18 Marking scheme: Worksheet (A2)
1 a θ = 360 30 × 2π = 6 π ≈ 0.52 rad [1] b θ = 360 210 × 2π ≈ 3.7 rad [1] c θ = 360 05 . 0 × 2π ≈ 8.7 × 10−4 rad [1] 2 a θ = π 2 0 . 1 × 360 = 57.3° ≈ 57° [1] b θ = π 2 0 . 4 × 360 ≈ 230° [1] c θ = π 2 15 . 0 × 360 ≈ 8.6° [1]3 a 88 days is equivalent to 2π radians.
θ = 88 44 × 2π = π rad [1] b θ = 88 1 × 2π ≈ 0.071 rad (4.1°) [1]
4 a Friction between the tyres and the road. [1]
b Gravitational force acting on the planet due to the Sun. [1]
c Electrical force acting on the electron due to the positive nucleus. [1]
d The (inward) contact force between the clothes and the rotating drum. [1]
5 a ω =
r
v
=20000
150
[1] ω = 7.5 × 105 rad s−1 [1] b a = r v2 [1] a = 000 20 1502 [1] a = 1.125 m s–2 ≈ 1.1 m s−2 [1] c F = ma = 80 × 1.125 [1] F = 90 N [1] 6 a i Time = 10 2 . 8 = 0.82 s [1]ii Distance = circumference of circle = 2π × 0.80 = 5.03 m ≈ 5.0 m [1]
iii speed = time distance = 82 . 0 03 . 5 [1] speed, v = 6.13 m s−1 ≈ 6.1 m s−1 [1] iv a = r v2 [1] a = 80 . 0 13 . 6 2 [1] a = 47 m s–2 [1]
v F = ma = 0.090 × 47 [1]
F ≈ 4.2 N [1]
b The tension in the string. [1]
c The stone describes a circle, therefore the angle between the velocity and the acceleration
(or centripetal force) must be 90°. [1]
7 a i speed = time distance speed v = 1.6 0.12 2π 2π × = T r [1] v = 0.471 m s−1 ≈ 0.47 m s−1 [1] ii r mv ma F = = 2 [1] 12 . 0 471 . 0 300 . 0 × 2 = F [1] frictional force ≈ 0.55 N [1] b Frictional force = 0.7mg [1] r mv mg 2 7 . 0 = [1] 12 . 0 81 . 9 7 . 0 7 . 0 = × × = gr v [1] speed = 0.908 m s−1 ≈ 0.91 m s−1 [1]
8 a Kinetic energy at B = loss of gravitational potential energy from A to B
2 1 mv2 = mgh or v = 2gh [1] v = 2×9.81×5.2 =10.1 m s–1 ≈ 10 m s−1 [1] b i a = r v2 [1] a = 16 1 . 10 2 [1] a = 6.38 m s−2 ≈ 6.4 m s−2 [1] ii Net force = ma R − mg = ma [1] R = mg + ma = m(a + g) = 70(6.38 + 9.81) [1] R ≈ 1.1 × 103 N [1]
9 a R cos 20° = W = 840 × 9.8 [1] R = ° × 20 cos 8 . 9 840 = 8760 N [1]
b R sin 20° = centripetal force =
r mv2 [1] r = R mv2 [1] r = ° × 20 sin 8760 32 840 2 [1] r = 287 m ≈ 290 m [1] 10 net force = r mv2 net force = 80 . 0 0 . 4 120 . 0 × 2 = 2.4 N [2] weight W of stone = mg = 0.120 × 9.81 = 1.18 N ≈ 1.2 N [1] At the top: W + TB= 2.4 so TB= 2.4 − 1.2 = 1.2 N [1] At the bottom: TA− W = 2.4 so TA = 2.4 + 1.2 = 3.6 N [1] ratio = B A T T = 2 . 1 6 . 3 = 3.0 [1]
1 Gravitational field strength at a point, g, is the force experienced per unit mass at that point. [1] 2 F = 2 r GMm − [1] Therefore: G = Mm Fr2 → ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 2 kg m N → [N m2 kg–2] [1] 3 g = mF F = mg = 80 × 1.6 [1]
F = 128 N ≈ 130 N (F is the ‘weight’ of the astronaut.) [1]
4 a F = 2 r GMm [1] F =
(
14)
2 27 27 11 10 0 5 10 7 1 10 7 1 10 67 6 − − − − × × × × × × . . . . [1] F ≈ 7.7 × 10–38 N [1] b F =(
12)
2 28 28 11 10 0 8 10 0 5 10 0 5 10 67 6 × × × × × × − . . . . [1] F = 2.61 × 1021 N ≈ 2.6 × 1021 N [1] c F = 2 2 11 0 . 2 1500 10 67 . 6 × − × [1] F = 6.00 × 10–6 N ≈ 6.00 × 10–6 N [1] 5 a g = 2 r GM − [1]b The field strength obeys an inverse square law with distance (g ∝ 12
r ). [1]
Doubling the distance decreases the field strength by a factor of four. [1]
c ratio =
( )
(
)
2 2 59 5 R GM R GM [1] ratio = 22 5 59 = 2 5 59 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ [1] ratio ≈ 140 [1] 6 g = 2 r GM − [1] g =(
7)
2 26 11 1.0 10 10 67 . 6 × × × × − (magnitude only) [1]8 a F = 2 r GMm − [1] F =
(
10)
2 24 11 10 9 . 3 10 0 . 6 1800 10 67 . 6 × × × × × − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = × = 3.9 10 m 2 10 8 . 7 10 10 r [1] F = 4.74 × 10–4 N ≈ 4.7 × 10–4 N [1] b F = 2 r GMm − F =(
10)
2 23 11 10 9 . 3 10 4 . 6 1800 10 67 . 6 × × × × × − [1] F = 5.05 × 10–5 N ≈ 5.1 × 10–5 N [1] c a = mF (F is the net force.) [1]
a =
4.74 10
45.05 10
51800
− −
×
−
×
[1] a ≈ 2.4 × 10–7 m s–2 (towards the centre of the Earth) [1]
9 a F = 2 r GMm − [1] F =
(
3)
2 24 11 10 6800 10 0 . 6 5000 10 67 . 6 × × × × × − (r = 6400 + 400 = 6800 km) [1] F = 4.33 × 104 N ≈ 4.3 × 104 N [1] b a = mF = 5000 10 33 . 4 × 4 [1] a = 8.66 ≈ 8.7 m s–2 [1] c a = r v2 [1] v2 = ar = 8.66 × 6800 × 103 [1] v = 7.67 × 103 m s–1 ≈ 7.7 km s–1 [1]10 a The work done in bringing unit mass [1]
from infinity to the point [1]
b 0 J [1] c Ep =− 6 24 11 10 4 6 10 0 6 10 67 6 × × × × − . . .
([1] mark only if minus sign missed) [2]
= −6.25 × 106 J [1]
d 6.25 × 106 J [1]
11 a Gravitational force on planet = 2
r GMm [1] Centripetal force = r mv2 [1] Equating these two forces, we have: 2
r GMm = r mv2 [1] Therefore: v2 = r GM or v = r GM [1]
b v = r GM = 11 11 30 10 5 . 1 10 0 . 2 10 67 . 6 × × × × − [1] v = 2.98 × 104 m s–1 ≈ 30 km s–1 [1]
12 The field strengths are the same at point P.
2 M x GM =
(
)
2 E x R GM − [1] R − x = x × M E M M [1] R − x = x × 81 so R − x = 9x [1] 10x = R so x = 10R [1]20 Marking scheme: Worksheet (A2)
1 a The period of an oscillator is the time for one complete oscillation. [1]
b The frequency of an oscillator is the number of oscillations completed per unit time
(or per second). [1]
2 a The gradient of a displacement against time graph is equal to velocity. [1]
The magnitude of the velocity (speed) is a maximum at 0 s or 0.4 s or 0.8 s. [1]
b For s.h.m., acceleration ∝ – displacement.
