JC2 H2 Physics ‘A’ levels 2007 Suggested Answers Paper 1: MCQ 1 C 11 D 21 A 31 A 2 B 12 B 22 B 32 B 3 A 13 C 23 D 33 D 4 B 14 A 24 B 34 D 5 A 15 D 25 A 35 C 6 A 16 A 26 D 36 D 7 C 17 C 27 B 37 D 8 B 18 D 28 B 38 A 9 C 19 C 29 D 39 B 10 D 20 B 30 B 40 C Reasoning for MCQs 1 Ans: C

Reasoning: Its obvious.

2 Ans: B

Reasoning: 2

2 LP kI

λ =

Equation must be homogenous. 2

Lλ

is unitless because the 2 quantities share the

same SI unit. As such, units of k = units of 2

PI

= units of R = ohm

3 Ans: A

Reasoning: Note that velocity of the object at its maximum height must be 0 and hence option D is out. In addition, the acceleration of the ball is at a constant value of g throughout the motion and hence only option A is correct.

4 Ans: B Reasoning: Air resistance is pointing to the left. Gravitational force is pointing vertical downwards. The resultant of these 2 forces will be in the direction of B.

5 Ans: A

Reasoning: Use v u at= + Now,

1

700030 50

500t= +

1 1.43t s=

2

700030 50

500 130t= +

+

2 1.8t s=

2 1 0.37t t s− = 6 Ans: A

Reason Using COM, Total initial momentum = total final momentum. Total initial momentum = m (2v) + 2m ( - v) = 0 So final momentum is zero. Since the 2 particles are sticking together, the final velocity of the particles will be zero as well. Hence, total kinetic energy after collision is 0.

7 Ans: C

Reasoning: Denoting final velocity of m as v’ and final velocity of M as v’’ Using COM, we have m + 0 = mv’ + Mv’’ -----(1) Using relative speed of separation, we have u – 0 = v’’ – v ------(2) Solving for v’’ , we will arrive at solution C.

8 Ans: B

Reasoning: That’s the definition of upthrust.

9 Ans: C

Reasoning: Weight is given as 1.6 x 10-3 N Electric force to the right is given by F = qE = 2 x 10-6 (80/0.1) = 1.6 x 10-3 N Since the resultant force to the right and downwards are equal in magnitude, the resultant force must be 45 degrees diagonal as in Option C.

10 Ans: D

Reasoning: Fnet = ma = 240 N Now, Fnet = Fdriving – Fresistance 240 = Fdriving – 160 Fdriving = 400 N Since P = Fv = 400 (10) = 4.0 kW

11 Ans: D

Reasoning: The case for PE = 12 KJ should be pretty obvious as it is halfway down. Now, for EPE, halfway down, EPE should be a rather low value. Remember that the rope will only stretch after a certain period of time. For KE, at the midst of motion, KE should be a rather substantial value. It cannot be too low, as not that much of KE has been transferred to EPE, if any.

. 12 Ans: B

Reasoning: Note that we have to use metres and radians. S = R (in km) θ (in deg)

= 1000 R (180

π) = 17.5 Rθ

13 Ans: C

Reasoning: The bob is at the lowest point of its circular bottom. Look at the figure on the right. The results follow.

14 Ans: A

Reasoning: Centripetal force is a resultant force so we do not draw it on a free body diagram. The answer follows.

15 Ans: D

Reasoning:

F = 24

2

(6 10 )(1)0.02

G x = 1 x 1018N

16 Ans: A

Reasoning: 12

100radsTπω −= =

since 0 0v xω=

0x = 0.030 m

T

W

17 Ans: C Reasoning:

212

mv mgh=

21(0,264) (9.81)

2m m h=

h = 0.0135 m r cosθ + h = r where r = 0.962 m. Solving, 9.8θ =

18 Ans: D

Reasoning: From definition.

19 Ans: C

Reasoning: PV = nRT

3 20(0.2 ) ( )(8.314)(100 273)

32P = +

P = 240 kPa

20 Ans: B

Reasoning: U Q W∆ = +

= 24 + (1.3 x 105) (1.3 – 3.6) x 10-4

= - 6 J 21. Ans. A Reason:

Sound produced from loudspeaker; sound – longitudinal waves, so option B and D wrong. Upon studying the diagram, = 10 cm. Since, v = f f = (330)/(0.10) = 3300 Hz = 3.3 kHz So, option A is correct.

