevant information coherently, employing structure, style and SP&G to render meaning clear. The text is legible.5 At least five of the likely

statements will be covered to a good standard including at least two from image formation and at least one from quality and detail.

4 At least three of the likely statements will be covered to a good standard. The response must include one of both image formation and factors affecting quality and detail.

The student presents relevant information and in a way which assists the communication of meaning. The text is legible. SP&G are sufficiently accurate not to obscure meaning.

3 At least two of the likely statements will be covered to a good standard. The response must include one of both image formation and factors affecting quality and detail.

2 At least two of the likely statements from image formation or quality and level of detail will be covered to a good standard. The other area (if covered) will have errors and omissions.

The student presents some relevant information in a simple form. The text is usually legible. SP&G allow meaning to be derived although errors are sometimes obstructive.

1 One of the likely statements will be covered to a good standard.

0 No relevant coverage of the likely statements.

The student’s presentation, SP&G seriously

obstruct understanding..

The following statements are likely to be present.

Process of Image formation•        Electrons through the middle of the lenses are

undeviated•        Electrons on the edges are deflected by

magnetic fields toward the axis of the TEM•        The condenser lens deflects the electrons into a

wide parallel beam incident uniformly on the sample.

•        The objective lens then forms an image of the sample.

•        The projector lens then casts a second image onto the fluorescent screen.

Factors affecting the quality and level of detail•        Wavelength depends on speed of the electrons•        Lower the wavelength gives greater the detail.•        Emitted electrons come from a heated cathode

and therefore have a speed distribution dependent on temperature.

•        The speed of the electrons is not always the same which causes different pathways through the lens and so aberration.

•        The sample thickness reduces the speed of the electrons increasing the wavelength and decreasing the detail.

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Q31.(a)     current heats the wire ✓

1

electrons (in filament) gain sufficient KE (to leave the filament) ✓1

(b)     electrons would collide (or be absorbed or scattered) by gas atoms (or molecules) ✓

1

(c)     Rearrange ½ m v2 = eV to give v = ( 2eV / m )1 / 2

1

or correct substitution in equation.1

v = = 4.1 × 107 m s-1

1

λ = ✓ = 1.8 × 10–11m✓1

(d)     Increasing the pd increases the speed (or kinetic energy or momentum) of the

electrons ✓1

which decreases their de Broglie wavelength ✓1

so they are diffracted less so the rings become smaller ✓1

[10]

Q32.(a)     At terminal speed (v)), the viscous force on the droplet = its weight

For weight: allow mg or the force of gravity on itFor viscous force: allow ‘drag’ or ‘resistance’ or ‘friction’Not upthrust.

6πηrv = 4π r3ρ g / 3 ✓1

Manipulation leading to r = (9 ηv / 2ρ g)1 / 2 ) ✓1

(b)     r (can be calculated as above then) used in the formula m = 4π r3ρ / 3 to find the droplet mass, m ✓ (WTTE) Alternative ; (from 6πηrv = mg : as all values are known use) m = 6πηrv / g ✓

1

(c)     electric force (or QV / d) = the droplet weight (or mg) ✓Do not give 1st mark if eV / d given instead of Q V / d

1

Q =

= 3.2 × 10-19 C ✓1

(d)     Millikan’s conclusion: Electron charge is (-)1.6 × 10-19 C (WTTE) ✓

The charge on each droplet is a whole number × 1.6 × 10-19 C which agrees with Millikan ✓

Student’s results suggest -3.2 x 10-19 C as smallest quantum of charge ✓allow multiple or n, where n is an integer

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