The magnitude of the acceleration is maximum when the displacement is equal to the
amplitude of the motion. [1]
The magnitude of the acceleration is a maximum at 0.2 s or 0.6 s or 1.0 s. [1]
3 a T = 12 2 . 13 [1] T = 1.1 s [1] b f = T 1 = 1 . 1 1 [1] f = 0.909 ≈ 0.91 Hz [1] 4 a Amplitude = 0.10 m [1] b Period = 4.0 × 10–2 s [1] c f = T 1 = 04 . 0 1 [1] f = 25 Hz [1] d ω = 2πf = 2π × 25 [1] ω = 157 rad s–1 ≈ 160 rad s–1 [1] e Maximum speed =
ω
A=157×0.10 [1] maximum speed = 15.7 m s−1 ≈ 16 m s−1 [1] 5 a Phase difference = 2π × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ T twhere T is the period and t is the time lag between the motions of the two objects. phase difference = 2π × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ T t = 2π × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 10 5 . 2 [1] phase difference = 2 π ≈ 1.6 rad [1] b Phase difference = 2π × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ T t = 2π × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 10 0 . 5 [1]
phase difference = π ≈ 3.1 rad [1]
6 a A = 16 cm [1] b ω = 2πf = T π 2 = 2.8 π 2 [1] ω = 2.24 rads–1 ≈ 2.2 rads–1 [1] c a = (2πf )2x (magnitude only) [1]
For maximum acceleration, the displacement x must be 16 cm.
a = 2 2 10 16 8 . 2 1 π 2 ⎟ × × − ⎠ ⎞ ⎜ ⎝ ⎛ × [1] a = 0.806 m s–2 ≈ 0.81 m s–2 [1] d Maximum speed =
ω
A=2.24×0.16 [1]7 a ω = 2πf = T π 2 = 2.0 π 2 [1] ω = 3.14 rad s–1 ≈ 3.1 rad s–1 [1] b a = –(2πf )2x or a = –ω2x [1] a = 3.142 × 3.0 × 10–2 [1] a ≈ 0.30 m s–2 [1] c x = A cos (2πft) = A cos (ωt) [1] x = 3.0 × 10–2 cos (3.14 × 6.7) [1] x ≈ –1.7 × 10–2 m [1]
8 a Gradient of x–t graph = velocity
[2]
b Gradient of v–t graph = acceleration
(for s.h.m. acceleration ∝ −displacement)
[2] c Kinetic energy = 2 1 mv2 ∝ v2 [2]
d Potential energy = total energy − kinetic energy
9 a a = −(2πf )2x [1] Therefore (2πf )2 = 6.4 × 105 [1] f = π × 2 10 4 . 6 5 = 127 Hz ≈ 130 Hz [1] b F = ma
Acceleration is maximum at maximum displacement, so magnitude of maximum force is given by:
F = ma = 0.700 × (6.4 × 105 × 0.08) [1]
F = 3.58 × 104 N ≈ 3.6 × 104 N [1]
10 a According to Hooke’s law, F = –kx [1]
(The minus sign shows that the force is directed towards the equilibrium position.)
From Newton’s second law: F = ma [1]
Equating, we have: ma = –kx [1] Hence: a = x m k ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − b For s.h.m. we have a = –(2πf )2x [1] Hence (2πf )2 = m k [1] Therefore f = m k π 2 1 c f = T 1 = 4 . 0 1 = 2.5 Hz [1] 2.5 = 850 2 1 k π [1] k = (2π × 2.5)2 × 850 ≈ 2.1 × 105 N m–1 [1]
1 a The atoms in a solid are arranged in a three-dimensional structure. [1]
There are strong attractive forces between the atoms. [1]
The atoms vibrate about their equilibrium positions. [1]
b The atoms in a liquid are more disordered than those in a solid. [1]
There are still attractive electrical forces between molecules but these are weaker than
those between similar atoms in a solid. [1]
The atoms in a liquid are free to move around. [1]
c The atoms in a gas move around randomly. [1]
There are virtually no forces between the molecules (except during collisions) because they
are much further apart than similar molecules in a liquid. [1]
The atoms of a gas move at high speeds (but no faster than those in a liquid at the same
temperature). [1]
2 The atoms move faster [1]
because their mean kinetic energy increases as the temperature is increased. [1]
The atoms still have a random motion. [1]
3 a The internal energy of a substance is the sum (of the random distribution) of the kinetic
and potential energies of its particles (atoms or molecules). [1]
b There is an increase in the average kinetic energy of the aluminium atoms as they vibrate
with larger amplitudes about their equilibrium positions. [1]
The potential energy remains the same because the mean separation between the atoms
does not change significantly. [1]
Hence, the internal energy increases because there is an increase in the kinetic energy of
the atoms. [1]
c As the metal melts, the mean separation between the atoms increases. [1]
Hence, the electrical potential energy of the atoms increases. [1] There is no change in the kinetic energy of the atoms because the temperature remains
the same. [1]
The internal energy of the metal increases because there is an increase in the electrical
potential energy of the atoms. [1]
4 Change in thermal energy = mass × specific heat capacity × change in temperature [1] 5 The specific heat capacity refers to the energy required to change the temperature of a substance. [1]
Specific latent heat of fusion is the energy required to melt a substance; there is no change in
temperature as the substance melts. [1]
6 E = mc∆θ [1]
E = 6.0 × 105 × 4200 × (24 – 21) [1]
E = 7.56 × 109 J ≈ 7.6 × 109 J [1]
9 a i T = 273 + 0 = 273 K [1] ii T = 273 + 80 = 353 K [1] iii T = 273 – 120 = 153 K [1] b i θ = 400 – 273 = 127 °C [1] ii θ = 272 – 273 = –1 °C [1] iii θ = 3 – 273 = –270 °C [1]
10 a The thermal energy E supplied and the specific heat capacity c remain constant.
The mass m is larger by a factor of 3. [1]
Since E = mc∆θ, we have: ∆θ = mc E ; ∆θ ∝ m 1 [1] Therefore ∆θ = 3 15 = 5.0 °C [1]
b The thermal energy E supplied is halved but the specific heat capacity c and the mass m
remain constant. [1] Since E = mc∆θ, we have: ∆θ = mc E ; ∆θ ∝ E [1] Therefore ∆θ = 2 15 = 7.5 °C [1] 11 a Melting point = 600 °C [1]
(There is no change in temperature during change of state.)
b The lead is being heated at a steady rate and therefore the temperature also increases
at a steady rate. [1]
c The energy supplied to the lead is used to break the atomic bonds and increase the
separation between the atoms of lead (and hence their potential energy increases). [1]
d E = mc∆θ [1]
E = 200 × 10–3 × 130 × (600 – 0) [1]
E = 1.56 × 104 J ≈ 1.6 × 104 J [1]
e In a time of 300 s, 1.56 × 104 J of energy is supplied to the lead.
Rate of heating = power power = 300 10 56 . 1 × 4 [1] power = 52 W [1] f Energy supplied = 52 × 100 = 5200 J [1] Lf
2
.
0
5200
=
∆
∆
=
m
E
[1] = 26 000 J kg−1 [1]12 The energy supplied per second is equal to the power of the heater.