22. Ans. B Reason: A YDS experiment – study bright fringe separation: x = D/a where a = distance between slits,

h

r

r

= wavelength of source, and D = distance from slit to screen.

To get maximum x, a has to be small, has to be long and D has to be large.

23. Ans. D Reason: Light of frequency 6.0 x 1014 Hz its is 5.0 x 10-7 m. d = (1.0 x 10-2 m)/(4.0 x 103 lines) = 2.5 x 10-6 m Use dsin = m : When m = 3, sin = (3)(5.0 x 10-7)/(2.5 x 10-6) = 0.6 = 36.9º

Therefore, the angle between the two three-order diffraction maxima (i.e., one on both sides of the zero-order bright fringe) = 2 x 36.9º = 73.8º ~ 74º

24. Ans. B Reason:

Given the magnitude of the charges on the top and bottom plates are the same, but the type of charge on the plates must be different so as to establish an electric field. So, option A and D are incorrect. Charged sphere Y stayed stationary in the uniform electric field set up by the top and bottom plates. There are two forces acting on the charged sphere Y its weight and the electric force. The direction of the electric force must be in the opposite direction of the weight, i.e., it must be directed upwards so as to balance the weight of the charge.

Option C is incorrect because the electric force on the charge is in the same direction as the weight. So, the answer is option B.

25. Ans. A Reason: E = -dV/dr

dV is the difference in potential between two adjacent equi-potential lines, dr is the distance between two adjacent equi-potential lines. From + 400 V to 0 V, – the magnitude of dV is the same but the sign is negative; – the magnitude of dr increases and the sign is positive; – the value of E is positive but its value becomes smaller with increasing distance away from P.

From 0 V to – 400 V, – the magnitude of dV is the same but the sign is negative; – the magnitude of dr increases and the sign is positive; – the value of E is positive but its value becomes larger with increasing distance away from P.

There is symmetry at the 0 V potential line so, answer is option A. 26. Ans. D Reason:

A recall question. The energy transferred by the cell in driving unit charge round the complete circuit is simply the electromotive force. Therefore, the answer is option D.

27. Ans. B Reason: Use R = L/A = 4L/d2

Manipulate: d = (4L/R)½()½ d = ()½

So, dAg = (Ag/Al)½ dAl = (0.5) ½ dAl = 0.71d 28. Ans. B Reason: The two 6.0 resistors are connected in parallel. The effective resistance of whole circuit is (3.0 + 3.0 ) = 6.0 . The current flowing in the effective circuit is 2.0 A. The current flowing in the ammeter will read 1.0 A. 29. Ans. D Reason: Maximum voltmeter reading will occur when variable resistor is set to 0.0 . The voltmeter will read 12 V since the cell has negligible internal resistance. So, the answer becomes very obvious option D. 30. Ans. B Reason: From information given, the electromagnetic force, FB = (0.020 x 10-3)(9.81)

= 1.96 x 10-4 N Since FB = BIL, 1.96 x 10-4 = B(3.0)(0.050) B = 1.3 x 10-3 T 31. Ans. A Reason: Change in magnetic flux linkage = NBA = (3000)(1.8)(0.010)2 = 1.7 Wb Average emf = change in magnetic flux linkage/time interval

= 1.7/0.060 = 28.3 V ~ 28 V

32. Ans. B Reason: Recall question again. Answer in lecture notes. 33. Ans. D Reason: 100% efficient transformer IpVp = IsVs Ns/Np = (Vs/Vp)rms 1000/200 = 80/Vp Vp, rms = 16 V <P> = (Vs,rms)2/Rs = (80)2/100 = 64 W (Ip, rms)(Vp, rms) = <P> = 64 Ip, rms = 4.0 A 34. Ans. D Reason: of violet light ~ 4.0 x 10-7 m E = hc/ = (6.63 x 10-34)(3.00 x 108)/(4.0 x 10-7) = 4.97 x 10-19 J ~ 5 x 10-19 J (1 s.f.) 35. Ans. C Reason: The vertical axis of the energy level diagram is calibrated in joules (J).

