1
The diagram below shows an apparatus used to demonstrate that electron beams possess
energy and momentum.
Electrons are accelerated by the electron gun and travel through the space in the tube which has
been evacuated to a low pressure. They collide with the mica wheel and cause it to rotate.
(a)
(i)
Explain how the electrons would lose energy if the tube were not evacuated.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(ii)
The pressure of the gas in the tube is 0.20 Pa when the room temperature is 300 K.
The volume of the tube is 3.5 × 10–2 m3. Determine the number of moles of gas in the
tube.
universal gas constant, R = 8.3 J mol–1 K–1
(2)
(iii)
The Avogadro constant is 6.0 × 1023 mol–1.
How many gas molecules are there in the tube?
...............................................................................................................
(1)
(b)
In the electron gun, the electrons of mass 9.1 × 10–31 kg and charge 1.6 × 10–19 C are
accelerated to a speed of 2.7 × 107 m s–1. The beam current is 5.0 mA.
(i)
Calculate, in J, the energy of each electron in the beam.
(1)
Page 1 of 7
(ii)
Use the definition of the volt to determine the potential difference through which the
electrons are accelerated.
(2)
(iii)
How many electrons are accelerated each second?
(2)
(iv)
What happens to the energy of the electrons when they strike the mica target?
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(v)
Determine the force exerted on the paddle wheel by the electrons. Assume that the
electrons are brought to rest when they collide.
(3)
(vi)
The moment of this force about the axle of the paddle causes rotational acceleration
of the paddle. What is the magnitude of this moment?
(2)
(Total 18 marks)
2
A 1.0 kΩ resistor is thermally insulated and a potential difference of 6.0 V is applied to it for 2.0
minutes. The thermal capacity of the resistor is 9.0 J K–1. The rise in temperature, in K, is
A
1.3 × 10–3
B
8.0 × 10–3
C
0.48
D
0.80
(Total 1 mark)
3
A metal wire is maintained at a constant temperature. Which one of the following graphs best
represents the relationship between the dissipated power P and the current I in the wire?
(Total 1 mark)
Page 2 of 7
4
The electrical energy for a small village of 155 houses is to be generated by a bank of solar cells.
The average power used by each house, taken over a year, is 800 W.
The average power per square metre arriving at the surface of the Earth from the Sun is 650 W.
The efficiency of the conversion of solar energy to electrical energy is 15%.
(a)
(i)
Calculate the average power the solar cells need to provide for the whole village.
(1)
(ii)
Calculate the total area of solar cells needed to provide the power in (i).
(3)
(iii)
Give one reason why, in practice, a greater area will need to be covered by solar
cells.
...............................................................................................................
...............................................................................................................
(1)
(iv)
Suggest two problems, other than the large area needed for solar cells, that occur
using solar power alone to provide the supply to the village.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(b)
The emf of the bank of solar cells is 230 V.
(i)
Calculate the supply current when the village is using 75 kW, assuming the cells have
no internal resistance.
(2)
(ii)
Calculate the potential difference delivered to the villagers' electrical equipment when
the current calculated in (i) is produced in a bank of cells with an internal resistance
of 0.050 Ω.
(3)
(Total 12 marks)
Page 3 of 7
Mark schemes
1
(a)
(i)
they would collide with atoms of gas
M1
losing energy by:
ionising the gas
A1
exciting gas molecules
A1
allow 1 for stating inelastic collisions
(3)
(ii)
pV = nRT
C1
n = 2.8 × 10–6
A1
(2)
(iii)
1.7 × 1018
B1
(their (ii) × 6.0 × 1023)
(1)
(b)
(i)
energy ½mv2 = 3.3 × 10–16 J
B1
(1)
(ii)
1 volt = 1 Joule per coulomb
C1
or p.d. = energy gained by electron / charge on electron
2070 V or 2100 V
A1
(2)
(iii)
number of electrons = current / e
C1
3.1 × 1016
A1
(2)
(iv)
it produces heating of the mica
B1
it causes (rotation) kinetic energy of the mica paddle
B1
(2)
Page 4 of 7
(v)
force = change in momentum per second
C1
change in momentum per collision = 2.46 × 10–23 (N s)
C1
force = their (iii) × their momentum (7.6 × 10–7 N)
A1
(3)
(vi)
moment = force × perpendicular distance
C1
1.9 × 10–8 N m
A1
(2)
[18]
2
3
4
C
[1]
C
[1]
(a)
(i)
124 kW
B1
(1)
(ii)
efficiency = useful power output/power input
C1
power needed = (100 / 15) × 124 = 827 kW
A1
area needed = 1270 m
(2)
B1
(3)
allow this mark only for 124 000 / 650 or (their power) / 650 (i.e. area for 100%
efficient system)
B1
(1)
(iii)
any good reason:
e.g. since the value is only an average so more needed sometimes
since little power is used at night more than the average is needed during the
day
since some parts may be in shade ( due to cloud)
since cloud may reduce intensity of radiation on the cells
since there will be less light in winter
(iv)
there is no power at night
B1
the power fluctuates according to the weather conditions / cloud
B1
seasonal changes
B1
Max 2
Page 5 of 7
(b)
(i)
power = VI or numerical substitution
C1
326A
A1
(2)
(ii)
resistance = V/I in any form
or V = E – Ir
C1
p.d. across internal resistance = 16.3 V (e.c.f.)
or correct substitution in V = E – Ir
C1
p.d. at consumer 214 V (e.c.f. from (b)(i)
A1
(3)
[12]
Page 6 of 7
Examiner reports
4
(a)
(b)
(i)
Most obtained 124 kW. A number of candidates incorrectly gave the unit as‘kW per
year’.
(ii)
The majority of the candidates obtained the correct answer. Many tried to determine
the input power to the solar cells that produced the required output. Such an
approach was often unsuccessful. The alternative approach of first determining the
useful power from the sun (97.5 W) was a method that led to more frequent success.
(iii)
There were many loose responses that simply referred to the fact that the sun may
not shine. If this were the case then having more cells would not help. It was
necessary to make clear that due to weather conditions some of the cells may not be
receiving sunlight ( due to partial cloud cover) or that the intensity of the sunlight may
be lower than the 650 W per square metre.
(iv)
Acceptable answers, which were commonly given, included the need for another
source to provide power at night and during adverse weather conditions. Many wrote
of cost being a problem. This was not accepted since any other supply would also
have cost implications and without comparison this was an unjustified comment.
(i)
The majority of the candidates completed this successfully.
(ii)
Again there were many correct answers but many candidates calculated the ‘lost
volts’ and then did not subtract the ‘lost volts’ from 230 V. Some candidates
incorrectly assumed that the resistance of the users' equipment was the same as in
part (i) which was not the case (0.706.Ω. in part (i) and 0.655 .Ω. in part (ii)). It was
the total resistance of the circuit that remained the same.
Page 7 of 7