WAEC SSCE - Physics - 2018 (Essay)
Ajụjụ 1 Ripọtì
State three materials used for making optical fibres.
Materials used for making optical fibres
- Silica (fused quartz / glass).
- Plastic (transparent polymer, e.g. acrylic/PMMA).
- Doped glass such as germanium-doped or fluoride glass used for the core and cladding.
Akọwa Nkọwa
Materials used for making optical fibres
- Silica (fused quartz / glass).
- Plastic (transparent polymer, e.g. acrylic/PMMA).
- Doped glass such as germanium-doped or fluoride glass used for the core and cladding.
Ajụjụ 2 Ripọtì
Name three classes of magnetic materials.
Three classes of magnetic materials
- Ferromagnetic materials (e.g. iron, nickel, cobalt) which are strongly attracted by a magnet.
- Paramagnetic materials (e.g. aluminium, platinum) which are weakly attracted by a magnet.
- Diamagnetic materials (e.g. copper, bismuth) which are weakly repelled by a magnet.
Akọwa Nkọwa
Three classes of magnetic materials
- Ferromagnetic materials (e.g. iron, nickel, cobalt) which are strongly attracted by a magnet.
- Paramagnetic materials (e.g. aluminium, platinum) which are weakly attracted by a magnet.
- Diamagnetic materials (e.g. copper, bismuth) which are weakly repelled by a magnet.
Ajụjụ 3 Ripọtì
(a) Define uniform acceleration.
(b) Forces act on a car in motion. List the
(i) horizontal forces and their directions;
(ii) vertical forces and their directions
(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\(^{-1}\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.
(i) Sketch a velocity-tune graph for the motion.
(ii) Use the graph to determine the (\(\alpha\)) total distance travelled by the car (\(\beta\)) deceleration of the car.
The figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.
(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.
(b) Forces acting on a moving car.
(i) Horizontal forces and their directions:
- The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
- Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.
(ii) Vertical forces and their directions:
- The weight of the car, acting vertically downward.
- The normal reaction from the road surface, acting vertically upward.
(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).
(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):
\[ s=\tfrac{1}{2}(OC+AB)\times v=\tfrac{1}{2}(60+30)\times 25=\tfrac{1}{2}\times 90\times 25=1125\text{ m.} \]
(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):
\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]
So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.
Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:
\[ W=\tfrac{1}{2}\,F\,e=\tfrac{1}{2}\times 12\times 0.005=0.03\text{ J.} \]
The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.
Akọwa Nkọwa
(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.
(b) Forces acting on a moving car.
(i) Horizontal forces and their directions:
- The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
- Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.
(ii) Vertical forces and their directions:
- The weight of the car, acting vertically downward.
- The normal reaction from the road surface, acting vertically upward.
(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).
(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):
\[ s=\tfrac{1}{2}(OC+AB)\times v=\tfrac{1}{2}(60+30)\times 25=\tfrac{1}{2}\times 90\times 25=1125\text{ m.} \]
(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):
\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]
So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.
Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:
\[ W=\tfrac{1}{2}\,F\,e=\tfrac{1}{2}\times 12\times 0.005=0.03\text{ J.} \]
The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.
Ajụjụ 4 Ripọtì
A black body radiates maximum energy when its surface temperature T and the corresponding wavelength \(\lambda\)max are related by the equation \(\lambda\)max T = constant. Given the values of the constant and surface temperature as 2.9 x 10\(^{-3}\) mK and 57°C respectively; Calculate the frequency of the energy radiated.
Wien's displacement law
Given \(\lambda_{\max}T = 2.9\times10^{-3}\ \text{m K}\), and \(T = 57^{\circ}\text{C} = 57 + 273 = 330\ \text{K}\).
\(\lambda_{\max} = \dfrac{2.9\times10^{-3}}{330} = 8.79\times10^{-6}\ \text{m}\).
Frequency: \(f = \dfrac{c}{\lambda_{\max}} = \dfrac{3.0\times10^{8}}{8.79\times10^{-6}}\).