Wavelength of photon emitted = 6.2 x 10-7 m corresponds to a electron transition from a higher energy level to a lower energy level. E = hc/ = 3.21 x 10-19 J so, option C is the answer. E = (- 1.6 + 4.8) x 10-19 J

36. Ans. D Reason:

From the information given (i.e., the two I-V graphs), we can see that photons having wavelength of 2 is more energetic than photons having wavelength of 1 it allows photoelectron emission from both E and F electrodes. So, 1 > 2 options A and B are incorrect. Looking at the first I-V graph, when both E and F electrodes are illuminated by light of wavelength 1, we are seeing photoelectrons emitted from electrode F reaching electrode E but not vice versa. We will only see photoelectrons emitted from electrode E reaching electrode F only when the electrodes are illuminated by photons that are more energetic, i.e., light having wavelength of 2.

From this information, we know that the work function of electrode E is greater than the work function of electrode F. Hence, the answer is option D.

37. Ans. D Reason:

Option A – incorrect because in semiconductors, electrons in valence bands are all used to form covalent bonds. Option B – incorrect because there is no energy (band) gap in a metal or the valence electrons do not completely fill up an energy band, like in the case of a mono-valent metal such as sodium (Na). Option C – incorrect because the presence of impurities is used to increase conductance and not resistance. Option D – correct because electrons and holes have opposite charges.

38. Ans. A Reason:

Conduction band is above the valence band in energy level diagram. Electrons always fill up the lowest energy states first. These two pieces of information will tell us that option C and D are incorrect. At absolute zero, there should not be any electrons in the conduction band for an intrinsic semiconductor. Therefore, option A is the best answer.

39. Ans. B Reason: The gradient of the graph gives the decay constant. So, = (7.5 – 4.0)/(800 x 365 x 24 x 3600) = 1.38 x 10-10 s-1 t½ = ln 2/ = 5.02 x 109 s = 159.3 years ~ 160 years. 40. Ans. C Reason: E = 2.13 x 10-13 J = (mass of excited nucleus – mass of nucleus at ground state)c2

1 uc2 = 931.5 MeV = 1.49 x 10-10 J So, 2.13 x 10-13 J = 0.00143 uc2 (mass of excited nucleus)c2 = 59.9308 uc2 + 0.00143 uc2 = 59.93223 uc2 mass of excited nucleus = 59.9322 u

2007 A-level H2 Paper 2 Answers 1 (a) (i) 2

2

-2

12

12.66 0 ( )(0.740)

29.72 m s

s ut at

g

g

= +

= +

=

(ii) 1 1

% x 100% 0.38%266

hh

∆ = =

2 0.005

% x 100% 0.68%0.740

tt

∆ = =

(b)

-2

% % 2(% ) 0.38 2(0.68) 1.74

1.74(0.0174)(9.72) 0.2 (1 . )

100

(9.7 0.2) m s

g h tg h t

gg s f

g

g

∆ ∆ ∆= + = + =

∆ = ∆ = =

= ±

(c) 1. There may be zero error in the timer. 2. Miscalibration of the timer. 2 (a) Electric field strength at a point is defined as the electric force per unit positive

charge experienced by a test charge placed at that point. (b) (i) Direction of the electric field along CX is towards the left. (ii) From graph, E = 2.4 kV m-1

-19 3 -16F = eE = (1.6 x 10 )(2.4 x 10 ) = 3.84 x 10 N

(c) (i) Based on the assumption that the electric force on electron is constant

along CX, then

3 -2CXCX

VF FE = = V = ( )CX = (2.4 x 10 )(4.0 x 10 ) = 96 V

e CX e

(ii) The magnitude of CXV calculated is an underestimate as the

average electric field or force was greater than the value used in part (i).

3 (a) A progressive wave is a periodic disturbance in a medium or in space, which transfers energy from one place to another by vibrations of the disturbance.