In a time of 1 s, water of mass 0.015 kg has its temperature changed from 15 °C to 42 °C. [1]
E = mc∆θ (where E is the energy supplied in 1 s) [1]
E = 0.015 × 4200 × (42 – 15) [1]
E = 1.7 × 103 J [1]
The power of the heater is therefore 1.7 kW. [1]
(You may use P = ( t m
)c∆θ )
13 The gas does work against atmospheric pressure. [1]
14 Heat ‘lost’ by hot water = heat ‘gained’ by cold water. [1]
0.3 × c × (90 – θ) = 0.2 × c × (θ – 10) [1]
where c is the specific heat capacity of the water and θ is the final temperature. The actual value of c is not required, since it cancels on both sides of the equation. Hence:
0.3 × (90 – θ) = 0.2 × (θ – 10) [1]
27 – 0.3θ = 0.2θ – 2.0 [1]
0.5θ = 29 so θ = 58 °C [1]
15 Heat ‘lost’ by metal = heat ‘gained’ by cold water [1]
0.075 × 500 × (θ – 48) = 0.2 × 4200 × (48 – 18) [1]
(θ is the initial temperature of the metal.) θ – 48 = 500 075 . 0 30 4200 2 . 0 × × × [1] θ – 48 = 672 [1] θ = 720 °C [1]
22 Marking scheme: Worksheet (A2)
1 a number of atoms = number of moles × NA
number of atoms = 1.0 × 6.02 × 1023 ≈ 6.0 × 1023 [1]
b Number of molecules = 3.6 × 6.02 × 1023 ≈ 2.2 × 1024 [1]
c Number of atoms = 0.26 × 6.02 × 1023 ≈ 1.6 × 1023 [1]
2 There are 6.02 × 1023 atoms in 4.0 g of helium. [1]
mass of atom = 23 10 02 . 6 004 . 0 × = 6.645 × 10 –27 kg ≈ 6.6 × 10–27 kg [1]
3 a There are 6.02 × 1023 atoms in 0.238 kg of uranium. [1]
mass of atom = 23 10 02 . 6 238 . 0 × = 3.95 × 10 −25 kg ≈ 4.0 × 10−25 kg [1] b i number of moles = uranium of mass molar uranium of mass [1] number of moles =
0.12
238
= 5.04 × 10 –4 ≈ 5.0 × 10–4 [1]ii number of atoms = number of moles × NA
number of atoms = 5.04 × 10–4 × 6.02 × 1023 = 3.06 × 1020 ≈ 3.1 × 1020 [1]
4 The absolute zero of temperature is –273.15 °C or 0 K. [1]
This is the lowest temperature any substance can have. [1]
At absolute zero of temperature, the substance has minimum internal energy. [1]
5 a Pressure × volume
= number of moles × universal gas constant × thermodynamic temperature [1]
b PV = nRT [1] P = V nRT = 020 . 0 293 31 . 8 0 . 1 × × [1] P = 1.22 × 105 Pa ≈ 1.2 × 105 Pa (120 kPa) [1] 6 a PV = nRT [1]
Comparing this equation with y = mx, we have:
y = PV, x = T, gradient, m = nR [1]
A graph of PV against T is a straight line through the origin.
Correct graph [1] n = R gradient [1] b PV = nRT [1]
At a constant temperature, the product
PV is a constant. [1]
Hence a graph of PV against P is a
7 a PV = nRT [1] n = 29 0 . 4 = 0.138 moles [1] P = V nRT = 030 . 0 ) 34 273 ( 31 . 8 138 . 0 × × + [1] P = 1.17 × 104 Pa ≈ 1.2 × 104 Pa (12 kPa) [1] b
TP is constant when the volume of the gas is constant. [1] The pressure is doubled, hence the absolute temperature of the gas is also doubled. [1] Therefore: temperature = 2 × (273 + 34) = 614 K [1] temperature in °C = 614 – 273 = 341 °C ≈ 340 °C [1] 8 a n = RT PV [1] n = ) 13 273 ( 31 . 8 10 0 . 2 10 180 3 2 − × × × × − + ) 13 273 ( 31 . 8 10 0 . 2 10 300 3 2 − × × × × − [1] n = 4.44 moles ≈ 4.4 moles [1] b Total volume, V = 4.0 × 10–2 m3, T = 273 – 13 = 260 K P = V nRT [1] P = 2 10 0 . 4 260 31 . 8 44 . 4 − × × × [1] P ≈ 2.4 × 105 Pa (240 kPa) [1] 9 a P = A F = 3 10 6 . 1 400 − × [1] P = 2.5 × 105 Pa [1] b n = RT PV [1] n = ) 0 . 5 273 ( 31 . 8 10 4 . 2 10 5 . 2 5 4 + × × × × − [1] n = 2.6 × 10–2 moles [1]
c i mass = number of moles × molar mass
mass = 2.6 × 10–2 × 29 = 0.754 g ≈ 0.75 g [1] ii density = volume mass density = 43 10 4 . 2 10 754 . 0 − − × × [1] density = 3.14 kg m–3 ≈ 3.1 kg m–3 [1]
10 Mean kinetic energy of atom ∝ absolute temperature [1] 2 2 1mv ∝ T or v2 ∝ m T 2 [1]
Since the mass m of the atom is constant, we have: v ∝ T [1]
The temperature of 0 °C in kelvin is T = 273 K The absolute temperature increases by a factor of
273 000 10
(= 36.6) [1]
Hence the speed will increase by a factor of 273
000
10 = 6.05
[1] The speed of the atoms at 10 000 K = 1.3 × 6.05 ≈ 7.9 km s–1 [1]
11 a The particles have a range of speeds and travel in different directions. [1]
b i Mean kinetic energy = 1.38 10 5400
2 3 2 3kT = × × −23× [1] = 1.118 × 10–19 J ≈ 1.1 × 10–19 J [1] ii mv kT 2 3 2 1 2 = [1] 27 23 10 7 . 1 5400 10 38 . 1 3 3 − − × × × × = = m kT v [1] speed = 1.147 × 104 m s–1 ≈ 11 km s–1 [1]
12 a Mean kinetic energy = 1.38 10 273
2 3 2 3 23 × × × = − kT [1] = 5.65 × 10–21 J ≈ 5.7 × 10–21 J [1]
b There are 6.02 × 1023 molecules of carbon dioxide. [1]
mass of molecule = 26 23 7.31 10 10 02 . 6 044 . 0 = × − × kg [1] 21 2 5.65 10 2 1 = × − mv J [1] 26 21 10 31 . 7 10 65 . 5 2 − − × × × = v [1] speed = 393 m s–1 ≈ 390 m s–1 [1]
c Total kinetic energy of one mole of gas = kT× 2 3 NA = RT 2 3 (Note: R = k × NA) [1]
For an ideal gas, the change in internal energy is entirely kinetic energy.
Change in internal energy = 8.31 100
2 3 ) 273 373 ( 2 3 × − = × × R [1]
change in internal energy = 1.2465 kJ ≈ 1.2 kJ [1]
13 a i The molecule has 3 degrees of freedom for translational motion and 2 degrees of freedom
for rotation – making a total of 5. [1]
Therefore, mean energy = 5 × kT 2
1 = kT
2 5
[1]
ii The molecule has 3 degrees of freedom for translational motion and 3 degrees of freedom
for rotation – making a total of 6. [1]
Therefore, mean energy = 6 × kT 2 1
= 3kT [1]
b Internal energy = 3kT×NA = RT3 (Note: R = k × NA) [1]
1 Electric field strength is the force per unit charge at that point. [1]
The potential at a point is the work that must be done to bring unit charge
from infinity to that point. [1]
2 a E =
d
V
, d =E
V
[1] d = 000 4005000 = 1.25 × 10 –2 m ≈ 1.3 cm [1] b F = EQ = 400 000 × 1.6 × 10–19 N [1] F = 6.4 × 10–14 N [1] 3 E = 2 0 π 4 r Q ε so k = 12 0 4 8.85 10 1 π 4 1 − × × π = ε k = 8.99 × 109 m F–1 ≈ 9.0 × 109 m F–1 [1]4 The force between the charges obeys an inverse square law with distance; that is, F ∝ 12
r [1] Point B: The distance is the same. The force between the charges is F. [1] Point C: The distance is doubled, so the force decreases by a factor of 4. [1] The force between the charges is
4 F
. [1]
Point D: The distance is trebled, so the force decreases by a factor of 32 = 9. [1]
The force between the charges is 9 F
. [1]
Point E: The distance between the charges is 8 R. [1]
The force between the charges decreases by a factor
of ( 8)2 = 8 [1]
The force between the charges is 8 F . [1] 5 a E = 2 0 π 4 r Q ε [1] E = 126 2 15 . 0 10 85 . 8 π 4 10 5 . 2 × × × × − − [1] E = 9.99 × 105 V m–1 ≈ 1.0 × 106 V m–1 [1]
b The distance from the centre of the dome increases by a factor of 3.
