WAEC SSCE - Physics - 1998

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The image distance of an object placed at a distance of 2f from a converging lens of focal length f can be found using the lens formula: 1/f = 1/v - 1/u where f is the focal length of the lens, v is the image distance, and u is the object distance. Since the object is placed at a distance of 2f, u = 2f. Substituting these values in the lens formula, we get: 1/f = 1/v - 1/2f Multiplying both sides by 2f, we get: 2 - f/f = v Simplifying this equation, we get: v = 2f Therefore, the image distance of an object placed at a distance of 2f from a converging lens of focal length f is 2f.

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An eclipse of the sun by the moon occurs when the sun, the moon and the earth are in a straight line and the moon is between the sun and the earth. The moon casts its shadow on the earth, blocking the light from the sun, and this causes a temporary darkness on the earth. This phenomenon is known as a solar eclipse, and it occurs when the new moon comes between the sun and the earth in its monthly orbit.

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Lenz's law of electromagnetism states that the direction of an induced current in a conductor is always such as to oppose the change that produced it. In other words, when there is a change in magnetic flux through a circuit, it produces an induced electromotive force (emf) which in turn generates a current. This induced current flows in such a direction that it opposes the change that produced it. This law is a consequence of the conservation of energy principle, as the induced current creates a magnetic field that opposes the change in the magnetic field that caused the current to be induced.

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The angle marked theta in the diagram is called the "dip" of the magnet. The dip is the angle between the magnetic field lines of the Earth and a horizontal plane at the location of the magnet. The dip angle is important in navigation, especially for compasses, because it helps determine the latitude of the location.

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When an ebonite rod is rubbed with fur, electrons are transferred from the fur to the ebonite rod. Electrons are negatively charged, so when electrons move from one object to another, it leaves the object with a positive charge, and the other with a negative charge. Therefore, the ebonite rod will be negatively charged while the fur will be positively charged. Thus, the correct answer is: "the ebonite rod will be negatively charged while the fur will be positively charged".

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The formula for calculating the wavelength of a wave is given by: \begin{equation*} \lambda = \frac{v}{f} \end{equation*} where - \(\lambda\) is the wavelength of the wave - v is the velocity of the wave in the medium - f is the frequency of the wave. The velocity of the electromagnetic wave in air is given as 3.0 x 10⁸ms^-1. The refractive index of water is given as \(\frac{4}{3}\). The refractive index is the ratio of the speed of light in a vacuum to the speed of light in a medium. Hence the speed of the wave in water is given as: \begin{equation*} v_{water} = \frac{v_{air}}{n} = \frac{3.0\times10^8}{4/3} = 2.25\times10^8ms^{-1} \end{equation*} Substituting the values of v and f in the formula above, we have: \begin{equation*} \lambda = \frac{v}{f} = \frac{2.25\times10^8}{5.0\times10^{14}} = 4.5\times10^{-7}m \end{equation*} Therefore, the wavelength of the electromagnetic wave in water is 4.5 x 10^-7m. Hence the correct option is: 4.5 x 10^-7m.

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The phenomenon which occurs when light changes direction as it passes from one medium to another is called refraction. Refraction is the bending of light as it passes from one medium to another, such as from air to water. The amount of bending depends on the angle at which the light strikes the surface, and the difference in the refractive indices of the two media. This phenomenon is the reason why objects can appear distorted when viewed through a curved surface like a lens, and it is also why a straw in a glass of water appears to bend at the water's surface.

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The relationship between tension (T) and frequency (f) for a string fixed at both ends, such as a sonometer wire, is given by the equation: f = (1/2L)√(T/μ) where L is the length of the string, μ is the linear mass density of the string. Since the length of the wire is kept constant, we can equate the two expressions for frequency to find the relationship between the two tensions: (1/2L)√(T1/μ) = f1 = 250 Hz (1/2L)√(T2/μ) = f2 = 350 Hz Dividing the two equations gives: (f2/f1) = √(T2/T1) (350/250) = √(T2/N10N) Squaring both sides of the equation gives: (350/250)^2 = T2/N10N Solving for T2, we get: T2 = (350/250)^2 x N10N T2 = 19.6 N (to 2 significant figures) Therefore, the new tension required to produce a frequency of 350 Hz is 19.6 N. Answer option (B) is correct.

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The difference between the first and second resonance lengths is equal to half the wavelength of the sound wave. Hence, the wavelength is: λ = 2 × (0.37m – 0.12m) = 0.5m The speed of sound is given as 340m/s. We can use the formula: v = fλ where v is the velocity of sound, f is the frequency of vibration, and λ is the wavelength of the sound wave. Substituting the values we get: 340 = f × 0.5 Solving for f, we get: f = 680 Hz Therefore, the frequency of vibration is 680Hz. So, the answer is 680Hz.

