WAEC SSCE - Physics - 2014 (Essay)
Tambaya 1 Rahoto
(a) Define electromotive force.
(b) State:
(i) the principle of operation of a potentiometer,
(ii) two advantages that a potentiometer has over a voltmeter in measuring potential difference.
(c)(i) Sketch and label a diagram of a gold-leaf electroscope.
(ii) Give one use of a gold-leaf electroscope.
(d)(i) Explain the action of a magnetic relay.
(ii) List two factors which determine the magnitude of an induced emf in a coil.
(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x10\(^{-3}\) T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.
(a) Electromotive force (emf)
The electromotive force of a source is the total energy supplied by the source in driving one coulomb of charge round the complete circuit; that is, the work done per unit charge by the source (equal to the terminal potential difference of the source on open circuit). Its SI unit is the volt (V).
(b) Potentiometer
(i) Principle of operation: A steady current is passed through a uniform resistance wire, so the potential difference across a portion of the wire is directly proportional to its length, giving a uniform potential gradient. An unknown emf or p.d. is balanced against the p.d. across a length of the wire; at balance no current flows through the galvanometer (null method), so \( V \propto L \), where \(V\) is the p.d. and \(L\) is the balance length.
(ii) Two advantages of a potentiometer over a voltmeter:
- At balance it draws no current from the cell being measured, so it measures the true emf and eliminates error due to the internal resistance of the cell (a voltmeter draws current and reads less than the emf).
- It is more sensitive and more accurate, since it uses a null method and needs no scale calibration that could introduce zero or scale errors.
(c)(i) Labelled sketch of a gold-leaf electroscope:
The metal cap collects charge and passes it down the metal rod to the metal plate and the gold leaf. Because the plate and leaf then carry like charges, the leaf is repelled and diverges from the plate. The rod is held by an insulating plug in the top of an earthed metal case that has glass windows for viewing.
(ii) Use of a gold-leaf electroscope: to detect the presence of electric charge on a body (it can also be used to test the sign of a charge and to compare the magnitudes of charges).
(d)(i) Action of a magnetic relay: A small current in the coil (the control circuit) magnetises a soft-iron core, turning it into an electromagnet. The electromagnet attracts a pivoted soft-iron armature, and the movement of the armature closes (or opens) a pair of contacts in a separate circuit. In this way a small current in the first circuit is used to switch on or off a much larger current in the second circuit.
(ii) Two factors which determine the magnitude of an induced emf in a coil:
- The rate of change of magnetic flux linkage (the speed at which the flux through the coil changes).
- The number of turns in the coil.
(iii) Force per unit length on the wire
Given: current \(I = 5\ \text{A}\), flux density \(B = 2.0\times10^{-3}\ \text{T}\), angle to the field \(\theta = 30^{\circ}\).
The force on a current-carrying conductor is \(F = BIL\sin\theta\), so the force per unit length is
\[ \frac{F}{L} = BI\sin\theta = (2.0\times10^{-3})\times 5 \times \sin 30^{\circ}. \] \[ \frac{F}{L} = (2.0\times10^{-3})\times 5 \times 0.5 = 5.0\times10^{-3}\ \text{N m}^{-1}. \]The force per unit length exerted on the wire is \(5.0\times10^{-3}\ \text{N m}^{-1}\).
Bayanin Amsa
(a) Electromotive force (emf)
The electromotive force of a source is the total energy supplied by the source in driving one coulomb of charge round the complete circuit; that is, the work done per unit charge by the source (equal to the terminal potential difference of the source on open circuit). Its SI unit is the volt (V).
(b) Potentiometer
(i) Principle of operation: A steady current is passed through a uniform resistance wire, so the potential difference across a portion of the wire is directly proportional to its length, giving a uniform potential gradient. An unknown emf or p.d. is balanced against the p.d. across a length of the wire; at balance no current flows through the galvanometer (null method), so \( V \propto L \), where \(V\) is the p.d. and \(L\) is the balance length.
(ii) Two advantages of a potentiometer over a voltmeter:
- At balance it draws no current from the cell being measured, so it measures the true emf and eliminates error due to the internal resistance of the cell (a voltmeter draws current and reads less than the emf).
- It is more sensitive and more accurate, since it uses a null method and needs no scale calibration that could introduce zero or scale errors.
(c)(i) Labelled sketch of a gold-leaf electroscope:
The metal cap collects charge and passes it down the metal rod to the metal plate and the gold leaf. Because the plate and leaf then carry like charges, the leaf is repelled and diverges from the plate. The rod is held by an insulating plug in the top of an earthed metal case that has glass windows for viewing.
(ii) Use of a gold-leaf electroscope: to detect the presence of electric charge on a body (it can also be used to test the sign of a charge and to compare the magnitudes of charges).
(d)(i) Action of a magnetic relay: A small current in the coil (the control circuit) magnetises a soft-iron core, turning it into an electromagnet. The electromagnet attracts a pivoted soft-iron armature, and the movement of the armature closes (or opens) a pair of contacts in a separate circuit. In this way a small current in the first circuit is used to switch on or off a much larger current in the second circuit.
