WAEC SSCE - Physics - 1998 (Essay)
Question 1 Report
(a) Explain the term photoelectric effect.
(b)
The diagram above represents a photocell with its associated electric circuit. Identify each of the physical quantities
represented by the letters A, B, C, D, E and F
(c) What factor determines the: (i) current produced by the photocell
(ii) maximum kinetic energy of the photoelectrons?
(d) State one similarity and one difference between photoemission and evaporation
(e) Name two methods by which a beam of free electrons may be produced other than photoemission
(f) State two applications of photoelectric effect.
(g) A light wavelength 5.0 x 10\(^{-7}\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\(^{-19}\)J, calculate the:
(i) frequency of the light
(ii) energy of the incident photon,
(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\(^{-1}\); Planck's constant = 6.6 x 10\(^{-34}\)Js).
(a) Photoelectric effect
The photoelectric effect is the emission of electrons from the surface of a metal when electromagnetic radiation (light) of sufficiently high frequency falls on it.
(b) The physical quantities
- A - the incident light (beam of photons/radiation) falling on the emitting surface.
- B - the photoelectrons (stream of electrons) emitted from the surface and travelling across the tube.
- C - the cathode, i.e. the photo-emissive metal surface (emitter).
- D - the microammeter (galvanometer) that measures the photoelectric current.
- E - the anode (collector electrode) which attracts the emitted electrons.
- F - the evacuated glass envelope (vacuum tube) enclosing the electrodes.
(The battery labelled 100 V supplies the accelerating p.d. between anode and cathode.)
(c)(i) The current produced by the photocell is determined by the intensity (brightness) of the incident light.
(c)(ii) The maximum kinetic energy of the photoelectrons is determined by the frequency of the incident light (together with the work function of the metal).
(d) Photoemission and evaporation
Similarity: both are surface phenomena in which particles are given enough energy to escape from the surface of a material.
Difference: in photoemission the energy is supplied by light (photons) and the particles emitted are electrons; in evaporation the energy is supplied as heat and the particles emitted are molecules (or atoms) of the liquid.
(e) Other ways of producing free electrons
- Thermionic emission (heating a metal filament).
- Field (cold) emission using a very strong electric field.
(f) Applications of the photoelectric effect
- Automatic (photoelectric) switches for doors, burglar alarms and street lights.
- Light/exposure meters in cameras and the reproduction of sound from film sound-tracks.
(g) Calculations
Given \(\lambda = 5.0 \times 10^{-7}\,\text{m}\), \(W = 3.04 \times 10^{-19}\,\text{J}\), \(c = 3.00 \times 10^{8}\,\text{ms}^{-1}\), \(h = 6.6 \times 10^{-34}\,\text{Js}\).
(i) Frequency of the light
\[ f = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}}{5.0 \times 10^{-7}} = 6.0 \times 10^{14}\,\text{Hz}. \](ii) Energy of the incident photon
\[ E = hf = 6.6 \times 10^{-34} \times 6.0 \times 10^{14} = 3.96 \times 10^{-19}\,\text{J}. \](iii) Maximum kinetic energy of the photoelectrons
\[ K_{max} = hf - W = 3.96 \times 10^{-19} - 3.04 \times 10^{-19} = 9.2 \times 10^{-20}\,\text{J}. \]Answer Details
(a) Photoelectric effect
The photoelectric effect is the emission of electrons from the surface of a metal when electromagnetic radiation (light) of sufficiently high frequency falls on it.
(b) The physical quantities
- A - the incident light (beam of photons/radiation) falling on the emitting surface.
- B - the photoelectrons (stream of electrons) emitted from the surface and travelling across the tube.
- C - the cathode, i.e. the photo-emissive metal surface (emitter).
- D - the microammeter (galvanometer) that measures the photoelectric current.
- E - the anode (collector electrode) which attracts the emitted electrons.
