Electron - General Physics - Solved Exam, Exams of Physics

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(i) What is thermionic emission? It is the emission of electrons from the surface of a hot metal. (ii) X-rays are produced when high-energy electrons collide with a target. Draw a labelled diagram of an X-ray tube. Vacuum Cathode Target / anode High (accelerating anode) voltage or H.T. / shielding / cooling / low (cathode) voltage or L.T. (iii) What are X-rays? X-rays are electromagnetic radiation of high frequency / short wavelength (iv) How do they differ from light rays? X-rays penetrate matter / cause ionization. (v) Give two uses of X-rays. (Medical) analysis of bone structure/ luggage scanners (at airports) / any specific medical, industrial or security use, etc. (vi) When electrons hit the target in an X-ray tube, only a small percentage of their energy is converted into X-rays. What happens to the rest of their energy. The energy gets converted to heat. (vii) How does this influence the type of target used? The target material must have (very) high melting point. (viii) A potential difference (voltage) of 40 kV is applied across an X-ray tube. Calculate the maximum energy of an electron as it hits the target. W = qV W = (1.6 × 10-19)( 40 × 10^3 ) W = 6.4 × 10-15^ J (ix) Calculate the frequency of the most energetic X-ray produced. E = hf f = E/h f = (6.4 × 10-15)/(6.6 × 10-34) f = 9.7 × 10^18 Hz

(i) What is a photon? A photon consists of a discrete (specific) amount of energy/electromagnetic radiation. (ii) Draw a labelled diagram of the structure of a photocell. See diagram; A = cathode, B = anode Also label glass case and vacuum inside (iii) Using the graph, calculate the work function of the metal. The graph indicates that current only flows when the frequency of the radiation reached 5.2 × 10^14 Hz, so this corresponds to the threshold frequency (f 0 ). φ = hf 0 = (6.6 × 10-34)(5.2 × 10^14 ) = 3.432 × 10-19^ J (iv) What is the maximum speed of an emitted electron when light of wavelength 550 nm is incident on the photocell? h(c/λ) = φ + ½mv^2 (6.6 × 10-34)(3 × 10^8 /550 × 10-9) = 3.432 × 10-19^ + ½ (9.1× 10-31)(v)^2  v = 1.922 × 10^5 m s- (v) Explain why a current does not flow in the photocell when the frequency of the light is less than 5. × 10^14 Hz****. Because the frequency is less than the threshold frequency so does not contain enough energy to cause an electron to be released from an atom. (vi) The relationship between the current flowing in a photocell and the intensity of the light incident on the photocell was then investigated. Readings were taken and a graph was drawn to show the relationship. Draw a sketch of the graph obtained. Current is directly proportional to Intensity, so a straight line graph with the line going through the origin is required. (vii) How was the intensity of the light varied? Vary the distance from the light source to the photocell. (viii) What conclusion about the nature of light can be drawn from these investigations? Light is made up of discrete amounts of energy called photons.

(i) The SI unit is named in honour of Lord Kelvin. What is the temperature of the boiling point of water in kelvin? 273.15 + 100 = 373.15 K (ii) Define the newton, the unit of force. A force of 1 N gives a mass of 1 kg an acceleration of 1 m s-1. (iii) A force of 9 kN is applied to a golf ball by a golf club. The ball and club are in contact for 0.6 ms. Using Newton’s laws of motion, calculate the change in momentum of the ball. From Newton II: Force ∝ rate of change of momentum F ∝ (mv – mu)/t F = (mv – mu)/t {proportional constant = 1} (mv – mu) = F × t = (9 × 10^3 )( 0.6 × 10-3) = 5.4 kg m s-1. (iv) Nam three different electromagnetic radiations. x-rays, microwaves, ultra-violet etc. (v) What is the photoelectric effect? The Photoelectric Effect is the emission of electrons from a metal due to light of a suitable frequency falling upon it. (vi) Why as the quantum theory of light revolutionary? light has a particle nature (as well as a wave nature) (vii) High-energy radiation of frequency 3.3 × 10^14 Hz is used in medicine. What is the energy of a photon of this radiation? E = h f E = (6.6 × 10–34)( 3.3 × 10^14 ) = 2.18 × 10–19^ J. (viii) 100 MJ of energy are released in a nuclear reaction. Calculate the loss of mass during the reaction. E = mc^2  m = E/c^2  m = (1 × 10^8 ) / (9 × 10^16 ) = 1.11 × 10-9^ kg.

2006 Question 12 (d) (i) What are X-rays? High frequency electromagnetic radiation. (ii) Who discovered them? Rontgen (iii)In an X-ray tube electrons are emitted from a metal cathode and accelerated across the tube to hit a metal anode. How are the electrons emitted from the cathode? By thermionic emission. (iv) How are the electrons accelerated? By the high voltage between the anode and cathode. (v) Calculate the kinetic energy gained by an electron when it is accelerated through a potential difference of 50 kV in an X-ray tube. Ek (= W) = q V = (1.6 × 10-19)(50 × 10^3 ) = 8.0 × 10-15^ J (vi) Calculate the minimum wavelength of an X-ray emitted from the anode. E = h c/λ λ = [6.6 × 10-34^ × 3.0 × 10^8 ] /(8.0 × 10-15)

2005 Question 12 (d) (i) One hundred years ago, Albert Einstein explained the photoelectric effect. What is the photoelectric effect? The photoelectric effect is the emission of electrons from the surface of a metal when light of suitable frequency / energy shines on it. (ii) Write down an expression for Einstein’s photoelectric law. h f = φ + ½ mv^2 (iii)Summarise Einstein’s explanation of the photoelectric effect. Light is composed of packets (or bundles) of energy which he called photons. All of energy from one photon is given to one electron. Energy must be greater than the work function of the metal for the photoelectric effect to occur. (iv) Give one application of the photoelectric effect. Sound track in film, photography, counters, photocell, burglar alarm, automatic doors, etc.