The electric field strength decreases by a factor of 32 = 9. [1]
Therefore E = 9 10 0 . 1 × 6 = 1.1 × 104 V m–1 [1]
6 a i E = 2 0 π 4 r Q ε [1] E = 126 2 40 . 0 10 85 . 8 π 4 10 20 × × × × − − (r = 2 80 = 40 cm) [1] E = 1.124 × 106 V m–1 ≈ 1.1 × 106 V m–1 [1] ii E = 12 2 6 40 . 0 10 85 . 8 π 4 10 40 × × × × − − [1] E = 2.248 × 106 V m–1 ≈ 2.2 × 106 V m–1 [1]
(The electric field doubles because the charge is doubled, E ∝ Q.)
b Net field strength, E = 2.2 × 106 – 1.1 × 106 = 1.1 × 106 V m–1 [1]
The direction of the electric field at X is to the left. [1]
7 a Q = V × 4πεor = 20 000 × 4 × π × 8.85 × 10−12 × 0.15 [1] Q = 3.3 × 10–7 C [1] b E = 2 r kQ = 9 2 7 15 . 0 10 3 . 3 10 0 . 9 × × × − [1] = 1.32 × 105 V m–1 ≈ 1.3 × 105 V m–1 [1] c F = eV = 1.6 × 10−19 × 1.32 × 105 [1] F = 2.11 × 10−14 N ≈ 2.1 × 10−14 N [1] 8 a V = r Q 0 π 4 ε = r kQ = 9 2 9 10 5 10 2000 10 9.0 − − × × − × ×
([1] mark only if minus sign omitted)
[2]
V = 3.6 × 105 J C–1 =360 kV [1]
9 Similarities
• Both produce radial fields. [1]
• Both obey an inverse square law with distance; that is, F ∝ 12
r . [1]
• The field strengths are defined as force per unit (positive) charge or mass. [1]
• Both produce action at a distance. [1]
Differences
• Electrical forces can be either attractive or repulsive, whereas gravitational forces are
always attractive. [1]
10 The electric field strength due to the charge +Q is equal in magnitude but opposite in direction
to the electric field strength due to the charge +3Q. [1] Therefore: 2 0 π 4 x Q ε = 2 0( ) π 4 3 x R Q −
ε (where R is the distance between the charges = 10 cm) [1]
2 1 x = ( )2 3 x R− [1] so x x R− = 3 [1] x(1 + 3 ) = R so x = 3 1+ R = 0.37 R [1] x = 0.37 × 10 = 3.7 cm [1] 11 ratio = 2 2 2 0 2 4π r Gm r e ε
(where m = mass of proton and r = separation) [2]
ratio = 2 0 2 π 4 Gm e ε [1]
The r2 terms cancel and so this ratio is independent of the separation. [1]
ratio = 12 19 112 27 2 ) 10 7 . 1 ( 10 67 . 6 10 85 . 8 π 4 ) 10 6 . 1 ( − − − − × × × × × × × [1] ratio ≈ 1.2 × 1036 [1] 12 a F = 2 0 2 1 π 4 r Q Q ε = 9 × 10 9 × 2 15 19 19 ) 10 ( 10 6 . 1 10 6 . 1 − − − × × ×
[
1] = 230 N [1] b V = r Q 0 π 4ε
= 9 × 10 9 × 15 19 10 10 6 1 − − × . [1] = 1.44 × 106 V [1] c W = VQ = 1.44 × 106 × 1.6 × 10–19 [1] = 2.3 × 10–13 J [1] d 2 1 mv2 = W ⇒ v2 =m
W
2
[1] v2 = 27 13 10 7 1 10 3 2 2 − − × × × . . = 2.7 × 10–14 [1] v = 2.7×10−14 = 1.6 × 107 m s–1 [1]24 Marking scheme: Worksheet (A2)
1 a Q = VC = 9.0 × 30 × 10–6 [1]
Q = 2.7 × 10–4 C (270 µC) [1]
b number of excess electrons =
e Q = 194 10 6 . 1 10 7 . 2 − − × × [1] number = 1.69 × 1015 ≈ 1.7 × 1015 [1]
2 a i The charge is directly proportional to the voltage across the capacitor.
Hence doubling the voltage will double the charge. [1]
charge = 2 × 150 = 300 nC [1]
ii Since Q ∝ V for a given capacitor, increasing the voltage by a factor of three
will increase the charge by the same factor. [1]
charge = 3 × 150 = 450 nC [1] b C = VQ = 30 10 150 9 . − × [1] C = 5.0 × 10–8 F [1] 3 a E = 2 1 V2C = 2 1 × 9.02 × 1000 × 10–6 [1] E = 4.05 × 10–2 J ≈ 4.1 × 10–2 J [1]
b For a given capacitor, energy stored ∝ voltage2. [1]
energy = 22 × 4.05 × 10–2 ≈ 0.16 J [1] 4 a Ctotal = C1 + C2 [1] Ctotal = 20 + 40 = 60 nF [1] b total 1 C = 1 1 C + 2 1 C [1] total 1 C = 100 1 + 5001 = 0.012 µF −1 [1] Ctotal = 012 . 0 1 ≈ 83 µF [1] c total 1 C = 1 1 C + 2 1 C + 3 1 C [1] total 1 C = 10 1 + 50 1 + 100 1 = 0.13 µF−1 [1] Ctotal = 13 . 0 1 ≈ 7.7 µF [1]
d Total capacitance of the two capacitors in parallel = 50 + 50 = 100 µF. [1]
total 1 C = 50 1 + 100 1 = 0.03 µF−1 [1] Ctotal = 03 . 0 1 ≈ 33 µF [1]
e Total capacitance of the two capacitors in series is 83 µF (from b). [1]
Ctotal = 83 + 50 = 133 µF ≈ 130 µF [1]
5 a Ctotal = C1 + C2 [1]
Ctotal = 100 + 500 = 600 µF [1]
c Q = VC = 1.5 × 600 × 10–6 [1] Q = 9.0 × 10–4 C (900 µC) [1] d E = 2 1QV = 2 1 × 9.0 × 10–4 × 1.5 [1] E = 6.75 × 10–4 J ≈ 6.8 × 10–4 J [1] 6 a E = 2 1 V2C = 2 1 × 322 × 10 000 × 10–6 [1] E = 5.12 J ≈ 5.1 J [1] b P = t E = 300 . 0 12 . 5 [1] P ≈ 17 W [1] 7 a Q = VC = 12 × 1000 × 10–6 [1] Q = 1.2 × 10–2 C (12 mC) [1] b i Ctotal = C1 + C2 [1] Ctotal = 1000 + 500 = 1500 µF [1] ii V = C Q
(The charge Q is conserved and C is the total capacitance.) [1]
V = 26 10 1500 10 2 . 1 − − × × = 8.0 V [1] 8 a ∆Q = I∆t = 50 10 225× −3 = f I = 4.5 × 10–3 C [1] b C = 0 . 9 10 5 . 4 × −3 = V Q [1] = 5.0 × 10–4 F = 500 µF [1]
c i The capacitors are in parallel, so the total capacitance = 2C and charge stored = 2Q;
current =2I = 2 × 225 = 450 mA [1]
ii The capacitors are in series, so the total capacitance =
2
1C and charge stored = 2 1 Q; current = 2 1 I = 2 1 × 225 = 113 mA [1] 9 a Q = CV = 200 × 10–6 × 200 = 0.040 C [1] b E = 2 1 CV2 = 2 1 × 200 × 10–6 × 2002 = 4.0 J [1]
c The two capacitors now make a pair of capacitors in parallel of total capacitance
= 100 µF + 200 µF = 300 µF [1]
The charge is conserved, therefore V = 6 10 300 040 0 − × = . C Q [1] = 133 V [1] d E = 1CV2 = 1 × 300 × 10–6 × 1332 [1]
10 The capacitors are in parallel, so the total capacitance = 3C. [1]
The total charge Q remains constant. [1]
The energy stored by a capacitor is given by E = C Q 2 2 . [1] Einitial = C Q 2 2 and Efinal = ) 3 ( 2 2 C Q [1] Fraction of energy stored =
initial final E E = C Q C Q 2 ) 3 ( 2 2 2 = 3 1 [1] Fraction of energy ‘lost’ as heat in resistor = 1 –
3 1 = 3 2 . [1] The resistance governs how long it takes for the capacitor to discharge. The final voltage
across each capacitor is independent of the resistance. Hence, the energy lost as heat is
1 The direction of the magnetic field should be clockwise (and not anticlockwise). [1]
The separation between adjacent circular field lines should increase further away from
the wire (see below). [1]
2 a The conductor is pushed to the left. [1]
b The conductor is pushed to the left. [1]
c The conductor is pushed out of the plane of the paper. [1]
3 B = Il F [1] [B] = m A N = N A−1 m−1 [1] 4 F = BIl [1] F = 0.12 × 3.5 × 0.01 (length = 1.0 cm) [1] F = 4.2 × 10–3 N [1] 5 a F = BIl sin θ F = 0.050 × 3.0 × 0.04 × sin 90° [1] F = 6.0 × 10–3 N [1] b F = 0.050 × 3.0 × 0.04 × sin 30° [1] F = 3.0 × 10–3 N [1] c F = 0.050 × 3.0 × 0.04 × sin 65° [1] F = 5.44 × 10–3 N ≈ 5.4 × 10–3 N [1]
6 Force experienced by PQ = force experienced by RS (but in opposite direction). [1]
No force experienced by QR and PS (since current is parallel to the field). [1]
torque = one of the forces × perpendicular distance between forces = (BIL)x [1]
torque = BI(Lx), Lx = area of loop = A [1]
torque = BIA ∝ A [1] The torque is directly proportional to the area of the loop.