(ii) Two factors which determine the magnitude of an induced emf in a coil:
- The rate of change of magnetic flux linkage (the speed at which the flux through the coil changes).
- The number of turns in the coil.
(iii) Force per unit length on the wire
Given: current \(I = 5\ \text{A}\), flux density \(B = 2.0\times10^{-3}\ \text{T}\), angle to the field \(\theta = 30^{\circ}\).
The force on a current-carrying conductor is \(F = BIL\sin\theta\), so the force per unit length is
\[ \frac{F}{L} = BI\sin\theta = (2.0\times10^{-3})\times 5 \times \sin 30^{\circ}. \] \[ \frac{F}{L} = (2.0\times10^{-3})\times 5 \times 0.5 = 5.0\times10^{-3}\ \text{N m}^{-1}. \]The force per unit length exerted on the wire is \(5.0\times10^{-3}\ \text{N m}^{-1}\).
Tambaya 2 Rahoto
An electron moves with a speed of 2.00 x 10\(^7\) ms\(^{-1}\) in an orbit in a uniform magnetic field of 1.20 x 10\(^{-3}\) T. Calculate the radius of the orbit. [Mass of an electron = 9.11 x 10\(^{-3}\) kg; charge on an electron = 1.61 x 10\(^{-19}\)C]
Principle. An electron moving at speed v perpendicular to a magnetic field B experiences a magnetic force \(qvB\) that provides the centripetal force for its circular orbit:
\[ qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} \]Substituting \(v = 2.00\times10^{7}\ \text{m/s}\), \(B = 1.20\times10^{-3}\ \text{T}\), \(q = 1.6\times10^{-19}\ \text{C}\), and the electron mass \(m = 9.11\times10^{-31}\ \text{kg}\) (the printed \(10^{-3}\) is a misprint for \(10^{-31}\)):
\[ r = \frac{9.11\times10^{-31}\times2.00\times10^{7}}{1.6\times10^{-19}\times1.20\times10^{-3}} \] \[ r = \frac{1.822\times10^{-23}}{1.92\times10^{-22}} \approx 0.095\ \text{m} \]The radius of the orbit is about 0.095 m (9.5 cm).
Bayanin Amsa
Principle. An electron moving at speed v perpendicular to a magnetic field B experiences a magnetic force \(qvB\) that provides the centripetal force for its circular orbit:
\[ qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} \]Substituting \(v = 2.00\times10^{7}\ \text{m/s}\), \(B = 1.20\times10^{-3}\ \text{T}\), \(q = 1.6\times10^{-19}\ \text{C}\), and the electron mass \(m = 9.11\times10^{-31}\ \text{kg}\) (the printed \(10^{-3}\) is a misprint for \(10^{-31}\)):
\[ r = \frac{9.11\times10^{-31}\times2.00\times10^{7}}{1.6\times10^{-19}\times1.20\times10^{-3}} \] \[ r = \frac{1.822\times10^{-23}}{1.92\times10^{-22}} \approx 0.095\ \text{m} \]The radius of the orbit is about 0.095 m (9.5 cm).
Tambaya 3 Rahoto
A tennis ball projected at an angle 0 attains a range R = 78. If the velocity imparted to the ball by the racket is 30 ms\(^{-1}\), calculate O. [ g = 10 ms\(^{-2}\)]
Using the range formula. For a projectile launched with speed u at angle \(\theta\), the horizontal range is
\[ R = \frac{u^{2}\sin2\theta}{g} \]With \(R = 78\ \text{m}\), \(u = 30\ \text{m/s}\), \(g = 10\ \text{m/s}^{2}\):
\[ 78 = \frac{30^{2}\sin2\theta}{10} = \frac{900\sin2\theta}{10} = 90\sin2\theta \] \[ \sin2\theta = \frac{78}{90} = 0.867 \] \[ 2\theta = \sin^{-1}(0.867) = 60^{\circ} \quad(\text{or } 120^{\circ}) \] \[ \theta = 30^{\circ} \quad(\text{or } 60^{\circ}) \]The ball is projected at \(\theta \approx \mathbf{30^{\circ}}\) (the complementary angle \(60^{\circ}\) gives the same range).
Bayanin Amsa
Using the range formula. For a projectile launched with speed u at angle \(\theta\), the horizontal range is
\[ R = \frac{u^{2}\sin2\theta}{g} \]With \(R = 78\ \text{m}\), \(u = 30\ \text{m/s}\), \(g = 10\ \text{m/s}^{2}\):
\[ 78 = \frac{30^{2}\sin2\theta}{10} = \frac{900\sin2\theta}{10} = 90\sin2\theta \] \[ \sin2\theta = \frac{78}{90} = 0.867 \] \[ 2\theta = \sin^{-1}(0.867) = 60^{\circ} \quad(\text{or } 120^{\circ}) \] \[ \theta = 30^{\circ} \quad(\text{or } 60^{\circ}) \]The ball is projected at \(\theta \approx \mathbf{30^{\circ}}\) (the complementary angle \(60^{\circ}\) gives the same range).
Tambaya 4 Rahoto
(a) State two conditions under which photo-electrons can be emitted from the surface of a metal.