(b) (i) 1. The waves from the two sources must be of roughly the same

amplitudes. 2. The two waves have a constant phase difference between them. 3. The two waves must overlap first in order to have interference.

(ii) 1 7 directions between AB and CD along which maximum amplitude

occurs. 2 Any line which does not connect the intersection points of the

wave-fronts from the two sources. 4 (a) B-field directed out of the page. (b) (i) 2 2

B

2B

mv p / mF = =

r r p = F mr = Bqvmr = B(2e)pr

p = 2Ber

(ii) 2 2 2(2 ) 2( )

2 2K

p Ber BerE

m m m= = =

(c) The alpha particle does not get deflected immediately when it enters the B-

field. The particle moves in a straight line path further into the B-field region before it gets deflected downwards. Out of the B-field region, the alpha particle move in a straight line path tangential to point of exit.

5 (a) The electrons move towards the p-type material. (b) In the p-type region there are holes from the acceptor impurities and in the n-

type region there are extra electrons from the donor impurities. When a p-n junction is formed, some of the electrons from the n-type region which have reached the conduction band are free to diffuse across the junction and combine with holes in p-type region. Filling a hole makes a negative ion at the p-type region and leaves behind a positive ion on the n-type region. A space charge builds up, creating a depletion region which inhibits any further electron transfer.

(c) Positive terminal of the cell is connected to the n-type region while the

negative terminal is connected to p-type region to increase the width of the depletion region.

6 (a) X is a hydrogen nucleus.

(b) (i) 2

-27 8 2

-13

( )

(14.007525 4.003860 17.004507 1.008142)(1.66 x 10 )(3.0 x 10 )

1.9 x 10 J

N He O HE m m m m c∆ = + − −

= + − −= −

(ii) The reaction produced an increase in mass. For this reaction to

occur, the alpha particle must have a large kinetic energy when it bombards the nitrogen nucleus to produce an increase in mass after the reaction.

(iii) The oxygen nucleus and nucleus X move with some amount of kinetic

energy. This means that the kinetic energy of the alpha particle must be very much higher than that of part (ii). As a result, part of the K.E of the alpha particle can be converted into the K.E of the products and part of it used to produce an increase in mass.

7 (a) (i) Work done, W is defined as the product of the force, F and the

displacement, r in the direction of the force. W Fr= If ion Y is displaced away from ion X by a distance, dr , then work done by the force, F to pull ion Y back will cause a change in potential energy, pdE of ion Y. The change in potential energy of ion Y is then

given by

since and are opposite in directionp

p

dE W Fdr F dr

dEF G

dr

= = −

= − = −

In terms of magnitude, F G=

(ii)

For -10 2.8 x 10 mr , pdE

dris negative.

Since

is positive is repulsive

pdEF

drF F

= −

For -10 2.8 x 10 mr , pdE

dris positive.

Since

is negative is attractive

pdEF

drF F

= −

(iii) 1.

At -10 = 2.8 x 10 mr , 0 NpdEF

dr= − =

2. At -10 = 5.0 x 10 mr ,

-19

-910

( 2.0 ( 7.5))(1.6 x 10 )1.76 x 10 N

(8.00 3.00)10pdE

Fdr −

− − −= − = =−

(b)

8

2 98

p

p

A BE

r rdE A B

Fdr r r

= − +

= − = − +

If r is very small, then 9 2 98B A B

Fr r r = is positive

1. This force only acts for a short range of distance. 2. This force is a repulsive force. (c) (i) 1. Assuming that K.E of the ions is zero, then 6.0 pE eV= − .

Draw a horizontal line of 6.0 pE eV= − . It will cut the curve at

Minimum value of r = -102.35 x 10 m and Maximum value of r = -104.25 x 10 m

2. At -103.5 x 10 mr = , 7.1 pE eV= −

6.0 ( 7.1) 1.1 K T pE E E eV= − = − − − =

(ii) If the motion is simple harmonic, then vs pE r graph should be

symmetrical. Since the vs pE r graph is not symmetrical, the motion is

not simple harmonic. (d) As the lattice is heated, the total energy of the ions increase due to increase in

its K.E and P.E. Since its P.E increases, the minimum and maximum value of r between which the ions vibrate increases. Therefore, the dimension of the whole lattice increase as it is heated.

