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1. Across what potential difference does an electron have to be accelerated in order to reach the speed v = 9e7 m/s? Non-relativistically V Relativistically V When should you use relativistic calculations? V + Answer: Required speed of electron v = 9 x 107 m/s Charge of electron = 1.6 x 10-19 C Rest mass of the electron m0 = 9.1 x 10-31 kg + + + Consider an arrangement in which an electron is placed in between two plates of potential difference of V volts. Then one of the plates is at negative potential and the other is at positive as shown in figure. As electron is a negatively charged particle, it will get accelerated towards the positive plate. In the potential zone of V volts, it gains an electric potential energy and this potential energy then gets converted into the kinetic energy. v – e– – – – – – – + + + Non-relativistic kinetic energy = ½ mv2. = ½ x 9.1 x 10-31 x (9 x 107 ) = ½ x 9.1 x 10-31 x 81 x 1014 = ½ x 9.1 x 10-31 x 81 x 1014 = 368.55 x 10-17 J Thus Electric Potential Energy = Kinetic energy i.e . ch arg e Potential ----- (1) 1 mass velocity2 2 (Non-relativistic case) 1 2 mv 2 m0 v 2 V 2e i.e. e V i.e. i.e . i.e. 9.1 10 31 9 10 7 V 2 1.6 10 19 i .e. V 2 9.1 10 31 81 1014 9.1 81 10 2 230.34375 volt 19 3.2 3.2 10 Relativistic Case: When an electron moves with the speed comparable to the speed of light, its rest mass can not be considered. The mass of the electron will be more than the its rest mass when it moves at such a high velocity, comparable to the speed of light. In this case, The total energy of the electron = Relativistic Kinetic Energy + Rest Mass Energy Relativistic Kinetic Energy= Total energy of the electron – Rest Mass Energy i.e. EK = mc2 – m0c2 ----- (2) When an electron moves with the speed comparable to the speed of light, its mass m to its rest mass m0 relates by the equation m m0 1 v2 --- (3) c2 Using equation (3) in equation (2), we get EK = mc2 – m0c2 m0c2 m0c2 2 v 1 2 c 1 m0c2 1 v2 1 2 c Substituting the values, we get 1 -31 8 2 E K 9.1 10 3 10 1 2 9 10 7 1 8 2 3 10 1 i.e. E K 9.1 10 -31 9 1016 1 81 1014 1 16 9 10 1 i.e. E K 81.9 10 -3116 1 -2 1 9 10 1 i.e. E K 81.9 10 -15 1 1 0.09 1 i.e. E K 81.9 10 -15 1 0.91 -15 i.e. E K 81.9 10 1.0482 1 i.e. E K 81.9 0.0482 10 -15 = 3.94758 x 10-15 J To achieve this much kinetic energy, the required accelerating potential then can be calculated by using equation (1) as follows. Electric Potential Energy = Relativistic Kinetic energy i.e. i.e. i.e. eV = 3.94758 3.94758 10 -15 3.94758 10 -15 V 2.4672 10 4 volt -19 e 1.6 10 V = 24672 volt Relativistically. When should you use relativistic calculations? Relativistic calculations should be used when the speed of a body or particle under consideration is comparable to that of light. 2. An electron entering Thomson's e/m apparatus (Figure 3.2 and 3.3) has an initial velocity (in horizontal direction only) of 0.7 107 m/s. Lying around the lab is a permanent horseshoe magnet of strength 1.3 10-2 T, which you would like to use. (a) What electric field will you need in order to produce zero deflection of the electrons as they travel through the apparatus? V/m Answer: Velocity of electrons v = 0.7 x 107 m/s Magnetic field induction B = 1.3 x 10 –2 T -19 Charge of electron = 1.6 x 10 C Rest mass of the electron m = 9.1 x 10-31 kg The deflection of the electrons on the screen will be zero when the forces acting on the electrons due to electric and magnetic fields will balance each other. i.e. Electric force i.e. charge x Electric field i.e. eE i.e. E i.e. E E = magnetic force. = charge x velocity x magnetic field induction = e vB = vB = 0.7 x 107 x 1.3 x 10 –2 = 0.91 x 105 volt/meter = 9.1 x 104 V/m [ since every quantity is in SI units and therefore electric field must be expressed in SI unit i.e. in volt/meter or newton/coulomb.] (b) When the magnetic field is turned off, but the same electric field remains, how large a deflection will occur if the region of nonzero E and B fields is 2 cm long? cm Figure 3.2 Figure 3.3 [Relativistic mass of the electron when it passes through the plates at the speed of 0.7 x 107 m/s m0 9.1 10 -31 9.1 10 -31 9.1 10 -31 9.1 10 -31 m 2 0.9997 v2 0.49 1014 1 0.0544 10 -2 0.7 10 7 1 2 1 1 2 c 9 1016 3 10 8 i.e. m 9.1 10 -31 9.10273 10 -31 kg m0 ] 0.9997 Here velocity v of the electron is quite less as compared to the speed of light c. Hence the non-relativistic calculations will also do. In all the calculations then, only rest mass of the electron is used. With