\(f = 3.41\times10^{13}\ \text{Hz}\).
The frequency of the radiated energy is \(3.4\times10^{13}\ \text{Hz}\).
Akọwa Nkọwa
Wien's displacement law
Given \(\lambda_{\max}T = 2.9\times10^{-3}\ \text{m K}\), and \(T = 57^{\circ}\text{C} = 57 + 273 = 330\ \text{K}\).
\(\lambda_{\max} = \dfrac{2.9\times10^{-3}}{330} = 8.79\times10^{-6}\ \text{m}\).
Frequency: \(f = \dfrac{c}{\lambda_{\max}} = \dfrac{3.0\times10^{8}}{8.79\times10^{-6}}\).
\(f = 3.41\times10^{13}\ \text{Hz}\).
The frequency of the radiated energy is \(3.4\times10^{13}\ \text{Hz}\).
Ajụjụ 5 Ripọtì
(a) Define:
(i) reactance;
(ii) impedance in an a.c.
The diagram here illustrates an a.c. generator. When the coil is rotated, an e.m.f is induced in the coil.
(i) Explain why an e.m.f. is induced.
(ii) State the purpose of the slip-rings.
(iii) Name and state the law used to determine the direction of the induced current.
(iv) State two ways to increase the induced e.m.f.
(c) A lamp is rated 12 V. 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.
(a) Definitions
- Reactance is the opposition offered to the flow of alternating current by an inductor or a capacitor, arising from the induced back-e.m.f (inductor) or stored charge (capacitor). It is measured in ohms and, unlike resistance, does not dissipate energy.
- Impedance is the total opposition to the flow of alternating current in a circuit, being the combined effect of resistance and reactance. It is measured in ohms, \(Z=\sqrt{R^2+X^2}\).
(b) The a.c. generator (shown with the coil between the N and S poles, the slip rings and the load)
- Why an e.m.f. is induced: As the coil rotates in the magnetic field between the N and S poles, the magnetic flux linking the coil changes continuously. By Faraday's law, a changing flux linkage induces an e.m.f. in the coil; the e.m.f. is greatest when the coil moves parallel to the field (cutting flux fastest) and zero when it moves perpendicular to it.
- Purpose of the slip-rings: They provide a continuous sliding electrical contact between the ends of the rotating coil and the external circuit (through the brushes), so that the induced alternating e.m.f. is delivered to the load without the connecting wires twisting.
- Law for the direction of the induced current: Fleming's right-hand (dynamo) rule. Holding the thumb, first finger and second finger of the right hand mutually at right angles, the First finger points along the Field (N to S), the Thumb along the Thrust (motion of the conductor), and the seCond finger then gives the direction of the induced Current.
- Two ways to increase the induced e.m.f.: increase the number of turns on the coil; rotate the coil faster (higher speed of rotation). (Others: use a stronger magnet, or a larger coil area / soft-iron core.)
(c) Energy transformed by a 12 V, 6 W lamp in 5 minutes
Power \(P=6\text{ W}\); time \(t=5\times60=300\text{ s}\).
\[E=Pt=6\times300=1800\text{ J}.\]The lamp transforms 1800 J of electrical energy in 5 minutes.
Akọwa Nkọwa
(a) Definitions
- Reactance is the opposition offered to the flow of alternating current by an inductor or a capacitor, arising from the induced back-e.m.f (inductor) or stored charge (capacitor). It is measured in ohms and, unlike resistance, does not dissipate energy.
- Impedance is the total opposition to the flow of alternating current in a circuit, being the combined effect of resistance and reactance. It is measured in ohms, \(Z=\sqrt{R^2+X^2}\).
(b) The a.c. generator (shown with the coil between the N and S poles, the slip rings and the load)
- Why an e.m.f. is induced: As the coil rotates in the magnetic field between the N and S poles, the magnetic flux linking the coil changes continuously. By Faraday's law, a changing flux linkage induces an e.m.f. in the coil; the e.m.f. is greatest when the coil moves parallel to the field (cutting flux fastest) and zero when it moves perpendicular to it.