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Diverging lenses are also known as concave lenses. These lenses are thinner at the centre and thicker at the edges. The images formed by diverging lenses are always diminished, virtual and erect. Diminished means that the image is smaller than the object. Virtual means that the image appears to be behind the lens and cannot be projected on a screen. Erect means that the image appears to be in the same orientation as the object. The reason for these characteristics is that the diverging lens spreads out the light rays that enter it, causing them to diverge or move apart. This creates an image that is smaller and appears to be located behind the lens.

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The power rating of the lamp is given as 60W, and the voltage rating is given as 240V. We can use the formula for power, which is P = V^2/R, to calculate the resistance of the filament. First, we rearrange the formula to solve for resistance R: R = V^2 / P Plugging in the given values, we get: R = (240V)^2 / 60W Simplifying, we get: R = 960\(\Omega\) Therefore, the resistance of the filament of the lamp is 960\(\Omega\).

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When the box is just about to slide down the plank, the force of static friction acting on the box is equal and opposite to the component of the weight of the box parallel to the surface of the plank. This component is given by mg sin\(\alpha\), where m is the mass of the box, g is the acceleration due to gravity and \(\alpha\) is the angle of inclination of the plank. Since \(\theta\) is equal to \(\alpha\), the angle of inclination of the plank is also \(\theta\). Therefore, the coefficient of static friction \(\mu\) is given by: \(\mu = \frac{\text{Force of friction}}{\text{Normal force}} = \frac{mg \sin \theta}{mg \cos \theta} = \tan \theta = \tan \alpha\) Hence, the correct option is tan\(\alpha\).

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The frictional effect between the layers of a moving fluid is called viscosity. Viscosity is a measure of a fluid's resistance to flow, which arises due to the friction between the layers of the fluid as they move past each other. A highly viscous fluid has a thick and sticky consistency and resists flowing, while a less viscous fluid flows more easily. Viscosity is an important property of fluids, and it has many practical applications in areas such as engineering, chemistry, and biology.

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When the pilot recorded the atmospheric pressure outside the plane as 63cm of Hg, it means that the pressure due to the atmosphere is balancing the pressure exerted by a column of mercury 63cm high in a barometer tube. Likewise, when the ground observer recorded a reading of 75cm of Hg, it means that the pressure due to the atmosphere is balancing the pressure exerted by a column of mercury 75cm high in a barometer tube. The pressure due to the atmosphere at any point is proportional to the height of the column of air above that point. This implies that the difference in the atmospheric pressure between the pilot's location and the ground observer's location is directly proportional to the height difference between the two points. Using the given data, the pressure difference between the two points is 75 - 63 = 12 cm of Hg. Let h be the height of the plane above the ground, then by the principle of hydrostatics, the pressure difference is also given by: h = (pressure difference) x (density of mercury/density of air) x (gravitational acceleration) Substituting the given values and solving for h, we have: h = 12 x (13.6/0.00136) x (9.81) = 1200m Therefore, the height of the plane above the ground is 1200m. Answer: 1200m

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The physical quantities that have both magnitude and direction are called vectors. Among the given options, the pair of physical quantities that comprise vectors are "force and velocity". Both force and velocity have a direction and magnitude associated with them. The direction of force is along the line of action of the force and the direction of velocity is the direction of motion. Therefore, the correct answer is "force and velocity".

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The concave meniscus of water in a clean glass tube is due to the adhesive forces between the water molecules and the glass tube being greater than the cohesive forces between the water molecules themselves. This causes the water to be attracted to the glass, resulting in the formation of a concave meniscus. Therefore, the correct option is "adhesion between water and glass molecules is greater than the cohesion between water molecules."

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The electrical method of magnetisation involves passing an electric current through a magnetic material to create a magnetic field. The polarity of the magnet that is produced depends on the direction of the current. The direction of the current can be determined by the direction of the flow of electrons in the wire, which is usually indicated by an arrow in a circuit diagram. Therefore, the correct answer is "direction of current."

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The magnifying power of a telescope in normal adjustment is given by the ratio of the focal length of the objective to that of the eyepiece: Magnifying power, M = fo/fe where fo is the focal length of the objective and fe is the focal length of the eyepiece. The separation of the lenses in a telescope is the sum of the focal lengths of the lenses: separation of lenses = fo + fe Given that fo = 100cm and fe = 10cm, we can find the magnifying power and the separation of the lenses: M = fo/fe = 100cm/10cm = 10 separation of lenses = fo + fe = 100cm + 10cm = 110cm = 1.10m Therefore, the separation of the lenses is 1.10m. Answer:.

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The process of increasing the energy of an atom via inelastic collision with an electron is known as excitation. When an electron collides with an atom, it can transfer some of its energy to the atom. This results in the atom's electrons jumping to higher energy levels, or excited states. Excitation can occur through various processes, such as collisions with other particles or the absorption of light. In this case, the excitation occurs through an inelastic collision with an electron. The energy transferred to the atom can be used for various purposes, such as to emit light or to initiate chemical reactions.