ii F = BIl ⇒ I = Bl F [1] I = ) 07 . 0 10 0 . 6 ( 10 0 . 1 3 3 × × × − − = 2.38 A ≈ 2.4 A [1]
8 a F = BIl × number of turns [1]
F = 0.19 × 2.8 × 0.07 × 25 = 0.93 N [1]
b Torque= Fd = 0.93 × 0.03 [1]
torque = 0.028 N m [1]
c The longest side always stays at 90º to the magnetic field as the coil turns
so the force is constant. [1]
The perpendicular distance between the forces changes as the coil turns
so the torque (moment) changes. [1]
9 a F = mg = (103.14 – 102.00) × 10–3 × 9.81 so F = 1.12 × 10–2 N ≈ 1.1 × 10–2 N [1] b B = Il F [1] B = 05 . 0 2 . 8 10 12 . 1 2 × × − [1] B = 2.73 × 10−2 T (27 mT) [1]
1 F = EQ = 5.0 × 105 × 3.2 × 10−19 [1] F = 1.6 × 10−13 N [1] 2 a E = d V = 2 10 0 . 3 600 − × [1] E = 2.0 × 104 V m−1 [1]
The field acts towards the negative plate. [1]
b The electric field is uniform between the plates (except at the ‘edges’). [1]
The electric field is at right angles to the plate. [1]
c i Since the droplet is stationary,
the electric force on the droplet
must be equal and opposite to its weight. [1] The electric force must act upwards,
so the charge on the droplet must
be negative. [1] ii E = QF Q = E F = 415 10 0 . 2 10 4 . 6 × × − [1] Q = 3.2 × 10−19 C [1] 3 F = BQv [1] F = 0.18 × 1.6 × 10–19 × 4.0 × 106 [1] F = 1.15 × 10–13 N ≈ 1.2 × 10−13 N [1] 4 a F = BQv [1] F = 0.004 × 1.6 × 10–19 × 8.0 × 106 [1] F = 5.12 × 10−15 N ≈ 5.1 × 10−15 N [1] b a = mF = 15 31
5.12 10
9.11 10
− −×
×
[1] a = 5.63 × 1015 m s−2 ≈ 5.6 × 1015 m s−2 [1]c From circular motion, the centripetal acceleration a is given by: a = r v2 r = a v2 = 15 2 6 10 63 . 5 ) 10 0 . 8 ( × × [1] r = 1.14 × 10−2 m ≈ 1.1 × 10−2 m (1.1 cm) [1]
5 a
Both arrows at A and B are towards the centre of the circle. [1]
b The force on the electron is at 90° to the velocity. Hence the path described by the
electron is a circle. [1]
c The magnetic force provides the centripetal force. [1]
Therefore: BQv = r mv2 [1] BQ = r mv or v = m BQr [1] v = 3 3119 2 10 1 . 9 10 0 . 5 10 6 . 1 10 0 . 2 − − − − × × × × × × [1] v = 1.76 × 107 m s–1 ≈ 1.8 × 107 m s−1 [1] d v = m
BQr , so the speed v is directly proportional to the radius r. [1] Radius is halved, so v = 2 10 76 . 1 × 7 = 8.8 × 106 m s−1 [1] 6 a Ek = 15 × 103 × 1.6 × 10−19 = 2.4 × 10−15 J (1 eV = 1.6 × 10−19 J) [1] 2 1 mv2 = 2.4 × 10−15 v = 27 15 10 7 . 1 10 4 . 2 2 − − × × × [1] v = 1.68 × 106 m s−1 ≈ 1.7 × 106 m s−1 [1] b F = ma = r mv2 [1] F = 05 . 0 ) 10 68 . 1 ( 10 7 . 1 × −27× × 6 2 [1] F = 9.60 × 10−14 N ≈ 9.6 × 10−14 N [1] c F = BQv [1] B = QvF = 19 6 14 10 68 . 1 10 6 . 1 10 60 . 9 × × × × − − [1] B ≈ 0.36 T [1] d speed = time distance time = speed nce circumfere = 6 10 68 . 1 05 . 0 π 2 × × [1] time = 1.87 × 10−7 s ≈ 1.9 × 10−7 s [1]
7 a In order for the positively charged ions to emerge from the slit,
the net force perpendicular to the velocity must be zero. [1]
electrical force on ion = magnetic force on ion [1]
EQ = BQv [1]
The charge Q cancels.
E = Bv [1]
The electric field strength is E = d
V . Therefore, the magnetic flux density is: B = v E = v d V = 6 3 10 0 . 6 024 . 0 / ) 10 0 . 5 ( × × [1] B = 3.47 × 10–2 T ≈ 35 mT [1] b v = m BQr so r = BQmv [1] ∆r = BQ v m m ) ( 1− 2 [1]
8 a The centripetal force is provided by the magnetic force. [1]
Therefore: Bev = r mv2 [1] Be = r mv or v = m Ber [1] T = speed nce circumfere = m Ber r π 2 [1] The radius r of the orbit cancels. Hence: T =
Be m π 2
The time T is independent of both the radius of the orbit r and the speed v. [1]
27 Marking scheme: Worksheet (A2)
1 a i Magnetic flux, Φ = BA [1]
ii Flux linkage = NΦ = NBA [1]
b Flux linkage = N(B cos θ)A = NBA cos θ [1]
(The component of B normal to the plane of the coil is B cos θ.)