- Purpose of the slip-rings: They provide a continuous sliding electrical contact between the ends of the rotating coil and the external circuit (through the brushes), so that the induced alternating e.m.f. is delivered to the load without the connecting wires twisting.
- Law for the direction of the induced current: Fleming's right-hand (dynamo) rule. Holding the thumb, first finger and second finger of the right hand mutually at right angles, the First finger points along the Field (N to S), the Thumb along the Thrust (motion of the conductor), and the seCond finger then gives the direction of the induced Current.
- Two ways to increase the induced e.m.f.: increase the number of turns on the coil; rotate the coil faster (higher speed of rotation). (Others: use a stronger magnet, or a larger coil area / soft-iron core.)
(c) Energy transformed by a 12 V, 6 W lamp in 5 minutes
Power \(P=6\text{ W}\); time \(t=5\times60=300\text{ s}\).
\[E=Pt=6\times300=1800\text{ J}.\]The lamp transforms 1800 J of electrical energy in 5 minutes.
Ajụjụ 6 Ripọtì
(a) List two factors each that affect heat loss by:
(i) radiation;
(ii) convection.
(b) State two factors that determine the quantity of heat in a body.
(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\(^5\) Jkg\(^{-1}\).
(d)A jug of heat capacity 250 Jkg\(^{-1}\) contains water at 28°C. An electric heater of resistance 35\(\Omega\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:
(i) mass of water in the jug before heating;
(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\(^{-1}\)K\(^{-1}\)]
(a) Factors affecting heat loss
(i) By radiation:
- The temperature of the body (and of its surroundings).
- The nature and colour of the surface (dull/black surfaces radiate more than shiny/white ones) and its surface area.
(ii) By convection:
- The temperature difference between the body and the surrounding fluid.
- The exposed surface area and the freedom of the fluid to circulate (draught/movement of the fluid).
(b) Factors determining the quantity of heat in a body
- The mass of the body.
- The temperature (and the nature/specific heat capacity of the material).
(c) The statement "the specific latent heat of vaporization of mercury is \(2.72\times10^{5}\ \text{J kg}^{-1}\)" means that \(2.72\times10^{5}\) joules of heat is required to change 1 kilogram of mercury from liquid to vapour at its boiling point without any change in temperature.
(d) The jug of water
Power of heater: \(P = \dfrac{V^{2}}{R} = \dfrac{220^{2}}{35} = \dfrac{48400}{35} = 1382.9\ \text{W}\).
(i) Mass of water before heating. In the first \(t_{1} = 4\text{ min} = 240\text{ s}\) the temperature rises from \(28^{\circ}\text{C}\) to \(100^{\circ}\text{C}\), so \(\Delta\theta = 72\ \text{K}\).
Heat supplied \(= P t_{1} = 1382.9\times240 = 3.319\times10^{5}\ \text{J}\).
Taking heat capacity of jug \(C = 250\ \text{J K}^{-1}\) and \(c_{water} = 4200\ \text{J kg}^{-1}\text{K}^{-1}\):
\((m c_{water} + C)\Delta\theta = P t_{1}\)
\((4200m + 250)\times 72 = 331886\)
\(4200m + 250 = 4609.5 \Rightarrow 4200m = 4359.5 \Rightarrow m = 1.04\ \text{kg}\).
The mass of water before heating is about 1.04 kg.
(ii) Specific latent heat of vaporization of steam. In the next \(t_{2} = 5\text{ min} = 300\text{ s}\) the water boils at constant temperature and \(0.3\ \text{kg}\) evaporates.
Heat supplied \(= P t_{2} = 1382.9\times300 = 4.149\times10^{5}\ \text{J}\).
\(L = \dfrac{P t_{2}}{m_{evap}} = \dfrac{414857}{0.3} = 1.38\times10^{6}\ \text{J kg}^{-1}\).