2004 Question 9 (i) Distinguish between photoelectric emission and thermionic emission. Photoelectric Effect: Emission of electrons when light of suitable frequency falls on a metal. Thermionic Emission: Emission of electrons from the surface of a hot metal. (ii) A freshly cleaned piece of zinc metal is placed on the cap of a negatively charged gold leaf electroscope and illuminated with ultraviolet radiation. Explain why the leaves of the electroscope collapse. Photoelectric emission occurs (electrons get emitted from the surface of the metal). The leaves become uncharged and therefore collapse. (iii) Explain why the leaves do not collapse when the zinc is covered by a piece of ordinary glass. Ordinary glass does not transmit UV light (iv) Explain why the leaves do not collapse when the zinc is illuminated with green light. The energy associated with photons of green light is too low for the photoelectric effect does not occur, so no electrons are emitted from the electroscope. (v) Explain why the leaves do not collapse when the electroscope is charged positively. Any electrons emitted are attracted back to the positive electroscope. (vi) The zinc metal is illuminated with ultraviolet light of wavelength 240 nm. The work function of zinc is 4.3 eV. Calculate the threshold frequency of zinc. φ = (4.3) eV = (4.3)( 1.6 × 10–19) J E = hf 0  f 0 = (4.3)( 1.6 × 10–19)/ 6.6 × 10–34^ = 1.04 × 10^15 Hz (vii) Calulate the maximum kinetic energy of an emitted electron. c = fλ  f = c/λ= (3.0 × 10^8 )/(240 × 10^9 ) = 1.25 × 10^15 Hz Ek = hf - φ Ek = (6.6 × 10–34)[( 1.25 × 10^15 ) - 1.04 × 10^15 )] = 1.39 × 10-

(i) List two properties of the electron. Negative charge, negligible mass , orbits nucleus, deflected by electric / magnetic field etc. (ii) Name the Irishman who gave the electron its name in the nineteenth century. George Stoney (iii) Give an expression for the force acting on a charge q moving at a velocity v at right angles to a magnetic field of flux density B****. F = Bqv (iv) An electron is emitted from the cathode and accelerated through a potential difference of 4kV in a cathode ray tube (CRT) as shown in the diagram. How much energy does the electron gain? The final kinetic energy gained by the electron equals the initial (electrical) potential energy = 4 KeV = (4000)(1.6 × 10–19) = 6.4×10−^16 J (v) What is the speed of the electron at the anode? (Assume that the speed of the electron leaving the cathode is negligible.) E = ½ mv^2  6.4 ×10-16^ = ½ (9.1 × 10-31)(v^2 )  v =3.75 × 10^7 m s- (vi) After leaving the anode, the electron travels at a constant speed and enters a magnetic field at right angles, where it is deflected. The flux density of the magnetic field is 5 × 10–2^ T. Calculate the force acting on the electron. F = Bev = (5 × 10–2)(1.6 × 10–19)( 3.75 × 10^7 ) = 3.0 ×10−^13 N (vii) Calculate the radius of the circular path followed by the electron, in the magnetic field. F = mv^2 /r  3.0 ×10−^13 = (9.1 × 10–31)( 3.75 × 10^7 )^2 /r  r = 4.3×10−^3 m (viii) What happens to the energy of the electron when it hits the screen of the CRT? It gets converted to light.

2002 Question 9 (i) Explain with the aid of a labelled diagram how X-rays are produced. A = Hot cathode B = target Also label vacuum, shield, cooling, window (ii) Justify the statement “X-ray production may be considered as the inverse of the photoelectric effect.” X-ray: Electrons in, electromagnetic radiation is emitted. Photoelectric: electromagnetic radiation in and electrons are emitted. (iii) Describe an experiment to demonstrate the photoelectric effect. Apparatus: gold leaf electroscope with zinc plate on top, ultraviolet light source Procedure: Charge the electroscope negatively. Shine ultraviolet light on the zinc plate. Observation: The leaves fall together (iv) Outline Einstein’s explanation of the photoelectric effect.

• Electromagnetic radiation consists of packets of energy which he called quanta.
• The amount of energy in each quantum could be calculated using the formula E = hf
• The incoming radiation needs to have sufficient energy to release an electron from the surface of a metal or it won’t be absorbed at all.
• If the incoming energy has more energy than an electron needs to be released (called its work function) then the excess energy goes to the kinetic energy of the electron. (v) Give two applications of a photocell. Burglar alarm, smoke alarms, safety switch. light meters, automatic lights, counters, automatic doors, control of central heating burners, sound track in films, scanner reading bar codes, stopping conveyer belt