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When an object is whirled round in a horizontal circle, it moves with a constant speed and acceleration towards the centre of the circle. This acceleration is known as the centripetal acceleration and is provided by the tension in the string. When the string is suddenly cut, the centripetal force acting on the stone is removed, and it will fly off in a direction tangential to the circular path. Therefore, the correct option is "fly off in a direction tangential to the circular path."

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When an object is immersed in a liquid, it displaces some volume of the liquid. According to Archimedes' principle, the weight of the liquid displaced is equal to the weight of the object. Since the object floats, the weight of the liquid displaced must be equal to the weight of the object. Let V be the volume of the rectangular block of wood and W its weight. Also, let the density of water be ρ1 and the density of the other liquid be ρ2. When the block is immersed in water, two-thirds of its volume is immersed. Therefore, the volume of water displaced is two-thirds of the volume of the block, which is 2V/3. According to Archimedes' principle, the weight of the water displaced is equal to the weight of the block, so: ρ1 x (2V/3) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ1 = W / [(2V/3) x g] Similarly, when the block is immersed in the other liquid, half of its volume is immersed. Therefore, the volume of the other liquid displaced is half of the volume of the block, which is V/2. According to Archimedes' principle, the weight of the other liquid displaced is equal to the weight of the block, so: ρ2 x (V/2) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ2 = W / [(V/2) x g] Dividing the second equation by the first equation gives: ρ2/ρ1 = [(V/2) x g] / [(2V/3) x g] Simplifying, we get: ρ2/ρ1 = 3/4 Therefore, the relative density of the liquid is 0.75. So the correct option is (E) 0.75.

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Heisenberg's uncertainty principle states that it is impossible to simultaneously determine the exact position and momentum of a particle. This principle is a fundamental concept in quantum mechanics, and it means that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The principle applies to all particles, including subatomic particles like electrons and photons. The uncertainty principle has important implications for the behavior of quantum systems and the interpretation of quantum mechanics. It is named after Werner Heisenberg, a German physicist who first described the principle in 1927.

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Hooke's Law states that the force applied to a spring is directly proportional to the extension or compression of the spring, provided the elastic limit is not exceeded. This can be expressed mathematically as F=kx, where F is the force, k is the spring constant, and x is the extension or compression. To determine the length L of the wire when a force of 30N is applied, we need to first find the spring constant k. We can do this by calculating the extension x for each force applied, using the formula: x = (L - L₀) where L₀ is the original length of the wire (500.0 mm). From the data given, we can calculate the extensions as follows: For 0 N: x = (500.0 - 500.0) = 0 mm For 5 N: x = (500.0 - 500.0) = 0 mm For 10 N: x = (501.0 - 500.0) = 1.0 mm Using Hooke's Law, we can find the spring constant k for the wire: k = F/x = 10 N / 1.0 mm = 10 N/mm Now we can use Hooke's Law again to find the extension x for a force of 30 N: 30 N = 10 N/mm * x x = 3.0 mm Finally, we can find the length L of the wire when a force of 30 N is applied: L = L₀ + x = 500.0 mm + 3.0 mm = 503.0 mm Therefore, the length of the wire when a force of 30N is applied is 503.0 mm. Answer is the correct answer.

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When a material is heated, it expands. The increase in volume can be calculated using the following formula: ΔV = VαΔT Where ΔV is the change in volume, V is the original volume, α is the coefficient of linear or cubic expansivity depending on the type of expansion, and ΔT is the change in temperature. For a cubic expansivity, the formula can be written as: ΔV = VβΔT Where Vβ is the coefficient of cubic expansivity. Using the given values, the increase in volume of the brass can be calculated as: ΔV = VβΔT = Vβ(Tf - Ti) Where Tf is the final temperature and Ti is the initial temperature. The initial volume of the brass can be calculated using its mass and density: V = m/ρ = 170/8.5 x 10^3 = 0.02 m^3 Substituting the values, we get: ΔV = Vβ(Tf - Ti) = 0.02 x 5.7 x 10^-3 x (30 - 0) = 3.42 x 10^-5 m^3 Therefore, the increase in volume of the brass is 3.42 x 10^-5 m^3. The correct option is (a) 3.4 x 10^-5 m^3.