2007 G.C.E. ‘A’ Level H2 Physics Paper 3 Suggested solutions 1(a) (i) The 2 nuclei are positively charged, and like charges repel one another. (ii) If v is sufficiently high, the two nuclei will have sufficient kinetic energy to come very close to

one another, overcoming electrostatic repulsion. There maybe a possibility of nuclear fusion.

(iii) Total initial momentum of the system = 3mv + (-2mv) = mv

According to the principle of conservation of linear momentum, since there is no external force acting on the system, the total momentum of the system must be conserved. In this instance, total momentum of the system can never be zero. Thus it is not possible for the nuclei to stop at the same instant otherwise the total momentum of the system would be zero at that instant. Or By Newton’s third law, both nuclei experience the same magnitude of force. Hence, both nuclei will have the same rate of change on momentum. Since the momentum of tritium is greater than that of deuterium (due to its greater mass), they will never come to a stop at the same time.

1(b) Let final velocity be v’

mv = 2mv’ + 3mv’ = 5mv’ v’ = v/5

1(c) (i) (ii) Taking motion to the right as positive,

For elastic collisions,

Relative speed of approach = Relative speed of separation Initial v of tritium – Initial v of deuterium = Final v of deuterium – Final v of tritium v – (-v) = vD – vT 2v = vD – vT ---- (1) By conservation of momentum,

velocity

time

Tritium nucleus

Closest approach

Tritium nucleus stops Deuterium nucleus stops

3mv + (-2mv) = 3mvT + 2mvD

v = 3vT + 2vD --- (2) [(1) x 3] + (2) : 7v = 5vD vD = 1.4 v Sub into (1): 2v = 1.4v - vT 0.6v = - vT vT = -0.6 v Final speed of tritium = 0.6 v Final speed of deuterium = 1.4 v 2(a) (i) E = |V/d| = |600 – (-600) / 2.5 x 10-2 | = 4.8 x 104 Vm-1 (ii) W = q∆V = (8.0 x 10-19)(500 – (-400)) = 7.2 x 10-16 J 2(b) The electric field strength between 2 parallel plates is uniform. Hence, in the absence of any other

charged bodies, electric potential changes uniformly from one plate to another. In this case the electric potential decreases from the left plate (+600 V) to the right plate (- 600V). The equipotential line in the centre of the plates, midway between +600 V and – 600 V would hence be 0 V. Alternative: Electric potential is defined as the work done by an external force in bringing a unit positive charge from infinity to a point in an electric field. Since the electric field strength between 2 parallel plates is uniform, the work done in bringing a positive test charge from infinity to the centre line from the +600 V plate is +600 J, while the work done in bringing a test charge from infinity to the centre line from the – 600 V plate is – 600 J. Hence the net work done is 0 J, which explains why the potential along the centre line between the plates is zero.

3(a) (i) From the graph, when V = 6.0 V, I = 1.43 A R = V/I = 4.2 Ω

(ii) The minimum value of R occurs at the point where the ratio of V to I is the smallest. The slope of the graph gives an indication of the variation of R. When the slope increases, it indicates that the R is decreasing in value.

Hence, minimum value of R occurs at the point where the slope is the greatest.

This occurs when V = 3.0 V and I = 0.95 A

R = 3.0 / 0.95 = 3.16 Ω

3(b) (i) PD across C = PD across 5.0 Ω = (5.0)(0.85) = 4.25 V

(ii) Referring to the graph, when V = 4.25 V, I = 1.3 A Current from the supply = 0.85 + 1.3 = 2.15 A (iii) E = V + Ir = 4.25 + (2.15)(0.80) = 5.97 V (iv) Power dissipated in C = IV = (1.3) (4.25) = 5.525 W Energy supplied in 20 minutes = P x (20 x 60) = 5.525 x 20 x 60 = 6.63 x 103 J 4(a) A mass suspended from a spring oscillating vertically undergoes changes between kinetic energy and

potential energy (which comprises of gravitational potential energy and elastic potential energy).

