WAEC SSCE - Physics - 1989 (Essay)
Question 1 Report
(a)(i) Explain the terms: photoelectric emission and threshold frequency; (ii) Einstein's photoelectric equation can be written as \(E = hf - hf_{o}\) What does each of the symbols used in the equation above represent?
(b) Calculate the frequency of the proton whose energy is required to eject a surface electron with a kinetic energy of \(1.97 x 10^{-16} eV\) if the work function of the metal is \(1.33 x 10^{-16}eV\). \((1 eV = 1.6 x 10^{-18}J; h = 6.60 x 10^{-34}JS)\).
(c) In a photoelectric cell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the numbers of emitted electrons.
(d) Explain what is meant by the duality of matter, illustrating your answer with observation phenomena.
(a)(i) Photoelectric emission and threshold frequency
- Photoelectric emission is the ejection of electrons from the surface of a metal when electromagnetic radiation (light) of sufficiently high frequency falls on it.
- Threshold frequency is the minimum frequency of the incident radiation below which no electrons are emitted, no matter how intense the light.
(a)(ii) Symbols in \( E = hf - hf_o \)
- \( E \) is the maximum kinetic energy of the emitted (photo)electron.
- \( hf \) is the energy of the incident photon (h = Planck's constant, f = its frequency).
- \( hf_o \) is the work function of the metal (\( f_o \) is the threshold frequency).
(b) Frequency of the incident radiation
By Einstein's equation, the photon energy equals the work function plus the kinetic energy given to the electron:
\[ hf = W + KE = (1.33 \times 10^{-16}) + (1.97 \times 10^{-16}) = 3.30 \times 10^{-16}\,\text{J} \]
\[ f = \frac{hf}{h} = \frac{3.30 \times 10^{-16}}{6.60 \times 10^{-34}} = 5.0 \times 10^{17}\,\text{Hz} \]
The frequency of the incident radiation is \( 5.0 \times 10^{17}\,\text{Hz} \) (treating the given energies in joules).
(c) Below the threshold frequency, each photon does not carry enough energy to free an electron; the light energy absorbed by the surface electrons is simply re-radiated or converted into heat, so no electron gains sufficient energy to overcome the work function and escape. The number of electrons emitted (once above threshold) depends on the intensity (brightness) of the incident light.
(d) Duality of matter
The duality of matter means that matter (such as electrons) can behave both as particles and as waves. It behaves as a particle in phenomena such as the photoelectric effect and collisions, and as a wave in phenomena such as electron diffraction and interference, where a beam of electrons produces a diffraction pattern like that of light. The de Broglie relation \( \lambda = \dfrac{h}{mv} \) links the two aspects.
Answer Details
(a)(i) Photoelectric emission and threshold frequency
- Photoelectric emission is the ejection of electrons from the surface of a metal when electromagnetic radiation (light) of sufficiently high frequency falls on it.
- Threshold frequency is the minimum frequency of the incident radiation below which no electrons are emitted, no matter how intense the light.
(a)(ii) Symbols in \( E = hf - hf_o \)
- \( E \) is the maximum kinetic energy of the emitted (photo)electron.
- \( hf \) is the energy of the incident photon (h = Planck's constant, f = its frequency).
- \( hf_o \) is the work function of the metal (\( f_o \) is the threshold frequency).
(b) Frequency of the incident radiation
By Einstein's equation, the photon energy equals the work function plus the kinetic energy given to the electron:
\[ hf = W + KE = (1.33 \times 10^{-16}) + (1.97 \times 10^{-16}) = 3.30 \times 10^{-16}\,\text{J} \]
\[ f = \frac{hf}{h} = \frac{3.30 \times 10^{-16}}{6.60 \times 10^{-34}} = 5.0 \times 10^{17}\,\text{Hz} \]
The frequency of the incident radiation is \( 5.0 \times 10^{17}\,\text{Hz} \) (treating the given energies in joules).
(c) Below the threshold frequency, each photon does not carry enough energy to free an electron; the light energy absorbed by the surface electrons is simply re-radiated or converted into heat, so no electron gains sufficient energy to overcome the work function and escape. The number of electrons emitted (once above threshold) depends on the intensity (brightness) of the incident light.
