= (Vs,rms)2/Rs = (80)2/100 = 64 W (Ip, rms)(Vp, rms) =
= 64 34.
Ip, rms = 4.0 A
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 energy level to a lower energy level.
E = hc/ = 3.21 x 10-19 J 36.
corresponds to a electron transition from a higher
so, option C is the answer. { E = (- 1.6 + 4.8) x 10-19 J}
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)
1 s = ut + at 2 2 1 2.66 = 0 + ( g )(0.740) 2 2 g = 9.72 m s -2
(ii)
1
2
(b)
%
∆h 1 = x 100% = 0.38% h 266
%
∆t 0.005 = x 100% = 0.68% t 0.740
∆g ∆h ∆t =% + 2(% ) = 0.38 + 2(0.68) = 1.74 g h t ∆g 1.74 = ∆g = (0.0174)(9.72) = 0.2 (1 s. f ) g 100
%
g = (9.7 ± 0.2) m s -2 (c)
2
1.
There may be zero error in the timer.
2.
Miscalibration of the timer.
(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
F = eE = (1.6 x 10-19 )(2.4 x 103 ) = 3.84 x 10-16 N (c)
(i)
Based on the assumption that the electric force on electron is constant along CX, then
E= (ii)
F VCX = e CX
VCX = (
F )CX = (2.4 x 103 )(4.0 x 10-2 ) = 96 V e
The magnitude of VCX 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)
4
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.
(a)
B-field directed out of the page.
(b)
(i)
mv 2 p2 / m FB = = r r 2 p = FB mr = Bqvmr = B(2e)pr p = 2Ber
(ii)
5
EK =
p 2 (2 Ber )2 2( Ber ) 2 = = 2m 2m m
(c)
The alpha particle does not get deflected immediately when it enters the Bfield. 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.
(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 ntype 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.
6
(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.
(a)
X is a hydrogen nucleus.
(b)
(i)
∆E = (mN + mHe − mO − mH )c 2 = (14.007525 + 4.003860 − 17.004507 − 1.008142)(1.66 x 10-27 )(3.0 x 108 ) 2 = −1.9 x 10-13 J
7
(a)
(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.
(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, dE p of ion Y. The change in potential energy of ion Y is then given by
dE p = W = − Fdr since F and dr are opposite in direction F =−
dE p dr
= −G
In terms of magnitude, F = G
(ii)
For r
2.8 x 10-10 m , Since F = −
dE p
dE p dr
dr F is positive
For r
2.8 x 10-10 m ,
Since F = −
dE p
dr F is negative
is negative.
F is repulsive dE p dr
is positive.
F is attractive
(iii)
1.
2.
At r = 2.8 x 10-10 m , F = −
dE p dr
=0N
At r = 5.0 x 10-10 m ,
dE p
(−2.0 − (−7.5))(1.6 x 10-19 ) F =− = = 1.76 x 10-9 N −10 dr (8.00 − 3.00)10 (b)
A B + r r8 dE p A B F =− = − 2 +8 9 dr r r
Ep = −
If r is very small, then
(c)
B r9
A r2
F =8
B is positive r9
1.
This force only acts for a short range of distance.
2.
This force is a repulsive force.
(i)
1.
Assuming that K.E of the ions is zero, then E p = −6.0 eV . Draw a horizontal line of E p = −6.0 eV . It will cut the curve at Minimum value of r = 2.35 x 10-10 m and Maximum value of r = 4.25 x 10-10 m
2.
At r = 3.5 x 10-10 m , E p = −7.1 eV
EK = ET − E p = −6.0 − (−7.1) = 1.1 eV (ii)
If the motion is simple harmonic, then E p vs r graph should be symmetrical. Since the E p vs 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) velocity Tritium nucleus stops
Deuterium nucleus stops
time Tritium nucleus Closest approach
(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,
3mv + (-2mv) = 3mvT + 2mvD v = 3vT + 2vD --- (2) [(1) x 3] + (2) : Sub into (1):
7v = 5vD vD = 1.4 v 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)
2(b)
(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
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
4(a)
(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
(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
= V + Ir = 4.25 + (2.15)(0.80) = 5.97 V
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.
Fig 1
Fig 2
Fig 3
x = + xo KE = 0 Total PE = max (GPE max, EPE min)
KE = max Total PE = min (GPE ↓, EPE ↑)
x = - xo
x=0
KE = 0 Total PE = max (GPE min, EPE max)
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).
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)
5(c)
(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
(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.
5(d)
(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.
(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
5(e)
(i)
g = GM/R2
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.
(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
6(d)
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
6(e)
(i)
For ideal gas, U
= 8.72 x 105 – 2.66 x 105 = 6.06 x 105 Pa, which is still above 3.23 x 105 Pa
= 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 Alternative: For one mol of ideal gas, U
(iii) 6(f)
= 6.16 x 10-21 x 6.02 x 1023 = 3.71 x 103 J
= 3/2 RT = 3/2 (8.314)(298.15) = 3.72 x 103 J
∆U = ∆n x U of 1 mole = (0.29)(3.71 x 103) = 1.08 x 103 J
(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
7(e)
(ii)
∆x = 3.0 x 103 m
(iii)
∆x∆p > h/4π ∆p = (h/4π ∆x) = 1.76 x 10-38 kgms-1
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.
Particle with energy E U
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.