2 Area A of coil = x2 [1]
flux linkage = NBA = NBx2 [1]
3 a There is no change to the magnetic flux linking the coil, hence according to Faraday’s law,
there is no induced e.m.f. [1]
b The magnetic flux density B increases as the magnet moves towards the coil. [1]
There is an increase in the magnetic flux linking the coil, hence an e.m.f. is induced across
the ends of the coil. [1]
4 a flux linkage = NBA so B =
NA linkage flux [1] B = 4 4 10 0 . 4 70 10 4 . 1 − − × × × [1] B = 5.0 × 10−3 T [1]
b flux linkage = NBA cos θ [1]
flux linkage = 70 × 0.050 × 4.0 × 10−4 × cos 60° [1]
flux linkage = 7.0 × 10−4 Wb [1]
5 a i magnetic flux, Φ = BA = 40 × 10–3 × (0.03 × 0.03) [1]
Φ = 3.6 × 10−5 Wb [1]
ii flux linkage = NΦ = 200 × 3.6 × 10−5 [1]
flux linkage = 7.2 × 10−3 Wb [1]
b Final flux linkage = 0, initial flux linkage = 7.2 × 10−3 Wb. [1]
Hence, change in magnetic flux linkage is 7.2 × 10−3 Wb. [1]
6 a Initial magnetic flux = BA = 0.15 × (π × [8.0 × 10−3]2) [1]
initial magnetic flux = 3.02 × 10−5 Wb [1]
final magnetic flux = 0 [1]
average magnitude of induced e.m.f. = rate of change of magnetic flux linkage E = t Φ N ∆ ∆ , so E = 1200 × 020 . 0 10 02 . 3 0− × −5 [1] E = 1.81 V ≈ 1.8 V (magnitude only) [1] b average current = resistance e.m.f. = 3 . 6 81 . 1 [1] I = 0.287 A ≈ 0.29 A [1]
7 a distance = speed × time = 2.0 × 1.0 = 2.0 m [1]
b Area swept = length × distance travelled = 0.10 × 2.0 = 0.20 m2 [1]
c Change in magnetic flux = area swept × magnetic flux density [1]
change in magnetic flux = 0.20 × 0.050 = 1.0 × 10−2 Wb [1]
d Magnitude of e.m.f. = rate of change of magnetic flux linkage [1]
E = t NΦ ∆ ∆( ) (N = 1) E = 0 . 1 10 0 . 1 × −2 = 1.0 × 10−2 V (1 Wb s−1 = 1 V) [1] e E = Blv = 0.050 × 2.0 × 0.10 = 1.0 × 10−2 V [1]
8 Initial magnetic flux = BA = 0.060 × π × (1.2 × 10−2)2 [1]
initial magnetic flux = 2.72 × 10−5 Wb [1]
final magnetic flux = –2.72 × 10−5 Wb (since the field is reversed) [1]
average magnitude of induced e.m.f. = rate of change of magnetic flux linkage E = t Φ N ∆ ∆ , so E = 2000 × 030 . 0 10 72 . 2 10 72 . 2 × −5− × −5 − [1] E = 3.62 V ≈ 3.6 V (magnitude only) [1]
9 a There is a current in the primary coil when the switch is closed. This current creates a
magnetic flux in the primary coil. [1]
Due to the soft iron ring, the magnetic flux created by the primary coil also links the secondary coil. With the switch closed, there is no change in the magnetic flux linkage
at the secondary and hence the lamp is not lit. [1]
When the switch is opened, the magnetic flux decreases to zero in a short period. The rapid change in magnetic flux at the secondary coil creates an e.m.f. and the lamp
illuminates for a short period. [1]
Eventually there is no magnetic flux at either the primary or the secondary coil and
hence there is no e.m.f. induced – the lamp stays off. [1]
10
distance = speed × time = vt [1]
area swept = length × distance travelled = Lvt [1]
change in magnetic flux = area swept × magnetic flux density [1]
change in magnetic flux = (Lvt) × B = BLvt [1] magnitude of e.m.f. = rate of change of magnetic flux linkage [1]
E = t
BLvt = BLv [1]
E = BLv = 40 × 10−6 × 0.20 × 0.30 [1]
E = 2.4 × 10−6 V (2.4 µV) [1]
11 a The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of
change of magnetic flux linkage. [1]
b Lenz’s law expresses the principle of conservation of energy. [1]
c i magnetic flux = magnetic flux density × cross-sectional area of coil
or Φ = BA [1]
ii Flux linkage = NBA [1]
Coil X: flux linkage = NBA = 200 × 0.10 × (π × 0.022) ≈ 2.5 × 10−2 Wb [1]
Coil Y: flux linkage = NBA = 4000 × 0.01 × (π × 0.032) ≈ 1.1 × 10−1 Wb [1]
1 a T = f 1 [1] 0.1 s [1] b i t = 0, 0.05 and 0.15 [1] ii t = 4 1 of a period [1] t = 0.025 s [1] iii t = 0.075 s [1] iv t = 2 1 = sin (20πt) [1] t = 0.0125 s [1]
2 a Corresponding parts of each wave occur at the same time. [1]
b Power P always positive. [1]
Two peaks for P in time taken for one peak of I and V. [1]
Sinusoidal shape above axis. [1]
3 a i I = VP = 230 690 [1] Irms = 3.0 A [1] ii Ipeak = 3 2 = 4.2 A [1] iii Vpeak= 230 2 = 325 V [1]
b Correct shape all above axis. [1]
Two cycles shown (i.e. one cycle of current waveform). [1]
4 a a.c. [1]
Current is positive and negative OR current flows one way, then the opposite way. [1]
b Voltage switches between +2 and −2 V. Power is
R V2
, so it has the same value for both
positive and negative values of the voltage. [1]
c i 2 V [1] ii 2 V [1] 5 Average power = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 5 . 2 × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 6 [1] = 7.5 W [1] 6 a i Period = 80 ms = 0.080 s [1] f = T 1 = 12. 5 Hz [1] ii Peak voltage = 4.5 V [1] r.m.s. voltage = 2 5 . 4 = 3.2 V [1] b V0 = 4.5 V [1] ω = 2πf = 2π × 12.5 = 78.5 s−1 ≈ 79 s−1 [1] 7 a I = V P = 6 24 [1] I = 4.0 A [1] b Vp = 6 × 30 1150 [1] Vp = 230 V [1] c Ip = 230 24 or Ip = 1150 4 30× [1] Ip = 0.10 A [1] d Maximum p.d. = 6 2 [1] maximum p.d. = 8.5 V [1]
e Heat is still produced inside the wires even if it cannot be conducted to the
outside. The wires may melt if they cannot lose the heat. [1]
8 a Number of turns on the primary =
230 115 × 500 = 1000 turns [1] b Is = R Vs = 5000 115 = 0.023 A r.m.s. [1] c Ip = 0.023 × 1000 500 = 0.0125 A ≈ 0.013 A r.m.s. [1] d Peak voltage = 115 2 = 162 V [1]
so the cables will break down. [1]
9 a I =
P
V
=1000
100
= 10 A r.m.s. [1]P = I 2R = 102 × 5 = 500 W [1]
b At high voltages the current is less for the same power. [1]
1 A ‘packet’, or quantum, of electromagnetic energy. [1]
2 The energy of a photon is proportional to the frequency of the radiation. [1]
Hence a γ-ray photon has greater energy than a photon of visible light (and therefore
is more harmful). [1]
3 a Electromagnetic radiation travels through space as waves and, as such, shows diffraction
and interference effects. [1]
b Electromagnetic radiation interacts with matter as ‘particles’. The photoelectric effect
provides strong evidence for the particle-like (photon) behaviour of electromagnetic
radiation. [1] 4 a c = fλ so f = 7 8 10 4 . 6 10 0 . 3 − × × = λ c [1] f = 4.69 × 1014 Hz ≈ 4.7 × 1014 Hz [1] b E = hf [1] E = 6.63 × 10−34 × 4.69 × 1014 [1] E = 3.1 × 10−19 J [1]
5 For an electron to escape from the surface of the metal, it must absorb energy from the photon
that is greater than the work function. [1]
The work function is the minimum energy required by the electron to escape from the surface
of the metal. [1]
The photon of visible light has energy less than the work function of the metal, whereas
the photon of ultraviolet radiation has energy greater than the work function. [1]
6 a The electron loses energy. [1]
This energy appears as a photon of electromagnetic radiation. [1]
b Energy of photon = E1 − E2 [1] Therefore: hf = E1 − E2 or h E E f = 1− 2 [1]
c The change in energy ∆E is greater. [1]
Hence the frequency of the radiation is greater (f ∝ ∆E). [1] The spectral line will be the right side of the line shown on the spectrum diagram.
8 ∆E=hf =hcλ [1] 9 8 34 10 670 10 0 . 3 10 63 . 6 − − × × × × = ∆E [1] ∆E = 2.97 × 10−19 J ≈ 3.0 × 10−19 J [1] 9 a Continuous spectrum [1] b Emission spectrum [1] c Absorption spectrum [1]
10 Electrons travel through space as waves. Evidence for this is provided by the diffraction of
electrons by matter (e.g. graphite). [1]
11 The electronvolt is the energy gained by an electron travelling through a potential difference
of one volt. [1]
12 The kinetic energy Ee of the electron is:
Ee = VQ = 6.0 × 1.6 × 10–19 [1]
Ee = 9.6 × 10–19 J [1]
The energy EUV of the ultraviolet photon is:
EUV = hf = λ hc [1] EUV = 7 8 34 10 5 . 2 10 0 . 3 10 63 . 6 − − × × × × [1] EUV = 7.96 × 10−19 J ≈ 8.0 × 10−19 J [1]
The energy of the ultraviolet photon is less than the kinetic energy of the electron. (The student is correct.)