- F - the evacuated glass envelope (vacuum tube) enclosing the electrodes.
(The battery labelled 100 V supplies the accelerating p.d. between anode and cathode.)
(c)(i) The current produced by the photocell is determined by the intensity (brightness) of the incident light.
(c)(ii) The maximum kinetic energy of the photoelectrons is determined by the frequency of the incident light (together with the work function of the metal).
(d) Photoemission and evaporation
Similarity: both are surface phenomena in which particles are given enough energy to escape from the surface of a material.
Difference: in photoemission the energy is supplied by light (photons) and the particles emitted are electrons; in evaporation the energy is supplied as heat and the particles emitted are molecules (or atoms) of the liquid.
(e) Other ways of producing free electrons
- Thermionic emission (heating a metal filament).
- Field (cold) emission using a very strong electric field.
(f) Applications of the photoelectric effect
- Automatic (photoelectric) switches for doors, burglar alarms and street lights.
- Light/exposure meters in cameras and the reproduction of sound from film sound-tracks.
(g) Calculations
Given \(\lambda = 5.0 \times 10^{-7}\,\text{m}\), \(W = 3.04 \times 10^{-19}\,\text{J}\), \(c = 3.00 \times 10^{8}\,\text{ms}^{-1}\), \(h = 6.6 \times 10^{-34}\,\text{Js}\).
(i) Frequency of the light
\[ f = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}}{5.0 \times 10^{-7}} = 6.0 \times 10^{14}\,\text{Hz}. \](ii) Energy of the incident photon
\[ E = hf = 6.6 \times 10^{-34} \times 6.0 \times 10^{14} = 3.96 \times 10^{-19}\,\text{J}. \](iii) Maximum kinetic energy of the photoelectrons
\[ K_{max} = hf - W = 3.96 \times 10^{-19} - 3.04 \times 10^{-19} = 9.2 \times 10^{-20}\,\text{J}. \]Question 2 Report
(a) Distinguish between heat and temperature.
(b) State two physical properties of substances which may be used to measure temperature.
(c) State two reasons why mercury is preferred to alcohol as a thermometric liquid.
(d)(i) Describe how a mercury-in-glass thermometer is calibrated.
(ii) State two precautions necessary to ensure an accurate result.
(e) Explain how land and sea breezes occur.
(a) Heat is a form of energy that flows from a body at a higher temperature to one at a lower temperature; it is measured in joules. Temperature is the degree of hotness or coldness of a body, which determines the direction of heat flow; it is measured in kelvin (or degrees Celsius). Heat depends on mass, while temperature does not.
(b) Physical properties used to measure temperature:
- Expansion of a liquid (e.g. mercury) or of a solid.
- Variation of electrical resistance of a metal with temperature (also e.m.f. of a thermocouple, or pressure/volume of a gas).
(c) Why mercury is preferred to alcohol:
- Mercury has a high boiling point (about 357 C), so it can measure high temperatures without vaporising.
- Mercury is opaque and silvery, so it is easily seen, and it does not wet (stick to) glass, giving accurate readings; it also expands uniformly and conducts heat well.
(d)(i) Calibration. The bulb of the thermometer is placed in pure melting ice; when the mercury thread is steady its position is marked as the lower fixed point (0 C). The bulb is then placed in the steam above boiling water at normal atmospheric pressure and, when steady, that position is marked as the upper fixed point (100 C). The distance between the two marks is divided into 100 equal parts (degrees).
(d)(ii) Precautions:
- Use pure melting ice and pure water so the fixed points are correct.
- For the steam point, ensure standard atmospheric pressure and expose the bulb to steam, not the boiling water, and allow the thread to become steady before marking.
(e) Land and sea breezes. By day the land heats up faster than the sea, so the air over the land becomes warmer, rises, and is replaced by cooler air blowing in from the sea: this is the sea breeze. By night the land cools faster than the sea, so the air over the sea is now warmer and rises, and cooler air blows out from the land to the sea: this is the land breeze.