(b) List two particle characteristics of electromagnetic waves.
(a) Conditions for photo-electron emission
- The frequency of the incident radiation must be equal to or greater than the threshold frequency of the metal (equivalently, the photon energy \(hf\) must be at least equal to the work function of the metal).
- The radiation must actually fall on (illuminate) the metal surface, and the surface must be a metal capable of emission (a clean, suitable emitting surface).
(b) Two particle characteristics of electromagnetic waves
- They are emitted and absorbed in discrete packets (quanta) called photons, each carrying energy \(E = hf\).
- Each photon carries momentum \(p = \dfrac{hf}{c}\) and can transfer energy to a single electron in a one-to-one collision (as in the photoelectric and Compton effects).
Bayanin Amsa
(a) Conditions for photo-electron emission
- The frequency of the incident radiation must be equal to or greater than the threshold frequency of the metal (equivalently, the photon energy \(hf\) must be at least equal to the work function of the metal).
- The radiation must actually fall on (illuminate) the metal surface, and the surface must be a metal capable of emission (a clean, suitable emitting surface).
(b) Two particle characteristics of electromagnetic waves
- They are emitted and absorbed in discrete packets (quanta) called photons, each carrying energy \(E = hf\).
- Each photon carries momentum \(p = \dfrac{hf}{c}\) and can transfer energy to a single electron in a one-to-one collision (as in the photoelectric and Compton effects).
Tambaya 5 Rahoto
(a) State the three characteristics of sound and the factor on which each of them depends.
(b) Explain resonance as applied to sound.
(c) What role does echo play in the construction of a concert hall?
(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10\(^{-7}\) in as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10\(^{-5}\) NM\(^{-2}\), calculate the:
(i) period of oscillation;
(ii) velocity;
(iii) acceleration;
(iv) force. [\(\pi\) = 3.14 ].
(a) Three characteristics of sound and their dependence
- Pitch depends on the frequency of the sound wave.
- Loudness depends on the amplitude (intensity) of the sound wave.
- Quality (timbre) depends on the number and relative amplitudes of the overtones (harmonics) present, i.e. the waveform.
(b) Resonance
Resonance in sound occurs when a body is set into vibration with large amplitude because it is acted on by a periodic force whose frequency equals the body's own natural frequency of vibration. At this matching frequency the body absorbs energy most effectively and vibrates strongly.
(c) Role of echo in concert-hall construction
Echoes (reflections of sound) must be controlled so that reflected sound does not arrive so late as to blur speech and music (excessive reverberation) nor die away too quickly. Sound-absorbing materials and suitably shaped, angled surfaces are used so that reverberation time is correct and the sound is spread evenly, giving good acoustics throughout the hall.
(d) The ear drum
Data: \(r = 2.1\text{ mm} = 2.1\times10^{-3}\text{ m}\), amplitude \(A = 0.8\times10^{-7}\text{ m}\), \(f = 2400\text{ Hz}\), pressure change \(p = 3.6\times10^{-5}\ \text{N m}^{-2}\), \(\pi = 3.14\).
(i) Period: \(T = \dfrac{1}{f} = \dfrac{1}{2400} = 4.17\times10^{-4}\ \text{s}\).
(ii) Maximum velocity: \(v = \omega A = 2\pi f A = 2\times3.14\times2400\times0.8\times10^{-7}\).
\(v = 15072 \times 0.8\times10^{-7} = 1.21\times10^{-3}\ \text{m s}^{-1}\).
(iii) Maximum acceleration: \(a = \omega^{2}A = (2\pi f)^{2}A = (15072)^{2}\times0.8\times10^{-7}\).
\(a = 2.27\times10^{8}\times0.8\times10^{-7} = 18.2\ \text{m s}^{-2}\).
(iv) Force: area of ear drum \(A_{d} = \pi r^{2} = 3.14\times(2.1\times10^{-3})^{2} = 1.39\times10^{-5}\ \text{m}^{2}\).
\(F = p \times A_{d} = 3.6\times10^{-5}\times1.39\times10^{-5} = 5.0\times10^{-10}\ \text{N}\).
Bayanin Amsa
(a) Three characteristics of sound and their dependence
- Pitch depends on the frequency of the sound wave.
- Loudness depends on the amplitude (intensity) of the sound wave.
- Quality (timbre) depends on the number and relative amplitudes of the overtones (harmonics) present, i.e. the waveform.
(b) Resonance
Resonance in sound occurs when a body is set into vibration with large amplitude because it is acted on by a periodic force whose frequency equals the body's own natural frequency of vibration. At this matching frequency the body absorbs energy most effectively and vibrates strongly.
(c) Role of echo in concert-hall construction
Echoes (reflections of sound) must be controlled so that reflected sound does not arrive so late as to blur speech and music (excessive reverberation) nor die away too quickly. Sound-absorbing materials and suitably shaped, angled surfaces are used so that reverberation time is correct and the sound is spread evenly, giving good acoustics throughout the hall.