Akọwa Nkọwa
(a) Factors affecting heat loss
(i) By radiation:
- The temperature of the body (and of its surroundings).
- The nature and colour of the surface (dull/black surfaces radiate more than shiny/white ones) and its surface area.
(ii) By convection:
- The temperature difference between the body and the surrounding fluid.
- The exposed surface area and the freedom of the fluid to circulate (draught/movement of the fluid).
(b) Factors determining the quantity of heat in a body
- The mass of the body.
- The temperature (and the nature/specific heat capacity of the material).
(c) The statement "the specific latent heat of vaporization of mercury is \(2.72\times10^{5}\ \text{J kg}^{-1}\)" means that \(2.72\times10^{5}\) joules of heat is required to change 1 kilogram of mercury from liquid to vapour at its boiling point without any change in temperature.
(d) The jug of water
Power of heater: \(P = \dfrac{V^{2}}{R} = \dfrac{220^{2}}{35} = \dfrac{48400}{35} = 1382.9\ \text{W}\).
(i) Mass of water before heating. In the first \(t_{1} = 4\text{ min} = 240\text{ s}\) the temperature rises from \(28^{\circ}\text{C}\) to \(100^{\circ}\text{C}\), so \(\Delta\theta = 72\ \text{K}\).
Heat supplied \(= P t_{1} = 1382.9\times240 = 3.319\times10^{5}\ \text{J}\).
Taking heat capacity of jug \(C = 250\ \text{J K}^{-1}\) and \(c_{water} = 4200\ \text{J kg}^{-1}\text{K}^{-1}\):
\((m c_{water} + C)\Delta\theta = P t_{1}\)
\((4200m + 250)\times 72 = 331886\)
\(4200m + 250 = 4609.5 \Rightarrow 4200m = 4359.5 \Rightarrow m = 1.04\ \text{kg}\).
The mass of water before heating is about 1.04 kg.
(ii) Specific latent heat of vaporization of steam. In the next \(t_{2} = 5\text{ min} = 300\text{ s}\) the water boils at constant temperature and \(0.3\ \text{kg}\) evaporates.
Heat supplied \(= P t_{2} = 1382.9\times300 = 4.149\times10^{5}\ \text{J}\).
\(L = \dfrac{P t_{2}}{m_{evap}} = \dfrac{414857}{0.3} = 1.38\times10^{6}\ \text{J kg}^{-1}\).
Ajụjụ 7 Ripọtì
a) What does the acronym LASER stand for?
b) What is a laser?
(a) LASER stands for Light Amplification by Stimulated Emission of Radiation.
(b) A laser is a device that produces an intense, narrow, highly directional beam of light that is monochromatic (single wavelength) and coherent (the waves are in phase), by the process of stimulated emission of radiation.
Akọwa Nkọwa
(a) LASER stands for Light Amplification by Stimulated Emission of Radiation.
(b) A laser is a device that produces an intense, narrow, highly directional beam of light that is monochromatic (single wavelength) and coherent (the waves are in phase), by the process of stimulated emission of radiation.
Ajụjụ 8 Ripọtì
(a) What is an intrinsic semiconductor?
(b) Distinguish between the p-type and n-type semi-conductors.
(a) Intrinsic semiconductor
An intrinsic semiconductor is a pure semiconductor (such as pure silicon or germanium) with no added impurity, in which the number of free electrons equals the number of holes, both being produced only by thermal breaking of covalent bonds.
(b) p-type versus n-type semiconductors
| p-type | n-type |
|---|---|
| Formed by doping with a trivalent impurity (e.g. boron, aluminium). | Formed by doping with a pentavalent impurity (e.g. phosphorus, arsenic). |
| Majority charge carriers are holes (positive). | Majority charge carriers are free electrons (negative). |
| The impurity is called an acceptor. | The impurity is called a donor. |
Akọwa Nkọwa
(a) Intrinsic semiconductor
An intrinsic semiconductor is a pure semiconductor (such as pure silicon or germanium) with no added impurity, in which the number of free electrons equals the number of holes, both being produced only by thermal breaking of covalent bonds.