When the mass-spring system is at its highest position (Fig 1), its kinetic energy is zero as it is instantaneously at rest. Its GPE is maximum, while the EPE is minimum as the spring has the least extension here. Its total potential energy is maximum at this point.

As it accelerates towards the equilibrium, GPE is converted into kinetic energy and EPE. Its kinetic energy increases, GPE decreases while EPE increases (as the extension increases). At equilibrium position (Fig 2), its kinetic energy is maximum, while the total potential energy is minimum.

As the mass-spring system moves below the equilibrium position, kinetic energy and GPE are converted into EPE. At the lowest point (Fig 3), kinetic energy and GPE are minimum, while EPE is a maximum (extension is greatest).

In all stages of the motion, the total energy of the mass-spring system (which comprises of kinetic energy, GPE and EPE) remains constant assuming there are no dissipative forces.

4(b) In radioactive decay, the binding energy per nucleon of the parent nucleus is less than the combined

binding energy per nucleon of the products. This means that energy is released during the process (in the form of kinetic energy of the products, electromagnetic radiation etc).

x = 0 x = + xo

x = - xo KE = 0 Total PE = max

(GPE max, EPE min)

KE = max Total PE = min (GPE ↓↓↓↓, EPE ↑↑↑↑) KE = 0

Total PE = max (GPE min, EPE max)

Fig 1 Fig 2 Fig 3

When the parent nucleus decays, the total mass of the products is less than that of the parent. Einstein’s mass-energy equivalence states that E = mc2. Hence, the decrease in mass provides for the energy released during the process.

5(a) f = 5000 rev min-1 = 5000/60 = 83.33 Hz (rev s-1) ω = 2πf = 2π(83.33) = 524 rad s-1 5(b) (i) ω = 2π/T = 2π/24 x 60 x 60 = 7.27 x 10-5 rad s-1 (ii) a = rω2 = (6.38 x 106) (7.27 x 10-5)2 = 3.37 x 10-2 m s-2 5(c) (i) The Earth spins on its axis. At the poles, a test mass placed at the poles also spins on the

Earth’s axis. There is no need for the gravitational force between a test mass placed at the poles and the Earth to assist in the rotation of the test mass. The acceleration of free fall g is then the gravitational strength at that point. At the Equator, part of the gravitational force between a test mass placed at the Equator and the Earth provides for the centripetal force to keep the test mass in orbit. As a result, the acceleration of free fall (gpole – ac) takes a smaller value at the Equator.

(ii) The difference in the acceleration of free fall is equal to the centripetal acceleration of a point at the Equator (3.37 x 10-2 ms-2). Therefore, the difference in the radius at the Equator and at the poles is small in comparison to the radius of the Earth itself. Hence, the difference in acceleration due to free fall is small.

5(d) (i) Newton’s Law of Gravitation states that the attractive force F between 2 masses, M and m is

directly proportional to the product of their masses and inversely proportional to the square of the distance between the centre of both masses, r.

F = GMm/r2

(ii) Consider a test mass placed at the surface of the Earth. The gravitational force of attraction on

the test mass m is then given by

F = GMm/r2 Definition of g = F/m g = GM/R2

5(e) (i) GMm/r2 = mrω2 GM/r2 = rω2

r3 = GM/ω2

= gR2 /ω2

= 9.81 x (6.38 x 106)2 / (7.27 x 10-5)2 r = 4.23 x 107 m

5(e) (ii) 1. satellite’s orbit lies in the plane of the Equator and is always directly above a point on the Equator.

2. rotates from West to East, following the direction of rotation of the Earth.

5(f) Polar orbit satellites

• Closer distance to earth means signals are received more strongly and have better resolution. • Reduced time delay between transmitting and receiving signals

Geostationary satellites

• Less energy required to launch and maintain. • Able to monitor a fixed place over a long period of time.

6(a) 77.30 K = 77.30 – 273.15 = - 195.85oC

6(b) This means that the temperature on such a scale is not dependent on the thermometric property of

any particular substance and has absolute zero as its minimum temperature. 6(c) (i) Internal energy of a gas is the sum of microscopic kinetic energies and microscopic potential

energies of all the gas molecules. Microscopic KE arises due to the motion of the gas molecules while microscopic PE arises due to the intermolecular forces between the gas molecules.