(d) The ear drum
Data: \(r = 2.1\text{ mm} = 2.1\times10^{-3}\text{ m}\), amplitude \(A = 0.8\times10^{-7}\text{ m}\), \(f = 2400\text{ Hz}\), pressure change \(p = 3.6\times10^{-5}\ \text{N m}^{-2}\), \(\pi = 3.14\).
(i) Period: \(T = \dfrac{1}{f} = \dfrac{1}{2400} = 4.17\times10^{-4}\ \text{s}\).
(ii) Maximum velocity: \(v = \omega A = 2\pi f A = 2\times3.14\times2400\times0.8\times10^{-7}\).
\(v = 15072 \times 0.8\times10^{-7} = 1.21\times10^{-3}\ \text{m s}^{-1}\).
(iii) Maximum acceleration: \(a = \omega^{2}A = (2\pi f)^{2}A = (15072)^{2}\times0.8\times10^{-7}\).
\(a = 2.27\times10^{8}\times0.8\times10^{-7} = 18.2\ \text{m s}^{-2}\).
(iv) Force: area of ear drum \(A_{d} = \pi r^{2} = 3.14\times(2.1\times10^{-3})^{2} = 1.39\times10^{-5}\ \text{m}^{2}\).
\(F = p \times A_{d} = 3.6\times10^{-5}\times1.39\times10^{-5} = 5.0\times10^{-10}\ \text{N}\).
Tambaya 6 Rahoto
a) Define heat capacity and state its unit.
(b) List two effects of heat on a substance.
(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.
(d) A quantity of pepper soup of mass 800 g poured into a plastic container with a tight-fitting lid has a temperature of 30°C. The container is then placed in a microwave oven, rated 1200 W and operated for 3 minutes.
(i) Calculate the final temperature attained by the soup. (Assuming no heat losses).
(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.
(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain. [Take specific heat capacity of the soup = 4000 Jkg\(^{-1}\) K\(^{-1}\)]
(a) Heat capacity
The heat capacity of a body is the quantity of heat required to raise the temperature of the whole body by one kelvin (or one degree Celsius). Its SI unit is the joule per kelvin (J K\(^{-1}\)).
(b) Two effects of heat on a substance
- It causes a rise in temperature.
- It causes expansion (increase in size) of the substance.
(Other acceptable effects: change of state, e.g. melting or boiling; change in physical properties such as electrical resistance; chemical change.)
(c) Removing a tightly fitted glass stopper
Warm the neck of the bottle gently (for example with warm water or by rubbing). The glass of the neck, being on the outside, heats up and expands first (before the stopper), so the neck widens slightly. The stopper, still at its original size, becomes loose and can be eased out.
(d) The microwave soup
Data: \(m = 800\text{ g} = 0.8\text{ kg}\), \(\theta_1 = 30^{\circ}\text{C}\), \(P = 1200\text{ W}\), \(t = 3\text{ min} = 180\text{ s}\), \(c = 4000\ \text{J kg}^{-1}\text{K}^{-1}\).
(i) Final temperature
Heat supplied: \(Q = Pt = 1200 \times 180 = 216000\ \text{J}\).
Temperature rise: \(\Delta\theta = \dfrac{Q}{mc} = \dfrac{216000}{0.8 \times 4000} = \dfrac{216000}{3200} = 67.5\ \text{K}\).
\(\theta_2 = 30 + 67.5 = \mathbf{97.5\ ^{\circ}C}\).
(ii) As the soup heats up, water in it turns to steam and the trapped air expands, so the pressure inside a tightly sealed container builds up greatly. Since the lid is tight-fitting, the gas cannot escape; the pressure can burst the container or blow off the lid violently. Hence tight-fitting lids are unsafe in microwave cooking.
(iii) On cooling, the hot vapour inside condenses and the trapped gas contracts, so the pressure inside the sealed container falls below the atmospheric pressure outside. The greater external atmospheric pressure then pushes the walls of the plastic container inward, producing the observed dent.
Bayanin Amsa
(a) Heat capacity
The heat capacity of a body is the quantity of heat required to raise the temperature of the whole body by one kelvin (or one degree Celsius). Its SI unit is the joule per kelvin (J K\(^{-1}\)).
(b) Two effects of heat on a substance
- It causes a rise in temperature.
- It causes expansion (increase in size) of the substance.
(Other acceptable effects: change of state, e.g. melting or boiling; change in physical properties such as electrical resistance; chemical change.)
(c) Removing a tightly fitted glass stopper
Warm the neck of the bottle gently (for example with warm water or by rubbing). The glass of the neck, being on the outside, heats up and expands first (before the stopper), so the neck widens slightly. The stopper, still at its original size, becomes loose and can be eased out.
(d) The microwave soup
Data: \(m = 800\text{ g} = 0.8\text{ kg}\), \(\theta_1 = 30^{\circ}\text{C}\), \(P = 1200\text{ W}\), \(t = 3\text{ min} = 180\text{ s}\), \(c = 4000\ \text{J kg}^{-1}\text{K}^{-1}\).
(i) Final temperature
Heat supplied: \(Q = Pt = 1200 \times 180 = 216000\ \text{J}\).