Answer Details
(a) Heat is a form of energy that flows from a body at a higher temperature to one at a lower temperature; it is measured in joules. Temperature is the degree of hotness or coldness of a body, which determines the direction of heat flow; it is measured in kelvin (or degrees Celsius). Heat depends on mass, while temperature does not.
(b) Physical properties used to measure temperature:
- Expansion of a liquid (e.g. mercury) or of a solid.
- Variation of electrical resistance of a metal with temperature (also e.m.f. of a thermocouple, or pressure/volume of a gas).
(c) Why mercury is preferred to alcohol:
- Mercury has a high boiling point (about 357 C), so it can measure high temperatures without vaporising.
- Mercury is opaque and silvery, so it is easily seen, and it does not wet (stick to) glass, giving accurate readings; it also expands uniformly and conducts heat well.
(d)(i) Calibration. The bulb of the thermometer is placed in pure melting ice; when the mercury thread is steady its position is marked as the lower fixed point (0 C). The bulb is then placed in the steam above boiling water at normal atmospheric pressure and, when steady, that position is marked as the upper fixed point (100 C). The distance between the two marks is divided into 100 equal parts (degrees).
(d)(ii) Precautions:
- Use pure melting ice and pure water so the fixed points are correct.
- For the steam point, ensure standard atmospheric pressure and expose the bulb to steam, not the boiling water, and allow the thread to become steady before marking.
(e) Land and sea breezes. By day the land heats up faster than the sea, so the air over the land becomes warmer, rises, and is replaced by cooler air blowing in from the sea: this is the sea breeze. By night the land cools faster than the sea, so the air over the sea is now warmer and rises, and cooler air blows out from the land to the sea: this is the land breeze.
Question 3 Report
(a) Explain the term uniform acceleration
(b)(i) Sketch and describe the velocity-time graph for the motion of a ball from the time it is projected vertically upwards until it returns to the point of projection.
(ii) Neglecting air resistance and using ycur sketch, explain how the acceleration of free fall due to gravity g, and the maximum height attained when the ball is projected vertically upwards can be determined.
(c) A stone is projected vertically upwards with a velocity of 20ms\(^{-1}\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\(^{-1}\). Neglecting air resistance, calculate the:
(i) length of time the second stone is in motion before they meet,
(ii) velocity of the first stone when they meet (Take g as 10ms\(^{-2}\))
(a) Uniform acceleration is motion in which the velocity changes by equal amounts in equal intervals of time. Thus, the rate of change of velocity is constant.
(b)(i) Taking upward velocity as positive, the velocity-time graph is as shown below. It is a straight line of constant negative gradient. The ball starts with velocity u, slows uniformly to zero velocity at the highest point, and then gains downward velocity uniformly until it returns to the point of projection with velocity −u.
(b)(ii) The gradient of a velocity-time graph gives acceleration:
\[\text{gradient}=\frac{\Delta v}{\Delta t}=-g.\]
Hence the acceleration due to gravity is the magnitude of the gradient:
\[g=\left|\frac{\Delta v}{\Delta t}\right|.\]
The maximum height attained is the area under the graph above the time axis, from projection to the instant the velocity becomes zero. Since this area is a triangle,
\[H=\frac{1}{2}\times u\times t_{\rm up}.\]
Since \(t_{\rm up}=u/g\),
\[H=\frac{1}{2}u\left(\frac{u}{g}\right)=\frac{u^2}{2g}.\]
The triangular area below the time axis during the downward journey has the same magnitude, showing that the ball falls through the same vertical distance before returning to the point of projection.
(c) Take upward as positive and \(g=10\,\text{m s}^{-2}\).
(i) Let \(t\) be the time for which the second stone is in motion before the stones meet. For the second stone, \(u=20\,\text{m s}^{-1}\) and \(v=10\,\text{m s}^{-1}\):
\[v=u-gt\]
\[10=20-10t\]
\[10t=10\]
\[t=1\,\text{s}.\]
Therefore, the second stone has been in motion for \(1\,\text{s}\).