13 a The threshold frequency is the minimum frequency of electromagnetic radiation that just
removes electrons from the surface of the metal. [1]
b At the threshold frequency, the energy of the photon is equal to the work function φ of the
metal. Hence: φ = hf0 (f0 = threshold frequency) [1] f0 = 34 19 10 63 . 6 10 6 . 1 9 . 1 − − × × × [1] f0 = 4.6 × 1014 Hz [1] 14 a E = hf = λ hc [1] E = 34 9 8 10 550 10 0 . 3 10 63 . 6 − − × × × × [1] E = 3.62 × 10–19 J ≈ 3.6 × 10–19 J [1]
b Power emitted as light = 0.05 × 60 = 3.0 W [1]
Number of photons emitted per second = 19 10 62 . 3 0 . 3 − × [1] = 8.3 × 1018 [1] 15 φ = 4.3 × 1.6 × 10–19 = 6.88 × 10–19 J [1] Energy of photon = λ hc = 7 8 34 10 1 . 2 10 0 . 3 10 63 . 6 − − × × × × = 9.47 × 10−19 J [1]
energy of photon = work function + maximum kinetic energy of electron [1] maximum kinetic energy of electron = (9.47 − 6.88) × 10−19 ≈ 2.6 × 10−19 J [1]
16 λ = mv h [1] ν = λ m h = 11 27 34 10 0 . 2 10 7 . 1 10 63 . 6 − − − × × × × [1] v = 1.95 × 104 m s−1 (20 km s−1) [1]
17 Energy lost by a single electron = energy of photon [1]
(The energy lost by a single electron travelling through the LED reappears as the energy of a single photon.) Therefore:
λ
hc
eV
=
[1] 19 7 8 34 10 6 . 1 10 8 . 5 10 0 . 3 10 63 . 6 − − − × × × × × × = = e hc V λ [1] V = 2.14 V ≈ 2.1 V [1]18 a Kinetic energy of electron = VQ = Ve
2 1 mev2 = Ve or e 2 2mp = Ve (where p = mev) [1] p = 2meVe [1] λ = p h v
mh =e (de Broglie equation) [1]
Therefore, λ = Ve m h e 2 b λ = Ve m h e 2 or V = m e h 2 e 2 2 λ [1] V = 31 341122 19 10 6 . 1 ) 10 0 . 4 ( 10 1 . 9 2 ) 10 63 . 6 ( − − − − × × × × × × × [1] V ≈ 940 V [1] 19 Using f = λ
c and Einstein’s photoelectric equation (hf = φ
+ ) 2 1 2 max mv : [1] Red light ⇒ 98 10 640 10 0 . 3 − × × × h = φ + (0.9 × 1.6 × 10−19) ⇒ 4.688 × 1014 h = φ + 1.440 × 10−19 (equation 1) [1] Blue light ⇒ 98 10 420 10 0 . 3 − × × × h = φ + (1.9 × 1.6 × 10−19) ⇒ 7.143 × 1014 h = φ + 3.040 × 10−19 (equation 2) [1]
Equations 1 and 2 are two simultaneous equations. (7.143 – 4.688) × 1014 h = (3.040 − 1.440) × 10−19
20 a External energy has to be supplied to excite or free an electron. [1]
(Allow: The electrons are trapped in an energy well.)
b An energy level of 0 eV means the electron is free from the atom.
The minimum energy is equal to 3.00 eV. [1]
Energy needed to free electron = 3.00 × 1.6 × 10−19 [1]
Energy needed to free electron = 4.80 × 10−19 J [1]
c i The difference between the energy levels −3.00 eV and −1.59 eV is equal to 1.41 eV. [1]
Hence, an electron jumps from −3.00 eV energy level to −1.59 eV energy level. [1]
ii λ hc hf E= = ∆ [1] 19 8 34 10 6 . 1 41 . 1 10 0 . 3 10 63 . 6 − − × × × × × = ∆ = E hc λ [1] λ = 8.82 × 10−7 m [1]
21 a This is the lowest energy level occupied by an electron in an atom. [1]
b The shortest wavelength corresponds to the change in energy between the two most widely
separated energy levels.
Hence, ∆E = 10.43 eV [1] λ hc hf E= = ∆ [1] 19 8 34 10 6 . 1 43 . 10 10 0 . 3 10 63 . 6 − − × × × × × = ∆ = E hc λ [1] λ = 1.19 × 10−7 m [1] 22 a E1 = 2 18 1 10 18 . 2 × − − = –2.18 × 10−18 J [1] E2 = 2 18 2 10 18 . 2 × − − = –5.45 × 10−19 J [1] b E2 − E1 = ∆E = hcλ [1] 19 8 34 10 ) 45 . 5 8 . 21 ( 10 0 . 3 10 63 . 6 − − × − × × × = ∆ = E hc λ [1] λ = 1.22 × 10−7 m [1]
This spectral line lies in the ultraviolet region of the spectrum. [1]
c ∆E = 2.18 × 10−18 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 2 7 1 6 1 = 1.607 × 10−20 J [1] 20 8 34 10 607 1 10 0 3 10 63 6 − − × × × × = ∆ = . . . E hc λ = 1.238 × 10−5 m [1] λ ≈ 1.24 × 10−5 m (12.4 µm) [1]
1 a change in energy = change in mass × (speed of light)2 or ∆E = ∆mc2 [1] b i ∆E = ∆mc2 ∆E = 0.001 × (3.0 × 108)2 [1] ∆E = 9.0 × 1013 J [1] ii ∆E = ∆mc2 ∆E = 9.1 × 10−31 × (3.0 × 108)2 = 8.19 × 10−14 J [1] ∆E ≈ 8.2 × 10−14 J [1] 2 a i Mass = 27 27 10 66 . 1 10 65 . 6 − − × × = 4.01 u [1] ii Mass = 27 26 10 66 . 1 10 16 . 2 − − × × = 13.01 u [1] b i Mass = 1.01 × 1.66 × 10−27 = 1.68 × 10−27 kg [1] ii Mass = 234.99 × 1.66 × 10−27 = 3.90 × 10−25 kg [1]
3 In all nuclear reactions the following quantities are conserved:
• charge (or proton number) • nucleon number
• mass–energy • momentum.
Any three of the above. [3]
4 a The nucleons within the nucleus are held tightly together by the strong nuclear force. [1] b The binding energy of a nucleus is the minimum energy required to separate the nucleus
into its constituent protons and neutrons. [1]
c binding energy per nucleon =
nucleons of
number
energy binding binding energy per nucleon =
16 128
[1]
binding energy per nucleon = 8.0 MeV [1]
5 a The half-life of a radioactive isotope is the mean time taken for the number of nuclei of the
isotope to decrease to half the initial number. [1]
b i 20 minutes is 1 half-life, so number of nuclei left =
2
0
N
[1]
ii 1.0 hour is 3 half-lives. [1]
Number of nuclei left =
8 2 1 0 0 3 N N = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ [1]
6 a i Activity is equal to the number of emissions (or decays of nuclei) per second.