Temperature rise: \(\Delta\theta = \dfrac{Q}{mc} = \dfrac{216000}{0.8 \times 4000} = \dfrac{216000}{3200} = 67.5\ \text{K}\).
\(\theta_2 = 30 + 67.5 = \mathbf{97.5\ ^{\circ}C}\).
(ii) As the soup heats up, water in it turns to steam and the trapped air expands, so the pressure inside a tightly sealed container builds up greatly. Since the lid is tight-fitting, the gas cannot escape; the pressure can burst the container or blow off the lid violently. Hence tight-fitting lids are unsafe in microwave cooking.
(iii) On cooling, the hot vapour inside condenses and the trapped gas contracts, so the pressure inside the sealed container falls below the atmospheric pressure outside. The greater external atmospheric pressure then pushes the walls of the plastic container inward, producing the observed dent.
Tambaya 7 Rahoto
(a) Write Einstein's photoelectric equation and identh: each component of the equation.
(b) For a photocell, star; one factor each that is responsible fo: the:
(i) emission (ii) rate of emission;
(iii) energy of photoelectrons.
(c)(i) Two nuclear equations are given below:
\(^{222}_{p}RN\) \(\to\) \(^{218}_{84}PO + ^q_2He\)...................A
\(^{214}_{83}RN\) \(\to\) \(^{214}_{84}PO + ^m_nX\)...................B
Determine the values of: (\(\alpha\)) p and q in equation A; (\(\beta\)) in and n in equation B and identify X.
(ii) Give a reason why it is important to dispose o radioactive waste safely.
(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\(^{-7}\)m. Calculate the energy of the photon:
----------------- - 6.0 x 10\(^{-19}\)J
---------------- - 8.2 x 10\(^{-19}\)J
---------------- - 8.8 x 10\(^{-19}\)J
---------------- - 16.7 x 10\(^{-19}\)J
(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\(^{-34}\) Js; c = 3.0 x 108 ms\(^{-1}\)].
(a) Einstein's photoelectric equation
\[ hf = W_{0} + \tfrac{1}{2}mv_{\max}^{2} \]
- \(hf\) = energy of the incident photon (\(h\) = Planck constant, \(f\) = frequency of the incident light).
- \(W_{0}\) = work function of the metal (minimum energy to free an electron from the surface).
- \(\tfrac{1}{2}mv_{\max}^{2}\) = maximum kinetic energy of the emitted photoelectron.
(b) Factors for a photocell
- (i) Emission: the frequency of the incident light must be at or above the threshold frequency (photon energy at least equal to the work function).
- (ii) Rate of emission: the intensity (brightness) of the incident light (number of photons arriving per second).
- (iii) Energy of photoelectrons: the frequency of the incident light (and the work function of the metal).
(c)(i) Nuclear equations
Equation A: \(^{222}_{p}\text{Rn} \to\ ^{218}_{84}\text{Po} +\ ^{q}_{2}\text{He}\). Balancing charge: \(p = 84 + 2 = 86\). Balancing mass: \(q = 222 - 218 = 4\). So \(p = 86\), \(q = 4\) (an alpha particle is emitted).
Equation B: \(^{214}_{83}\text{Bi} \to\ ^{214}_{84}\text{Po} +\ ^{m}_{n}\text{X}\). Balancing mass: \(m = 214 - 214 = 0\). Balancing charge: \(n = 83 - 84 = -1\). So \(m = 0\), \(n = -1\), and X is a beta particle (electron), \(^{0}_{-1}e\).
(ii) Radioactive waste emits ionising radiation that is harmful to living cells (it can cause cancer, genetic damage and death) and can persist for a long time, so it must be disposed of safely to protect people and the environment from radiation exposure.
(d)(i) Energy of the photon
\(E = \dfrac{hc}{\lambda} = \dfrac{6.6\times10^{-34}\times3.0\times10^{8}}{2.4\times10^{-7}} = \dfrac{19.8\times10^{-26}}{2.4\times10^{-7}} = 8.25\times10^{-19}\ \text{J}\).
This corresponds to the option \(8.2\times10^{-19}\ \text{J}\).
(ii) On the energy-level diagram, the photon of energy \(8.2\times10^{-19}\ \text{J}\) is emitted when the atom makes a downward transition (electron falling) between the two levels whose energy difference equals \(8.2\times10^{-19}\ \text{J}\); the arrow is drawn pointing downwards from the higher to the lower of those two levels. The exact levels can only be marked on the actual figure provided.
Bayanin Amsa
(a) Einstein's photoelectric equation
\[ hf = W_{0} + \tfrac{1}{2}mv_{\max}^{2} \]
- \(hf\) = energy of the incident photon (\(h\) = Planck constant, \(f\) = frequency of the incident light).
- \(W_{0}\) = work function of the metal (minimum energy to free an electron from the surface).
- \(\tfrac{1}{2}mv_{\max}^{2}\) = maximum kinetic energy of the emitted photoelectron.
(b) Factors for a photocell
- (i) Emission: the frequency of the incident light must be at or above the threshold frequency (photon energy at least equal to the work function).