(b) p-type versus n-type semiconductors
| p-type | n-type |
|---|---|
| Formed by doping with a trivalent impurity (e.g. boron, aluminium). | Formed by doping with a pentavalent impurity (e.g. phosphorus, arsenic). |
| Majority charge carriers are holes (positive). | Majority charge carriers are free electrons (negative). |
| The impurity is called an acceptor. | The impurity is called a donor. |
Ajụjụ 9 Ripọtì
A missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms\(^{-1}\). [g = 10ms\(^{-2}\)]
Maximum height for maximum range
A projectile attains its maximum range when it is launched at an angle of \(45^{\circ}\) to the horizontal.
Maximum height: \(H = \dfrac{u^{2}\sin^{2}\theta}{2g}\), with \(u = 200\ \text{m s}^{-1}\), \(\theta = 45^{\circ}\), \(g = 10\ \text{m s}^{-2}\).
\(\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\), so \(\sin^{2}45^{\circ} = 0.5\).
\(H = \dfrac{200^{2}\times 0.5}{2\times 10} = \dfrac{40000\times 0.5}{20} = \dfrac{20000}{20} = 1000\ \text{m}\).
The maximum height attained is 1000 m.
Akọwa Nkọwa
Maximum height for maximum range
A projectile attains its maximum range when it is launched at an angle of \(45^{\circ}\) to the horizontal.
Maximum height: \(H = \dfrac{u^{2}\sin^{2}\theta}{2g}\), with \(u = 200\ \text{m s}^{-1}\), \(\theta = 45^{\circ}\), \(g = 10\ \text{m s}^{-2}\).
\(\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\), so \(\sin^{2}45^{\circ} = 0.5\).
\(H = \dfrac{200^{2}\times 0.5}{2\times 10} = \dfrac{40000\times 0.5}{20} = \dfrac{20000}{20} = 1000\ \text{m}\).
The maximum height attained is 1000 m.
Ajụjụ 10 Ripọtì
(a) Define strain.
(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.
(a)
Strain is a measure of the deformation or elongation of a material when subjected to an external force or stress. It is defined as the ratio of the change in length or shape of an object to its original length or shape.
(b)
In this case, the rubber band has been stretched to twice its original length. Therefore, the change in length is equal to twice the original length (2L), where L is the original length.
To calculate the strain, we use the formula:
Strain = (Change in length) / (Original length)
Substituting the values, we get:
Strain = (2L - L) / L = L / L = 1
So, the strain on the rubber band is 1, which means that the rubber band has elongated by 100% of its original length. This is because the change in length is equal to the original length, so the strain is equal to 1 (or 100%).
Akọwa Nkọwa
(a)
Strain is a measure of the deformation or elongation of a material when subjected to an external force or stress. It is defined as the ratio of the change in length or shape of an object to its original length or shape.
(b)
In this case, the rubber band has been stretched to twice its original length. Therefore, the change in length is equal to twice the original length (2L), where L is the original length.
To calculate the strain, we use the formula:
Strain = (Change in length) / (Original length)
Substituting the values, we get:
Strain = (2L - L) / L = L / L = 1
So, the strain on the rubber band is 1, which means that the rubber band has elongated by 100% of its original length. This is because the change in length is equal to the original length, so the strain is equal to 1 (or 100%).
Ajụjụ 11 Ripọtì
(a) Define binding energy in an atom.
(b) List three evidence to support the claim that X-rays are electromagnetic waves.
(c) List three peaceful uses of nuclear energy.
(d) Light of wavelength 4.5 x 10\(^{-7}\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\(^{-9}\) J, calculate the:
(i) frequency of the incident light;
(ii) energy of the incident light;
(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\(^8\) ms\(^{-1}\), h = 6.6 x 10\(^{-34}\) Js]
(a) Binding energy
The binding energy of a nucleus is the energy required to completely separate the nucleus into its individual constituent protons and neutrons (equivalently, the energy released when the separate nucleons combine to form the nucleus). It equals the mass defect times \(c^{2}\).