only electric field and no magnetic field, electron will experience a force only due to electric field. This electric force eE will produce an acceleration a in the electron in the upward direction. Electron is a negatively charged particle and hence it will be attracted towards the positively charged plate, i.e. it gets deflected upward. This acceleration in the vertical upward direction can be calculated by using Newton’s second law of motion, (i.e. force equation F=ma) as follows. Electric Force = Force according to Newton’s Law i.e. eE = ma eE 1.6 10 19 0.91 10 5 1.6 9.1 10 4 10 19 i.e. a 1.6 1016 m / s 2 31 31 m 9.1 10 9.1 10 The electron will experience this acceleration for time t i.e. for the time till it is in the region of both the plates i.e. in the region of L=2 cm = 2 x 10-2 m. This time t can be obtained as follows. Electron passes through this region at a constant initial speed along the length of the tube. This speed is v = 0.7 x 107 m/s distance 2 10 2 2.8571 10 -9 second Then time required t 7 speed 0.7 10 vy + v0y =0 a vx _ When the electron enters the region of the electric field, its vertical velocity v0y is zero. But the acceleration a in the vertical direction lifts the electron and it starts moving towards positive plate. As a result, vertical velocity of the electron goes on increasing and finally, when it exits the region of the electric field, i.e. the region in between the plates, its vertical velocity becomes vy. vy This velocity vy can be obtained as follows. vy = v0y + at = 0 + 1.6 x 1016 x 2.8571 x 10-9 = 4.5714 x 107 m/s. v vx Then the deflection () experienced by the electron is decided by the values of vy and vx. As a result of the vertical component of the velocity, electron deflects along the angle and moves with the resultant velocity v. tan θ From the figure, vy θ tan 1 vx = 81.29420 vy vx 4.5714 10 7 tan 1 0.7 10 7 4.5714 10 7 tan 1 0.7 10 7 tan 1 6.5306 3.A photon of wavelength 2.2 nm Compton scatters from an electron at an angle of 90°. What is the modified wavelength? (Enter your answer correct to 5 significant figures.) nm What is the percentage change, / ? % Answer: Wavelength Angle of deflection = 2.2 nm = 2.2 x 10-9 m = 900 The modified wavelength ’ relates to the original wavelength of the photon by the relation h 1 cosθ λ'λ ---- (1) m0c where h – Planck’s constant, h = 6.627 x 10-34 Js m0 – rest mass of electron m0 = 9.1 x 10-31 kg c – speed of light c = 3 x 108 m/s – Angle of scattering = 900 Then cos 90 = 0 h 1 0 λ'λ m0c i.e. λ' λ h m0c i.e. Modified wavelength λ' λ h m0c 2.2 10 9 i.e. 6.627 10 34 9.1 10 31 3 10 8 2.2 10 9 0.24275 10 11 2.2 10 9 0.0024275 10 9 2.2024275 10 9 ’ = 2.2024 x 10-9 meter = 2.2024 nm Δλ 100 λ λ'-λ 2.202427 10 9 - 2 10 9 0.202427 100 100 100 9 λ 2 2 10 20.2427 10.12135 % 2 %change in the wavelength = 10.12135% %change = 4. A photon having 27 keV scatters from a free electron at rest. What is the maximum energy that the electron can obtain? keV Answer: Energy of the incident photon = 27 keV = 27 x 103 eV = 27 103 1.6 10 19 joule = 43.2 x 10-16 J E = 4.32 x 10-15 J The energy gained by the electron = Energy of the incident photon – energy of the scattered photon. = h – h’ = h ( –’) ----- (1) Thus for the maximum energy gain by the electron, –’ must be maximum. But is constant ’ must be minimum. From the relation c = ’’, the wavelength of scattered photon ’ must be maximum.. Wavelength of the scattered photon ’ is given by h 1 cosθ ----- (2) m0c Thus wavelength ’ can be maximum only if the term that is to be added to is maximum h 1 cosθ is maximum i.e if m0c i.e. if cos = 0 i.e. if = 900. λ' λ Then max. wavelength of the scattered photon is h λ max λ ----- (3) m0c The frequency () of the photon can be obtained from E = h E 4.32 10 15 i.e. 0.6519 1019 6.519 1018 Hz 34 h 6.627 10 From relation c = , the corresponding wavelength () of the incident photon is c 3 10 8 λ 0.4602 10 10 m 18 ν 6.519 10 Then maximum wavelength of the scattered photon h λ max λ m0c 0.4602 10 10 0.024275 10 10 (the second term taken directly from previous problem calculations) max = 0.484475 x 10-10 m Then the minimum frequency of the scattered radiation c 3 10 8 ν min 6.19227 1018 Hz λ max 0.484475 10 -10 The maximum energy gained by the electron can be obtained from equation (1). E = h ( –’) = h ( –min) = 6.627 x 10-34 (6.519 x 1018 – 6.19227 x 1018) = 6.627 x 10-34 x 0.32673 x 10-18 = 2.1652 x 10-16 joule ---------------- 2.1652 10 16 19 i.e. 1.6 10 E = 1 . 35327 keV OR E = 1353.27 eV 1.35327 10 3 eV