- (ii) Rate of emission: the intensity (brightness) of the incident light (number of photons arriving per second).
- (iii) Energy of photoelectrons: the frequency of the incident light (and the work function of the metal).
(c)(i) Nuclear equations
Equation A: \(^{222}_{p}\text{Rn} \to\ ^{218}_{84}\text{Po} +\ ^{q}_{2}\text{He}\). Balancing charge: \(p = 84 + 2 = 86\). Balancing mass: \(q = 222 - 218 = 4\). So \(p = 86\), \(q = 4\) (an alpha particle is emitted).
Equation B: \(^{214}_{83}\text{Bi} \to\ ^{214}_{84}\text{Po} +\ ^{m}_{n}\text{X}\). Balancing mass: \(m = 214 - 214 = 0\). Balancing charge: \(n = 83 - 84 = -1\). So \(m = 0\), \(n = -1\), and X is a beta particle (electron), \(^{0}_{-1}e\).
(ii) Radioactive waste emits ionising radiation that is harmful to living cells (it can cause cancer, genetic damage and death) and can persist for a long time, so it must be disposed of safely to protect people and the environment from radiation exposure.
(d)(i) Energy of the photon
\(E = \dfrac{hc}{\lambda} = \dfrac{6.6\times10^{-34}\times3.0\times10^{8}}{2.4\times10^{-7}} = \dfrac{19.8\times10^{-26}}{2.4\times10^{-7}} = 8.25\times10^{-19}\ \text{J}\).
This corresponds to the option \(8.2\times10^{-19}\ \text{J}\).
(ii) On the energy-level diagram, the photon of energy \(8.2\times10^{-19}\ \text{J}\) is emitted when the atom makes a downward transition (electron falling) between the two levels whose energy difference equals \(8.2\times10^{-19}\ \text{J}\); the arrow is drawn pointing downwards from the higher to the lower of those two levels. The exact levels can only be marked on the actual figure provided.
Tambaya 8 Rahoto
The diagram above represents the graph of electron energy against the frequency of the radiation incident on a metal surface. Interpret the: (a) slope of the graph; (b) intercept, OC; (c) intercept, OK.
The straight-line graph is a plot of the maximum kinetic energy \(E_k\) of the emitted electrons (vertical axis) against the frequency \(f\) of the radiation falling on the metal surface (horizontal axis). It obeys Einstein's photoelectric equation:
\[E_k = hf - W_0 = hf - hf_0,\]where \(h\) is Planck's constant, \(W_0\) is the work function of the metal and \(f_0\) is the threshold frequency. Comparing this with the equation of a straight line \(y = mx + c\) lets each feature of the graph be interpreted directly, as annotated below.
(a) Slope of the graph
The line has the form \(E_k = hf - hf_0\). Matching term by term with \(y = mx + c\), the gradient \(m\) corresponds to \(h\). Therefore
\[\text{slope} = \frac{\Delta E_k}{\Delta f} = h \approx 6.6 \times 10^{-34}\ \text{J s}.\]The slope of the graph represents Planck's constant \(h\). Because \(h\) is a universal constant, the graph has the same gradient for every metal.
(b) Intercept, OC
Point C is where the line, produced backwards, cuts the energy axis at \(f = 0\). Substituting \(f = 0\) into the equation gives
\[E_k = h(0) - hf_0 = -hf_0 = -W_0.\]The intercept OC is therefore negative, and its magnitude equals the work function \(W_0 = hf_0\) of the metal, i.e. \(OC = hf_0 = W_0\). This is the minimum energy needed to free an electron from the metal surface.
(c) Intercept, OK
Point K is where the line crosses the frequency axis at \(E_k = 0\). Setting \(E_k = 0\):
\[0 = hf - hf_0 \quad\Rightarrow\quad f = f_0.\]The intercept OK represents the threshold (cut-off) frequency \(f_0\) of the incident radiation: the minimum frequency below which no electrons are emitted from the surface, however intense the radiation.
Bayanin Amsa
The straight-line graph is a plot of the maximum kinetic energy \(E_k\) of the emitted electrons (vertical axis) against the frequency \(f\) of the radiation falling on the metal surface (horizontal axis). It obeys Einstein's photoelectric equation:
\[E_k = hf - W_0 = hf - hf_0,\]where \(h\) is Planck's constant, \(W_0\) is the work function of the metal and \(f_0\) is the threshold frequency. Comparing this with the equation of a straight line \(y = mx + c\) lets each feature of the graph be interpreted directly, as annotated below.
(a) Slope of the graph
The line has the form \(E_k = hf - hf_0\). Matching term by term with \(y = mx + c\), the gradient \(m\) corresponds to \(h\). Therefore
\[\text{slope} = \frac{\Delta E_k}{\Delta f} = h \approx 6.6 \times 10^{-34}\ \text{J s}.\]The slope of the graph represents Planck's constant \(h\). Because \(h\) is a universal constant, the graph has the same gradient for every metal.