(b) Three evidences that X-rays are electromagnetic waves
- They travel in straight lines with the speed of light and are undeflected by electric or magnetic fields (so they are uncharged).
- They undergo diffraction and interference when passed through crystals.
- They can be polarised.
(c) Three peaceful uses of nuclear energy
- Generation of electricity in nuclear power stations.
- Medical uses: radiotherapy to treat cancer and use of radioisotopes as tracers in diagnosis.
- Industrial and agricultural uses: sterilisation of equipment, food preservation, and thickness/level gauging.
(d) Photoelectric calculation
Data: \(\lambda = 4.5\times10^{-7}\ \text{m}\), work function \(W_{0} = 3.0\times10^{-19}\ \text{J}\) (taking the stated value as \(3.0\times10^{-19}\ \text{J}\)), \(c = 3.0\times10^{8}\ \text{m s}^{-1}\), \(h = 6.6\times10^{-34}\ \text{J s}\).
(i) Frequency: \(f = \dfrac{c}{\lambda} = \dfrac{3.0\times10^{8}}{4.5\times10^{-7}} = 6.67\times10^{14}\ \text{Hz}\).
(ii) Energy of the incident light: \(E = hf = 6.6\times10^{-34}\times6.67\times10^{14} = 4.4\times10^{-19}\ \text{J}\).
(iii) Energy (max KE) of the photoelectrons: \(E_{k} = E - W_{0} = 4.4\times10^{-19} - 3.0\times10^{-19} = 1.4\times10^{-19}\ \text{J}\).
Akọwa Nkọwa
(a) Binding energy
The binding energy of a nucleus is the energy required to completely separate the nucleus into its individual constituent protons and neutrons (equivalently, the energy released when the separate nucleons combine to form the nucleus). It equals the mass defect times \(c^{2}\).
(b) Three evidences that X-rays are electromagnetic waves
- They travel in straight lines with the speed of light and are undeflected by electric or magnetic fields (so they are uncharged).
- They undergo diffraction and interference when passed through crystals.
- They can be polarised.
(c) Three peaceful uses of nuclear energy
- Generation of electricity in nuclear power stations.
- Medical uses: radiotherapy to treat cancer and use of radioisotopes as tracers in diagnosis.
- Industrial and agricultural uses: sterilisation of equipment, food preservation, and thickness/level gauging.
(d) Photoelectric calculation
Data: \(\lambda = 4.5\times10^{-7}\ \text{m}\), work function \(W_{0} = 3.0\times10^{-19}\ \text{J}\) (taking the stated value as \(3.0\times10^{-19}\ \text{J}\)), \(c = 3.0\times10^{8}\ \text{m s}^{-1}\), \(h = 6.6\times10^{-34}\ \text{J s}\).
(i) Frequency: \(f = \dfrac{c}{\lambda} = \dfrac{3.0\times10^{8}}{4.5\times10^{-7}} = 6.67\times10^{14}\ \text{Hz}\).
(ii) Energy of the incident light: \(E = hf = 6.6\times10^{-34}\times6.67\times10^{14} = 4.4\times10^{-19}\ \text{J}\).
(iii) Energy (max KE) of the photoelectrons: \(E_{k} = E - W_{0} = 4.4\times10^{-19} - 3.0\times10^{-19} = 1.4\times10^{-19}\ \text{J}\).
Ajụjụ 12 Ripọtì
(a) Define dffraction.
(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \(\alpha\) Determine the value of the critical angle. \(\beta\) Calculate the refractive index of the block.
(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:
(i) fundamental frequency of the open pipe;
(ii) length of the closed pipe;
(iii) length of the open pipe. [Speed of sound in air = 330 ms\(^{-1}\)]
(a) Diffraction
Diffraction is the spreading of a wave into the region behind an obstacle, or the bending of a wave round the edges of an obstacle or through an aperture, as the wave passes them. It is most noticeable when the width of the aperture or obstacle is comparable with the wavelength of the wave.