(ii) An ideal gas is a hypothetical gas which obeys that equation of state of an ideal gas, PV =

nRT, for all values of pressure, temperature and volume. 6(d) (i) Since V and T are constant, n ∝ P ni/nf = Pi/Pf = 2.62 x 105 / 3.23 x 105 nf = 1.23 ni

ni = PiVi/RTi = (2.62 x 105 x 0.0120)/[8.314 x (25 + 273.15)] = 1.27 nf = 1.23 (1.27) = 1.56 mol Amt of air that needs to be supplied = 1.56 – 1.27 = 0.29 mol

(ii) To supply 4 tyres, amount of air needed to be transferred from supply = 4 x 0.29 = 1.16 mol

Fall in supply’s pressure , ∆P = ∆n RT/V = (1.16 x 8.314 x 298.15)/(0.0108) = 2.66 x 105 Pa Final pressure of supply = 8.72 x 105 – 2.66 x 105

= 6.06 x 105 Pa, which is still above 3.23 x 105 Pa 6(e) (i) For ideal gas, U = 3/2 NKT = 3/2 (1)(1.38 x 10-23)(298.15) = 6.17 x 10-21 J

(ii) No. of molecules in 1 mol = 6.02 x 1023 Hence, internal energy for one mole of air = 6.16 x 10-21 x 6.02 x 1023 = 3.71 x 103 J Alternative: For one mol of ideal gas, U = 3/2 RT = 3/2 (8.314)(298.15) = 3.72 x 103 J (iii) ∆U = ∆n x U of 1 mole = (0.29)(3.71 x 103) = 1.08 x 103 J 6(f) (i) Internal energy is a function of state of the system and the increase in internal energy of a

system is equal to the sum of the work done on the system and the heat supplied to the system. (ii) Since there is no heat supplied to or removed from the system, ∆Q = 0 Work done on the system = Increase in internal energy = 1.08 x 103 J 7(a) - Refer to notes for description of photoelectric effect.

- State the observations of the photoelectric effect and explain how these observations provide evidence for the particulate nature of electromagnetic radiation.

7(b) Equation :

Maximum kinetic energy of photoelectrons = Energy of incident photons – Work Function of metal

Terms: Maximum kinetic energy of the photoelectrons = Kinetic energy of the most energetic electron that is released from the metal plate Energy of incident photons = energy of electromagnetic radiation incident on metal plate, given by the product of the Planck’s constant and frequency of radiation, f (E=hf) Work function of metal = minimum energy required to release a photoelectron from the metal plate

Note: Cambridge examiners commented that many students simply stated the algebraic form of the photoelectric equation and explained what each of the symbols meant. This was NOT what they were looking for. Instead, they required students to explain the meaning of the terms, as shown above.

7(c) KEmax = hf - φ φ = (hc/λ) - KEmax

= (6.63 x 10-34)(3.0 x 108)/(3.82 x 10-7) – [1/2 x 9.11 x 10-31 x (6.87 x 105)2] = 3.06 x 10-19 J 7(d) (i) Length of pulse = speed x time = 3.0 x 108 x 1.0 x 10-5 = 3.0 x 103 m

(ii) ∆x = 3.0 x 103 m (iii) ∆x∆p > h/4π ∆p = (h/4π ∆x) = 1.76 x 10-38 kgms-1 7(e) A potential barrier is a region in space where there exists a maximum potential U, which prevents a

particle on side of it to pass through.

Classically, if E > U, the particle is able pass through the barrier. If E < U, the particle is not able to pass through the barrier. According to quantum physics, particles exhibit wave nature and are defined by a continuous wave function ψ. The square of the amplitude of this function |ψ|2 indicates the probability of finding a particle at a particular position and time. The wave function must be continuous on both sides of the potential barrier, indicating a finite probability that the particles can tunnel through the barrier, regardless of the value of E. Hence, there is a probability that the particle can be found on the other side of the barrier, and |ψ|2 is not zero at the region on the right side of the potential barrier.

U

Particle with energy E