(b) Intercept, OC
Point C is where the line, produced backwards, cuts the energy axis at \(f = 0\). Substituting \(f = 0\) into the equation gives
\[E_k = h(0) - hf_0 = -hf_0 = -W_0.\]The intercept OC is therefore negative, and its magnitude equals the work function \(W_0 = hf_0\) of the metal, i.e. \(OC = hf_0 = W_0\). This is the minimum energy needed to free an electron from the metal surface.
(c) Intercept, OK
Point K is where the line crosses the frequency axis at \(E_k = 0\). Setting \(E_k = 0\):
\[0 = hf - hf_0 \quad\Rightarrow\quad f = f_0.\]The intercept OK represents the threshold (cut-off) frequency \(f_0\) of the incident radiation: the minimum frequency below which no electrons are emitted from the surface, however intense the radiation.
Tambaya 9 Rahoto
(a) Explain how a gas can be made to conduct electricity.
(b) Name the electric charge carriers in gases.
(a) How a gas can be made to conduct electricity
Under normal conditions a gas is a very good insulator because it contains almost no free charge carriers; its molecules are neutral. A gas is made to conduct by ionising it, that is, by knocking electrons out of some of its molecules so that free electrons and positive ions are produced. This is achieved by:
- Applying a very high potential difference (strong electric field) across the gas so that stray electrons are accelerated hard enough to knock electrons out of neutral molecules on collision, causing further ionisation (an ionisation avalanche).
- Reducing the gas pressure (as in a discharge tube). At low pressure the molecules are far apart, so accelerated electrons travel a long free path and gain enough energy between collisions to ionise the gas.
- Ionising the gas by other agents such as X-rays, ultraviolet light, or radiation from radioactive sources, which free electrons from the molecules.
Once ions and free electrons exist, they drift under the applied field and constitute a current, so the gas conducts.
(b) The electric charge carriers in gases
- Negatively charged free electrons (and negative ions).
- Positively charged positive ions (molecules that have lost electrons).
Bayanin Amsa
(a) How a gas can be made to conduct electricity
Under normal conditions a gas is a very good insulator because it contains almost no free charge carriers; its molecules are neutral. A gas is made to conduct by ionising it, that is, by knocking electrons out of some of its molecules so that free electrons and positive ions are produced. This is achieved by:
- Applying a very high potential difference (strong electric field) across the gas so that stray electrons are accelerated hard enough to knock electrons out of neutral molecules on collision, causing further ionisation (an ionisation avalanche).
- Reducing the gas pressure (as in a discharge tube). At low pressure the molecules are far apart, so accelerated electrons travel a long free path and gain enough energy between collisions to ionise the gas.
- Ionising the gas by other agents such as X-rays, ultraviolet light, or radiation from radioactive sources, which free electrons from the molecules.
Once ions and free electrons exist, they drift under the applied field and constitute a current, so the gas conducts.
(b) The electric charge carriers in gases
- Negatively charged free electrons (and negative ions).
- Positively charged positive ions (molecules that have lost electrons).
Tambaya 10 Rahoto
(a) Give two examples each of:
(i) rotational motion;
(ii) linear motion.
(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.
(c) State Newton's second law of motion
(d) Explain the term inertia.
(e)
The diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\(^{-2}\) and experiences a frictional force of 5 N, calculate the:
(i) net force on it;
(ii) magnitude of F;
(iii) coefficient of kinetic friction. [ g = 10 ms\(^{-2}\)]
(a) Examples of motion
(i) Rotational motion: a spinning wheel or tyre; the blades of a rotating fan; a spinning top.
(ii) Linear motion: a car moving along a straight road; a ball falling freely under gravity; an object sliding along a straight track.
(b) Experiment to determine the density of an irregularly shaped solid
- Find the mass \(m\) of the solid using a beam or electronic balance.
- Partly fill a measuring cylinder with water and note the initial volume reading \(V_1\).
- Tie the solid with a thin thread and lower it gently until it is fully immersed in the water. Note the new volume reading \(V_2\).
- The volume of the solid is \(V = V_2 - V_1\) (volume of water displaced).
- Calculate the density from \(\rho = \dfrac{m}{V_2 - V_1}\).
(c) Newton's second law of motion
The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(d) Inertia
Inertia is the property of a body by virtue of which it resists any change to its state of rest or of uniform motion in a straight line. The greater the mass of a body, the greater its inertia.
(e) Calculations (mass \(m = 5.0\) kg, acceleration \(a = 2.0\ \text{m s}^{-2}\), friction \(f = 5\) N, \(g = 10\ \text{m s}^{-2}\); F is horizontal)
(i) Net force
\[ F_{net} = ma = 5.0 \times 2.0 = 10\ \text{N} \]
(ii) Magnitude of F
The horizontal pull provides the net force after overcoming friction:
\[ F - f = F_{net} \;\Rightarrow\; F = F_{net} + f = 10 + 5 = 15\ \text{N} \]
(iii) Coefficient of kinetic friction
The normal reaction equals the weight: \(R = mg = 5.0 \times 10 = 50\ \text{N}\).
\[ \mu = \frac{f}{R} = \frac{5}{50} = 0.1 \]
Bayanin Amsa
(a) Examples of motion
(i) Rotational motion: a spinning wheel or tyre; the blades of a rotating fan; a spinning top.