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This is a question about the relationship between the temperature and volume of a gas when pressure is held constant, known as Charles' Law. According to Charles' Law, the volume of a gas is directly proportional to its temperature when the pressure is constant. In this question, the length of the column of oxygen in the capillary tube is directly proportional to its volume. This means that if we increase the temperature of the oxygen while keeping the pressure constant, the length of the column of oxygen will increase proportionally. We can use the formula for Charles' Law, V1/T1 = V2/T2, to solve for the final temperature T2. Let V1 be the initial volume of the oxygen when the length of the column is 50cm, T1 be the initial temperature of the oxygen, V2 be the final volume of the oxygen when the length of the column is 60cm, and T2 be the final temperature of the oxygen that we want to find. We know that V1 = V2 since the pressure is constant, and we can substitute the given values into the formula to get: V1/T1 = V2/T2 V1 = 50 cm V2 = 60 cm T1 = 27°C + 273.15 = 300.15 K Solving for T2, we get: T2 = (V2/V1) x T1 = (60/50) x 300.15 = 360.18 K Converting back to Celsius, we get: T2 = 360.18 K - 273.15 = 87.03°C Therefore, the answer is option (D) 87.0^oC.

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As the end Q of the plank is raised, the component of W, normal to the plank will decrease. This is because the weight, W, acting vertically downwards on the plank can be resolved into two components: one perpendicular to the plank and the other parallel to the plank. The component perpendicular to the plank is Wcosθ, where θ is the angle between the weight vector and the normal to the plank. As Q is raised, the angle θ increases, causing the value of cosθ to decrease, hence decreasing the component of W perpendicular to the plank.

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The correct answer is **radiation**. Heat can be transferred through vacuum by radiation, which is the emission of electromagnetic waves from a body due to its temperature. A vacuum flask consists of two walls of glass with a vacuum in between them, and this vacuum is a poor conductor of heat, which reduces heat loss by conduction and convection. However, the inner wall of a vacuum flask can still lose heat by radiation. Hence, to reduce heat loss by radiation, the inside of a vacuum flask is usually coated with a reflective surface, typically silver, which reflects heat back into the flask, keeping the contents warmer for longer.

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Weight is the force experienced by a body due to gravity. It is given by the product of mass and gravitational acceleration, which is 9.8m/s² on earth and 1.6m/s² on the moon. Let's assume the mass of the body to be m kg. On the moon, weight, W_m = mg_m = 6.0N, where g_m = 1.6m/s² is the gravitational acceleration on the moon. On the earth, weight, W_e = mg_e, where g_e = 9.8m/s² is the gravitational acceleration on the earth. We are required to find W_e, given that the gravitational force of the moon is one-sixth that of the earth, which means the gravitational acceleration on the moon is one-sixth that of the earth. So, g_m = (1/6)g_e Substituting this value of g_m in the equation for weight on the moon, we get: W_m = mg_m = m(1/6)g_e = (1/6)(mg_e) Hence, weight of the body on earth, W_e = 6W_m = 6*(1/6)(mg_e) = mg_e Therefore, W_e = 6.0N * 9.8m/s² = 58.8N Hence, the weight of the body on earth is 58.8N. Therefore, the answer is 36.0N.

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The impulse experienced by an object is equal to the change in momentum of the object. Momentum is defined as the product of mass and velocity, i.e., p=mv. When the ball hits the wall, it experiences an impulse that causes a change in its momentum. Initially, the ball is moving towards the wall with a velocity of 2m/s, and the velocity changes direction after the collision, so the final velocity is -2m/s. The change in velocity, Δv = (-2) - (2) = -4m/s. Using the formula for impulse, J = Δp = mΔv, we can calculate the impulse experienced by the ball: J = 5.0kg × (-4m/s) = -20.0kgm/s. Therefore, the impulse experienced by the ball is 20.0 kgm/s. Note that the negative sign indicates that the impulse is in the opposite direction to the initial velocity of the ball.

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The motion of the object can be divided into two components, horizontal and vertical. The horizontal component of the velocity will remain constant, while the vertical component will change due to the acceleration due to gravity acting in the downward direction. We can use the vertical component of the velocity to determine the time taken by the object to reach the highest point. The initial vertical velocity of the object can be calculated using the formula: v₀sinθ = (100m/s)sin60^o = 86.6m/s where v₀ is the initial velocity, and θ is the angle of the velocity vector with the horizontal. At the highest point, the vertical component of the velocity will be zero. Using the formula: v = v₀ + at where v is the final velocity, a is the acceleration due to gravity (-10m/s²), and t is the time taken to reach the highest point, we can calculate the time taken as follows: 0 = 86.6 - 10t t = 8.66s Therefore, the time taken by the object to reach the highest point is approximately 8.7 seconds. The closest option to this answer is (b) 8.7s.

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(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.

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(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.

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(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

1. Open the key when readings are not being taken, so that the wire does not heat up and change its resistance.
2. Ensure that all electrical connections are clean and tight.
3. Read the ammeter and voltmeter at eye level to avoid parallax error.
4. 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

\[V_{\rm lost}=Ir=\frac{8}{9}\times1.0=\boxed{0.89\text{ V}}\]

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(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

(ii) Energy of the incident photon

(iii) Maximum kinetic energy of the photoelectrons