(d) Duality of matter
The duality of matter means that matter (such as electrons) can behave both as particles and as waves. It behaves as a particle in phenomena such as the photoelectric effect and collisions, and as a wave in phenomena such as electron diffraction and interference, where a beam of electrons produces a diffraction pattern like that of light. The de Broglie relation \( \lambda = \dfrac{h}{mv} \) links the two aspects.
Question 2 Report
(a) What is meant by the statement: The linear expansivity of a solid is \(1.0 \times 10^{-5}K^{-1}\)?
(b)(i) Describe an experiment to determine the linear expansivity of a steel rod. (ii) Steel bars, each of length 3m at 29°C are to be used for constructing a rail line. If the linear expansivity of steel is \(1.0 \times 10^{-5}K^{-1}\), calculate the safety gap that must be left between successive bars if the highest temperature expected is 41°C.
(c) State three advantages and two disadvantages of thermal expansion of solids.
(a) Meaning of a linear expansivity of \( 1.0 \times 10^{-5}\,\text{K}^{-1} \)
It means that each unit length of the solid increases by \( 1.0 \times 10^{-5} \) of its original length for every 1 K (1 °C) rise in temperature. For example, a 1 m length would lengthen by \( 1.0 \times 10^{-5}\,\text{m} \) for each degree rise.
(b)(i) Experiment to determine the linear expansivity of a steel rod
Measure the original length \( l_0 \) of the steel rod and its initial temperature \( \theta_1 \). Place the rod in a jacket through which steam is passed, with one end fixed and the other end pressing against a micrometer/spherometer (or an optical lever). Record the increase in length \( \Delta l \) as the rod is heated to the steam temperature \( \theta_2 \). The linear expansivity is
\[ \alpha = \frac{\Delta l}{l_0(\theta_2 - \theta_1)} \]
(b)(ii) Safety gap between rails
Expansion of one bar: \( \Delta l = l_0 \, \alpha \, \Delta\theta \)
\[ \Delta l = 3 \times (1.0 \times 10^{-5}) \times (41 - 29) \]
\[ \Delta l = 3 \times 1.0 \times 10^{-5} \times 12 = 3.6 \times 10^{-4}\,\text{m} = 0.36\,\text{mm} \]
A gap of about 0.36 mm must be left between successive bars.
(c) Advantages and disadvantages of thermal expansion of solids
Three advantages:
- Fitting a metal tyre or ring onto a wheel (heat to expand, then cools and grips tightly).
- Hot riveting of metal plates (rivet contracts on cooling and clamps the plates firmly).
- Use in bimetallic strips for thermostats, thermometers and fire alarms.
Two disadvantages:
- Railway lines and bridges may buckle or bend if no expansion gaps are provided.
- Overhead telephone/electric cables sag in hot weather and may snap when they contract in cold weather; tight-fitting parts may jam or crack.
Answer Details
(a) Meaning of a linear expansivity of \( 1.0 \times 10^{-5}\,\text{K}^{-1} \)
It means that each unit length of the solid increases by \( 1.0 \times 10^{-5} \) of its original length for every 1 K (1 °C) rise in temperature. For example, a 1 m length would lengthen by \( 1.0 \times 10^{-5}\,\text{m} \) for each degree rise.
(b)(i) Experiment to determine the linear expansivity of a steel rod
Measure the original length \( l_0 \) of the steel rod and its initial temperature \( \theta_1 \). Place the rod in a jacket through which steam is passed, with one end fixed and the other end pressing against a micrometer/spherometer (or an optical lever). Record the increase in length \( \Delta l \) as the rod is heated to the steam temperature \( \theta_2 \). The linear expansivity is
\[ \alpha = \frac{\Delta l}{l_0(\theta_2 - \theta_1)} \]
(b)(ii) Safety gap between rails
Expansion of one bar: \( \Delta l = l_0 \, \alpha \, \Delta\theta \)
\[ \Delta l = 3 \times (1.0 \times 10^{-5}) \times (41 - 29) \]
\[ \Delta l = 3 \times 1.0 \times 10^{-5} \times 12 = 3.6 \times 10^{-4}\,\text{m} = 0.36\,\text{mm} \]
A gap of about 0.36 mm must be left between successive bars.