7 a The nuclide Fe5626 is the most stable. [1]
It has the maximum value for the binding energy per nucleon. [1]
b Binding energy = binding energy per nucleon × number of nucleons
binding energy ≈ 12.3 × 10−13 × 12 [1]
binding energy ≈ 1.5 × 10−11 J [1]
c From the graph, the binding energies per nucleon of H21 and He42 are approximately
1.0 × 10−13 J and 11.2 × 10−13 J. [1]
energy released = difference in binding energy per nucleon × number of nucleons [1]
energy released = [11.2 × 10−13 – 1.0 × 10−13] × 4 [1]
energy released = 4.08 × 10−12 J ≈ 4.1 × 10−12 J [1]
d High temperatures (~108 K) and pressures. [2]
8 uranium92 11proton +14301neutron → 23592 [1]
mass defect = [(143 × 1.009) + (92 × 1.007)]u – (234.992)u [1]
mass defect = 1.939 u = 1.939 × 1.66 × 10−27 kg [1]
mass defect = 3.219 × 10−27 kg [1]
binding energy = mass defect × (speed of light)2 [1]
binding energy = 3.219 × 10−27 × (3.0 × 108)2 = 2.897 × 10−10 J [1]
binding energy per nucleon =
nucleons of
number
energy binding binding energy per nucleon =
235 10 897 . 2 × −10 = 1.233 × 10−12 ≈ 1.2 × 10−12 J [1] 9 a Fission is the splitting of a heavy nucleus like 23592U into two approximately equal fragments.
The splitting occurs when the heavy nucleus absorbs a neutron. [1]
b i All particles identified on the diagram. [1]
ii In the reaction above, there is a decrease in the mass of the particles. [1]
According to ∆E = ∆mc2, a decrease in mass implies that energy is released
in the process. [1]
iii The change in mass is ∆m given by:
∆m = [1.575 × 10−25 + 2.306 × 10−25 + 2(1.675 × 10−27)] – [3.902 × 10−25 + 1.675 × 10−27][1]
∆m = –4.250 × 10−28 kg [1]
(The minus sign means a decrease in mass and hence energy is released in this reaction.)
∆E = ∆mc2 [1]
∆E = 4.250 × 10−28 × (3.0 × 108)2 [1]
10 Binding energy of ‘reactant’ = 236 × 7.59 = 1791 MeV (binding energy of neutron = 0) [1]
Total binding energy of ‘products’ = (146 × 8.41) + (87 × 8.59) ≈ 1975 MeV [1]
Therefore energy released = 1975 – 1791 = 184 MeV [1]
11 a t1/2 = λ 693 . 0 so λ = 1/2 693 . 0 t [1] λ = 56 693 . 0 [1] λ = 1.238 × 10−2 s−1 ≈ 1.2 × 10−2 s−1 [1] b A = λN [1] A = 56 693 . 0 × 6.0 × 1010 [1] A ≈ 7.4 × 108 Bq [1] 12 a A = λN so λ = N A [1] λ = 149 10 0 . 8 10 0 . 5 × × [1] λ = 6.25 × 10−6 s−1 ≈ 6.3 × 10−6 s−1 [1] b t1/2 = λ 693 . 0 [1] t1/2 = 6 10 25 . 6 693 . 0 − × [1] t1/2 = 1.11 × 105 s ≈ 1.1 × 105 s [1] c N = N0 e−λt [1] N = 8.0 × 1014 (6.25106 40 3600)
e
− × −× × [1] N = 3.25 × 1014 ≈ 3.3 × 1014 [1]13 a The decay constant is the probability that an individual nucleus will decay per unit time. [1] b i t1/2 = λ 693 . 0 so λ = 1/2 693 . 0 t [1] λ = 3600 24 18 693 . 0 × × [1] λ = 4.46 × 10−7 s−1 ≈ 4.5 × 10−7 s−1 [1] ii A = λN [1] A = 4.46 × 10−7 × 4.0 × 1012 [1] A = 1.78 × 106 Bq ≈ 1.8 × 106 Bq [1]
iii 36 days is equal to 2 half-lives. [1]
activity = 2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × 1.78 × 106 = 4.45 × 105 Bq ≈ 4.5 × 105 Bq [1] 14 a number of nuclei = number of moles × NA
15 According to Einstein’s equation: ∆E = ∆mc2 [1]
In this case, ∆E is the energy of two photons and ∆m is the mass of two protons. [1] Hence: 2 ×
λ
hc
= (2 × mp) c2 [1] 8 27 34 p 2 p 1.7 10 3.0 10 10 63 . 6 × × × × = = = − − c m h c m hc λ [1] λ = 1.3 × 10−15 m [1]16 For fusion, we have:
energy released per kg = number of ‘pairs’ of H2
1 in 1 kg × 4.08 × 10−12 J (from 7 c) [1] energy per kg = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × ×6.02×1023 2 1000 2 1 × 4.08 × 10−12 [1] energy per kg = 6.14 × 1014 J ≈ 6.1 × 1014 J [1]
For fission, we have:
energy released per kg = number of nuclei in 1 kg × 3.83 × 10−11 J (from 9 b) [1]
energy per kg = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×6.02×1023 235 1000 × 3.83 × 10−11 [1] energy per kg = 9.8 × 1013 J [1]
There is less energy released per fusion than per fission. However, there are many more nuclei per kg for fusion. Hence fusion produces more energy per kg than fission. [1]
17 N = N0 e−λt and λ = 1/2 693 . 0 t [1] fraction left = (0.693 / ) 0 1/2 e e t t t N N = −λ = − [1] fraction left = (0.693 5.0 109/4.5109)
e
− × × × [1] fraction left = 0.463 ≈ 0.46 [1]1 Processor (or processing device) [1]
Output device [1]
2 a Axis labels [1]
Graph correct shape [1]
b Thermistor, strain gauge, microphone, etc. [1]
3 a Length of fine/thin wire [1]
sealed in a plastic case [1]
b i A l R= ρ so R l A= ρ = 150 15 . 0 10 0 . 5 × −7× [1] A = 5.0 × 10−10 m2 [1] ii Extends by 150 1
so resistance increases by a factor of 150
1 , which is 1 Ω. [1]
4 a Fraction of the output of a device is fed back to the input [1]
reducing any changes in the input to the system. [1]
b Increased bandwidth, less distortion, greater stability in gain (less affected by temperature
changes). [2]
5 a Inverting amplifier [1]
When Vin is positive then Vout is negative, or vice versa. [1] b The output voltage reaches the power supply voltage and cannot exceed this value. [1]
c 10 V [1]
d i Gain becomes −10 rather than −5, i.e. sloping line has twice the gradient. [1]
Vout is +10 V for Vin below −1 V or Vout is −10 V for Vin above +1 V. [1] ii Gain becomes −2.5 rather than −5, i.e. sloping line has half the gradient. [1]
iii The sloping line increases in length before flattening off. [1]
e Output voltage peaks at ±5 V. [1]
6 a The output is not saturated so the potential at both the inverting and non-inverting inputs
is very nearly the same. [1]
Since the non-inverting input is earthed, this value is 0 V. [1]
b 1.0 V [1] c I = = = 2000 1 R V 5.0 × 10−4 A [1]
d Negligible current goes into the (–) terminal of the op-amp since it has a very high input
resistance (impedance). [1] e Size of gain = in f RR = 5 so Rf = 5Rin Rf = 5 × 2 = 10 kΩ [1]
7 a Any two from:
infinite open-loop voltage gain, infinite input resistance (or impedance),
infinite bandwidth, zero output resistance (or impedance) [2]
b i I = 4 3 10 10− = R V [1] I = 1.0 × 10−7 A [1] ii V = IR = −1.0 × 10−7 × 100 × 103 = −1.0 × 10−2 V [1] iii Gain = 3 2 in out 10 0 . 1 10 0 . 1 − − × × − = V V = −10 [1]
iv 1.0 × 10−2 V (with Q being more positive) [1]
8 a For [2] marks, any two from the following points. [2]
With an inverting amplifier the output is half a cycle out of phase with the input, whereas with a non-inverting amplifier the input and output are in phase.
The input goes to the (−) terminal on the op-amp for an inverting amplifier and to the (+) terminal for the non-inverting amplifier.
The input resistance (impedance ) is higher for an op-amp used as a non-inverting amplifier. For an inverting amplifier the gain is
in f
R R −
but for an inverting amplifier the gain is 2 1 1 R R + . b G = 2 1 1 R R + = 20 40 1+ [1] G = 3 [1] c in out V V G= ⇒ = = = 3 0 . 8 out in G V V 2.67 V ≈ 2.7 V [1] d