(b)(i) Critical angle
The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is exactly \(90^\circ\). For any angle of incidence greater than the critical angle the light undergoes total internal reflection.
(\(\alpha\)) Value of the critical angle
From the diagram the ray travels along (grazes) the top surface of the block, so its angle of incidence in air is \(90^\circ\). It is refracted into the block, and the refracted ray makes \(44^\circ\) with the surface. Its angle measured from the normal is therefore
\[ r = 90^\circ - 44^\circ = 46^\circ. \]At grazing incidence the angle of refraction is equal to the critical angle of the block, hence
\[ c = 46^\circ. \](\(\beta\)) Refractive index of the block
\[ n = \frac{1}{\sin c} = \frac{1}{\sin 46^\circ} = \frac{1}{0.7193} = 1.39. \](c) Pipes
A pipe closed at one end sounds only the odd harmonics. Its fundamental is \(f_c = 200\,\text{Hz}\), so its first overtone is the third harmonic:
\[ 3 \times 200 = 600\,\text{Hz}. \](i) Fundamental frequency of the open pipe
An open pipe sounds all harmonics, so its first overtone is the second harmonic, \(2f_o\). Given that the two first overtones are equal:
\[ 2f_o = 600 \Rightarrow f_o = 300\,\text{Hz}. \](ii) Length of the closed pipe
For a closed pipe \(f_c = \dfrac{v}{4L_c}\), so
\[ L_c = \frac{v}{4f_c} = \frac{330}{4 \times 200} = 0.41\,\text{m}. \](iii) Length of the open pipe
For an open pipe \(f_o = \dfrac{v}{2L_o}\), so
\[ L_o = \frac{v}{2f_o} = \frac{330}{2 \times 300} = 0.55\,\text{m}. \]Akọwa Nkọwa
(a) Diffraction
Diffraction is the spreading of a wave into the region behind an obstacle, or the bending of a wave round the edges of an obstacle or through an aperture, as the wave passes them. It is most noticeable when the width of the aperture or obstacle is comparable with the wavelength of the wave.
(b)(i) Critical angle
The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is exactly \(90^\circ\). For any angle of incidence greater than the critical angle the light undergoes total internal reflection.
(\(\alpha\)) Value of the critical angle
From the diagram the ray travels along (grazes) the top surface of the block, so its angle of incidence in air is \(90^\circ\). It is refracted into the block, and the refracted ray makes \(44^\circ\) with the surface. Its angle measured from the normal is therefore
\[ r = 90^\circ - 44^\circ = 46^\circ. \]At grazing incidence the angle of refraction is equal to the critical angle of the block, hence
\[ c = 46^\circ. \](\(\beta\)) Refractive index of the block
\[ n = \frac{1}{\sin c} = \frac{1}{\sin 46^\circ} = \frac{1}{0.7193} = 1.39. \](c) Pipes
A pipe closed at one end sounds only the odd harmonics. Its fundamental is \(f_c = 200\,\text{Hz}\), so its first overtone is the third harmonic:
\[ 3 \times 200 = 600\,\text{Hz}. \](i) Fundamental frequency of the open pipe
An open pipe sounds all harmonics, so its first overtone is the second harmonic, \(2f_o\). Given that the two first overtones are equal:
\[ 2f_o = 600 \Rightarrow f_o = 300\,\text{Hz}. \](ii) Length of the closed pipe
For a closed pipe \(f_c = \dfrac{v}{4L_c}\), so
\[ L_c = \frac{v}{4f_c} = \frac{330}{4 \times 200} = 0.41\,\text{m}. \](iii) Length of the open pipe
For an open pipe \(f_o = \dfrac{v}{2L_o}\), so
\[ L_o = \frac{v}{2f_o} = \frac{330}{2 \times 300} = 0.55\,\text{m}. \]