(ii) Linear motion: a car moving along a straight road; a ball falling freely under gravity; an object sliding along a straight track.
(b) Experiment to determine the density of an irregularly shaped solid
- Find the mass \(m\) of the solid using a beam or electronic balance.
- Partly fill a measuring cylinder with water and note the initial volume reading \(V_1\).
- Tie the solid with a thin thread and lower it gently until it is fully immersed in the water. Note the new volume reading \(V_2\).
- The volume of the solid is \(V = V_2 - V_1\) (volume of water displaced).
- Calculate the density from \(\rho = \dfrac{m}{V_2 - V_1}\).
(c) Newton's second law of motion
The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(d) Inertia
Inertia is the property of a body by virtue of which it resists any change to its state of rest or of uniform motion in a straight line. The greater the mass of a body, the greater its inertia.
(e) Calculations (mass \(m = 5.0\) kg, acceleration \(a = 2.0\ \text{m s}^{-2}\), friction \(f = 5\) N, \(g = 10\ \text{m s}^{-2}\); F is horizontal)
(i) Net force
\[ F_{net} = ma = 5.0 \times 2.0 = 10\ \text{N} \]
(ii) Magnitude of F
The horizontal pull provides the net force after overcoming friction:
\[ F - f = F_{net} \;\Rightarrow\; F = F_{net} + f = 10 + 5 = 15\ \text{N} \]
(iii) Coefficient of kinetic friction
The normal reaction equals the weight: \(R = mg = 5.0 \times 10 = 50\ \text{N}\).
\[ \mu = \frac{f}{R} = \frac{5}{50} = 0.1 \]
Tambaya 11 Rahoto
A particle is projected horizontally at 10 ms\(^{-2}\) from the top of a tower 20 M high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g= 10 ms\(^{-2}\)
Vertical motion (to find time of flight). The particle is projected horizontally, so its initial vertical velocity is zero. Falling a height \(h = 20\ \text{m}\):
\[ h = \tfrac{1}{2}g t^{2} \;\Rightarrow\; 20 = \tfrac{1}{2}\times10\times t^{2} \] \[ t^{2} = 4 \;\Rightarrow\; t = 2\ \text{s} \]Horizontal motion (constant velocity). Horizontal speed \(u = 10\ \text{m/s}\), so the horizontal distance (range) is
\[ x = u\,t = 10\times2 = 20\ \text{m} \]The particle strikes the ground 20 m from the foot of the tower.
Bayanin Amsa
Vertical motion (to find time of flight). The particle is projected horizontally, so its initial vertical velocity is zero. Falling a height \(h = 20\ \text{m}\):
\[ h = \tfrac{1}{2}g t^{2} \;\Rightarrow\; 20 = \tfrac{1}{2}\times10\times t^{2} \] \[ t^{2} = 4 \;\Rightarrow\; t = 2\ \text{s} \]Horizontal motion (constant velocity). Horizontal speed \(u = 10\ \text{m/s}\), so the horizontal distance (range) is
\[ x = u\,t = 10\times2 = 20\ \text{m} \]The particle strikes the ground 20 m from the foot of the tower.
Tambaya 12 Rahoto
A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the value of Young's modulus
Definition. Young's modulus is the ratio of tensile stress to tensile strain within the elastic limit:
\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{F\,L}{A\,e} \]Convert the data.
- Original length \(L = 50\ \text{cm} = 0.50\ \text{m}\)
- Cross-sectional area \(A = 4.0\ \text{cm}^{2} = 4.0\times10^{-4}\ \text{m}^{2}\)
- Force \(F = 35\ \text{kN} = 35\,000\ \text{N}\)
- Extension \(e = 0.25\ \text{mm} = 2.5\times10^{-4}\ \text{m}\)
Substitute.
\[ Y = \frac{35\,000\times0.50}{4.0\times10^{-4}\times2.5\times10^{-4}} = \frac{17\,500}{1.0\times10^{-7}} \] \[ Y = 1.75\times10^{11}\ \text{Pa} \]Young's modulus of the bar is \(1.75\times10^{11}\ \text{Pa}\) (N m\(^{-2}\)).
Bayanin Amsa
Definition. Young's modulus is the ratio of tensile stress to tensile strain within the elastic limit:
\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{F\,L}{A\,e} \]Convert the data.
- Original length \(L = 50\ \text{cm} = 0.50\ \text{m}\)
- Cross-sectional area \(A = 4.0\ \text{cm}^{2} = 4.0\times10^{-4}\ \text{m}^{2}\)
- Force \(F = 35\ \text{kN} = 35\,000\ \text{N}\)
- Extension \(e = 0.25\ \text{mm} = 2.5\times10^{-4}\ \text{m}\)
Substitute.
\[ Y = \frac{35\,000\times0.50}{4.0\times10^{-4}\times2.5\times10^{-4}} = \frac{17\,500}{1.0\times10^{-7}} \] \[ Y = 1.75\times10^{11}\ \text{Pa} \]Young's modulus of the bar is \(1.75\times10^{11}\ \text{Pa}\) (N m\(^{-2}\)).