(c) Advantages and disadvantages of thermal expansion of solids
Three advantages:
- Fitting a metal tyre or ring onto a wheel (heat to expand, then cools and grips tightly).
- Hot riveting of metal plates (rivet contracts on cooling and clamps the plates firmly).
- Use in bimetallic strips for thermostats, thermometers and fire alarms.
Two disadvantages:
- Railway lines and bridges may buckle or bend if no expansion gaps are provided.
- Overhead telephone/electric cables sag in hot weather and may snap when they contract in cold weather; tight-fitting parts may jam or crack.
Question 3 Report
(a) What is a wave motion?
The equation \(y = A \sin \frac{2\pi}{\lambda}(Vt-X)\) represents a wavetrain in which y is the vertical displacement of a particle at distance X from the origin in the medium through which the wave is travelling. Explain, with the aid of a diagram, what A and \(\lambda\) represent.
(b) (i) Describe an experiment to determine the frequency of a note emitted by a source of sound
(ii) A pipe closed at one end is 1 m long. The air in the pipe is set into vibration and a fundamental note is produced. If the velocity of sound in air is 340ms\(^{-1}\), calculate the frequency of the note
(c) State two differences between a sound wave and a radio wave.
(a)
Wave motion is the propagation of a disturbance from one point to another, transmitting energy without any net transfer of matter.
For the wave equation
\[y=A\sin\left[\frac{2\pi}{\lambda}(Vt-X)\right],\]
0 is the amplitude, that is, the maximum displacement of a particle of the medium from its equilibrium position.
\(\lambda\) is the wavelength, that is, the distance between two successive points vibrating in the same phase, such as two successive crests.
(b)(i)
Set up a resonance tube containing water and a movable vertical tube. Strike the sound source, such as a tuning fork, gently and hold it just above the open end of the tube. Raise the tube slowly so that the length \(l\) of the enclosed air column increases.
The sound becomes loudest at resonance. At the first resonance, the air column has a node at the water surface and an antinode at the open end; hence, neglecting end correction,
\[l=\frac{\lambda}{4}.\]
Therefore,
\[\lambda=4l\]
and, if \(v\) is the velocity of sound in air, the frequency \(f\) of the source is
\[f=\frac{v}{\lambda}=\frac{v}{4l}.\]
Alternatively, locate two successive resonance lengths \(l_1\) and \(l_2\). Then \(l_2-l_1=\lambda/2\), so that
\[f=\frac{v}{2(l_2-l_1)}.\]
(b)(ii)
For the fundamental note of a pipe closed at one end,
\[L=\frac{\lambda}{4}.\]
Thus,
\[\lambda=4L=4(1)=4\ \text{m}.\]
\[f=\frac{v}{\lambda}=\frac{340}{4}=85\ \text{Hz}.\]
Frequency of the note = \(85\ \text{Hz}\).
(c)
| Sound wave | Radio wave |
|---|---|
| It is a mechanical wave and requires a material medium for propagation. | It is an electromagnetic wave and can travel through a vacuum. |
| It is longitudinal in air. | It is transverse. |
Answer Details
(a)
Wave motion is the propagation of a disturbance from one point to another, transmitting energy without any net transfer of matter.
For the wave equation
\[y=A\sin\left[\frac{2\pi}{\lambda}(Vt-X)\right],\]
0 is the amplitude, that is, the maximum displacement of a particle of the medium from its equilibrium position.
\(\lambda\) is the wavelength, that is, the distance between two successive points vibrating in the same phase, such as two successive crests.
(b)(i)
Set up a resonance tube containing water and a movable vertical tube. Strike the sound source, such as a tuning fork, gently and hold it just above the open end of the tube. Raise the tube slowly so that the length \(l\) of the enclosed air column increases.