(ii) The first stone has been in motion for \(2+1=3\,\text{s}\). Therefore,
\[v=u-gt=20-(10\times3)=-10\,\text{m s}^{-1}.\]
The velocity of the first stone is \(-10\,\text{m s}^{-1}\), that is, \(10\,\text{m s}^{-1}\) vertically downward.
Answer Details
(a) Uniform acceleration is motion in which the velocity changes by equal amounts in equal intervals of time. Thus, the rate of change of velocity is constant.
(b)(i) Taking upward velocity as positive, the velocity-time graph is as shown below. It is a straight line of constant negative gradient. The ball starts with velocity u, slows uniformly to zero velocity at the highest point, and then gains downward velocity uniformly until it returns to the point of projection with velocity −u.
(b)(ii) The gradient of a velocity-time graph gives acceleration:
\[\text{gradient}=\frac{\Delta v}{\Delta t}=-g.\]
Hence the acceleration due to gravity is the magnitude of the gradient:
\[g=\left|\frac{\Delta v}{\Delta t}\right|.\]
The maximum height attained is the area under the graph above the time axis, from projection to the instant the velocity becomes zero. Since this area is a triangle,
\[H=\frac{1}{2}\times u\times t_{\rm up}.\]
Since \(t_{\rm up}=u/g\),
\[H=\frac{1}{2}u\left(\frac{u}{g}\right)=\frac{u^2}{2g}.\]
The triangular area below the time axis during the downward journey has the same magnitude, showing that the ball falls through the same vertical distance before returning to the point of projection.
(c) Take upward as positive and \(g=10\,\text{m s}^{-2}\).
(i) Let \(t\) be the time for which the second stone is in motion before the stones meet. For the second stone, \(u=20\,\text{m s}^{-1}\) and \(v=10\,\text{m s}^{-1}\):
\[v=u-gt\]
\[10=20-10t\]
\[10t=10\]
\[t=1\,\text{s}.\]
Therefore, the second stone has been in motion for \(1\,\text{s}\).
(ii) The first stone has been in motion for \(2+1=3\,\text{s}\). Therefore,
\[v=u-gt=20-(10\times3)=-10\,\text{m s}^{-1}.\]
The velocity of the first stone is \(-10\,\text{m s}^{-1}\), that is, \(10\,\text{m s}^{-1}\) vertically downward.
Question 4 Report
(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.
(ii) State two precautions necessary to obtain an accurate result.
(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.
(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:
(i) current in the circuit
(ii) potential difference across the parallel combination of the resistors
(iii) lost volts of the battery.
(a)(i) Measurement of the resistance of the wire
Connect the battery, key K, rheostat, ammeter A and the test wire X in series. Connect the high-resistance voltmeter V in parallel across the test wire, as shown below.
Close the key and set the rheostat initially to give a small current. Record the current I from the ammeter and the potential difference V across the wire. Adjust the rheostat to obtain several other pairs of readings. Plot a graph of V against I. The gradient of the straight-line graph gives the resistance, since
\[V=IR\qquad\therefore\qquad R=\frac{\Delta V}{\Delta I}.\]
(a)(ii) Precautions
- Open the key when readings are not being taken, so that the wire does not heat up and change its resistance.
- Ensure that all electrical connections are clean and tight.
- Read the ammeter and voltmeter at eye level to avoid parallax error.
- Use small currents to minimise heating of the wire.
(b) Determination of resistivity
Measure the length \(L\) of the test wire with a metre rule. Measure its diameter \(d\) at several positions and in perpendicular directions using a micrometer screw gauge, and obtain the mean diameter. Its cross-sectional area is
\[A=\frac{\pi d^2}{4}.\]
If \(R\) is the resistance obtained from the gradient of the \(V\)-against-\(I\) graph, then
\[R=\frac{\rho L}{A}.\]
Hence the resistivity of the material of the wire is
\[\boxed{\rho=\frac{RA}{L}=\frac{R\pi d^2}{4L}}\]
in \(\Omega\,\text{m}\).
(c) Circuit and calculation
The two cells are in series, so
\[E=2+2=4.0\text{ V},\qquad r=0.5+0.5=1.0\ \Omega.\]
The equivalent resistance of the two \(3\ \Omega\) resistors in parallel is
\[R_p=\frac{3\times3}{3+3}=1.5\ \Omega.\]
Thus, the external resistance is
\[R_{\rm ext}=2+1.5=3.5\ \Omega,\]
and the total resistance, including internal resistance, is
\[R_T=3.5+1.0=4.5\ \Omega.\]
(i) Current in the circuit
\[I=\frac{E}{R_T}=\frac{4}{4.5}=\frac{8}{9}=\boxed{0.89\text{ A}}\]
(ii) Potential difference across the parallel combination
\[V_p=IR_p=\frac{8}{9}\times1.5=\frac{4}{3}=\boxed{1.33\text{ V}}\]
(iii) Lost volts of the battery
\[V_{\rm lost}=Ir=\frac{8}{9}\times1.0=\boxed{0.89\text{ V}}\]
Answer Details
(a)(i) Measurement of the resistance of the wire
Connect the battery, key K, rheostat, ammeter A and the test wire X in series. Connect the high-resistance voltmeter V in parallel across the test wire, as shown below.
Close the key and set the rheostat initially to give a small current. Record the current I from the ammeter and the potential difference V across the wire. Adjust the rheostat to obtain several other pairs of readings. Plot a graph of V against I. The gradient of the straight-line graph gives the resistance, since
\[V=IR\qquad\therefore\qquad R=\frac{\Delta V}{\Delta I}.\]
(a)(ii) Precautions
- Open the key when readings are not being taken, so that the wire does not heat up and change its resistance.
- Ensure that all electrical connections are clean and tight.
- Read the ammeter and voltmeter at eye level to avoid parallax error.
- Use small currents to minimise heating of the wire.
(b) Determination of resistivity
Measure the length \(L\) of the test wire with a metre rule. Measure its diameter \(d\) at several positions and in perpendicular directions using a micrometer screw gauge, and obtain the mean diameter. Its cross-sectional area is
\[A=\frac{\pi d^2}{4}.\]
If \(R\) is the resistance obtained from the gradient of the \(V\)-against-\(I\) graph, then
\[R=\frac{\rho L}{A}.\]
Hence the resistivity of the material of the wire is
\[\boxed{\rho=\frac{RA}{L}=\frac{R\pi d^2}{4L}}\]
in \(\Omega\,\text{m}\).
(c) Circuit and calculation
The two cells are in series, so
\[E=2+2=4.0\text{ V},\qquad r=0.5+0.5=1.0\ \Omega.\]
The equivalent resistance of the two \(3\ \Omega\) resistors in parallel is
\[R_p=\frac{3\times3}{3+3}=1.5\ \Omega.\]
Thus, the external resistance is
\[R_{\rm ext}=2+1.5=3.5\ \Omega,\]
and the total resistance, including internal resistance, is
\[R_T=3.5+1.0=4.5\ \Omega.\]
(i) Current in the circuit
\[I=\frac{E}{R_T}=\frac{4}{4.5}=\frac{8}{9}=\boxed{0.89\text{ A}}\]
(ii) Potential difference across the parallel combination
\[V_p=IR_p=\frac{8}{9}\times1.5=\frac{4}{3}=\boxed{1.33\text{ V}}\]
(iii) Lost volts of the battery
\[V_{\rm lost}=Ir=\frac{8}{9}\times1.0=\boxed{0.89\text{ V}}\]