The sound becomes loudest at resonance. At the first resonance, the air column has a node at the water surface and an antinode at the open end; hence, neglecting end correction,
\[l=\frac{\lambda}{4}.\]
Therefore,
\[\lambda=4l\]
and, if \(v\) is the velocity of sound in air, the frequency \(f\) of the source is
\[f=\frac{v}{\lambda}=\frac{v}{4l}.\]
Alternatively, locate two successive resonance lengths \(l_1\) and \(l_2\). Then \(l_2-l_1=\lambda/2\), so that
\[f=\frac{v}{2(l_2-l_1)}.\]
(b)(ii)
For the fundamental note of a pipe closed at one end,
\[L=\frac{\lambda}{4}.\]
Thus,
\[\lambda=4L=4(1)=4\ \text{m}.\]
\[f=\frac{v}{\lambda}=\frac{340}{4}=85\ \text{Hz}.\]
Frequency of the note = \(85\ \text{Hz}\).
(c)
| Sound wave | Radio wave |
|---|---|
| It is a mechanical wave and requires a material medium for propagation. | It is an electromagnetic wave and can travel through a vacuum. |
| It is longitudinal in air. | It is transverse. |
Question 4 Report
(a) State the laws of electromagnetic induction.
(b) (i) Describe a simple experiment to show how an induced e.m.f, can be produced; (ii) State two factors on which the magnitude of the induced e.m.f. depends
(c) Explain what is meant by the r.m.s. value of an alternating current
(d) (i) If the alternating current is represented by \(I = l_{o} \sin \omega t\), state what the symbol \(I, I_{o}, \omega\) and \(\omega t\)represent.
(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.
(a) Laws of electromagnetic induction
- Faraday's law: whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linkage.
- Lenz's law: the direction of the induced current is always such that it opposes the change producing it.
(b)(i) Simple experiment
Connect a coil of wire to a sensitive galvanometer. When a bar magnet is pushed into (or pulled out of) the coil, the galvanometer deflects, showing that an e.m.f. (and current) is induced. When the magnet is held still there is no deflection.
(b)(ii) Two factors on which the induced e.m.f. depends
- The rate at which the magnetic flux changes (the speed of the magnet).
- The number of turns on the coil (and the strength of the magnet).
(c) r.m.s. value of an alternating current
The root-mean-square value of an alternating current is the value of the steady direct current that would dissipate heat in a given resistor at the same rate as the alternating current does.
(d)(i) Meaning of the symbols in \( I = I_o \sin \omega t \)
- \( I \) is the instantaneous value of the current at time t.
- \( I_o \) is the peak (maximum) value of the current.
- \( \omega \) is the angular frequency, \( \omega = 2\pi f \).
- \( \omega t \) is the phase angle of the current at time t.
(d)(ii) Instantaneous value
Peak value: \( I_o = I_{rms}\sqrt{2} = 15.0 \times 1.414 = 21.2\,\text{A} \).
At phase angle \( \omega t = 30^\circ \):
\[ I = I_o \sin 30^\circ = 21.2 \times 0.5 = 10.6\,\text{A} \]
The instantaneous current is about 10.6 A.
Answer Details
(a) Laws of electromagnetic induction
- Faraday's law: whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linkage.
- Lenz's law: the direction of the induced current is always such that it opposes the change producing it.
(b)(i) Simple experiment
Connect a coil of wire to a sensitive galvanometer. When a bar magnet is pushed into (or pulled out of) the coil, the galvanometer deflects, showing that an e.m.f. (and current) is induced. When the magnet is held still there is no deflection.
(b)(ii) Two factors on which the induced e.m.f. depends
- The rate at which the magnetic flux changes (the speed of the magnet).
- The number of turns on the coil (and the strength of the magnet).
(c) r.m.s. value of an alternating current
The root-mean-square value of an alternating current is the value of the steady direct current that would dissipate heat in a given resistor at the same rate as the alternating current does.
(d)(i) Meaning of the symbols in \( I = I_o \sin \omega t \)
- \( I \) is the instantaneous value of the current at time t.
- \( I_o \) is the peak (maximum) value of the current.
- \( \omega \) is the angular frequency, \( \omega = 2\pi f \).
- \( \omega t \) is the phase angle of the current at time t.
(d)(ii) Instantaneous value
Peak value: \( I_o = I_{rms}\sqrt{2} = 15.0 \times 1.414 = 21.2\,\text{A} \).
At phase angle \( \omega t = 30^\circ \):
\[ I = I_o \sin 30^\circ = 21.2 \times 0.5 = 10.6\,\text{A} \]
The instantaneous current is about 10.6 A.
