The electron beam inside a television picture tube is 0.40 mm in diameter and ca
ID: 2005227 • Letter: T
Question
The electron beam inside a television picture tube is 0.40 mm in diameter and carries a current of 50 A. This electron beam impinges on the inside of the picture tube screen.How many electrons strike the screen each second?
The electrons move with a velocity of 4.0 X 10^7 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0mm?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)
Explanation / Answer
a) charge of electron e = 1.6*10-19 C 1C = electron/1.6*10-19 C = 6.25*1018 electrons number of electrons strike the screen each second ps n = (6.25*1018 electrons)(50 A) = 31.25*1019 electrons/s b) energy = (1/2) mv2 = (0.5)(9.11*10-31 kg)(4*107 m/s)2 = 72.88*10-17 J = [(72.88*10-17) /(1.6*10-19)] eV = 4555 eV so each electron uses the voltage V = 4555 V to its acceleration and get velocity v = 4*107 m/s. electric field E = voltage/ separation = (4555)/(5*10-3) = 911*103 V/m c) power P = IV = (50)(4555) = 2.27*105 W = (0.5)(9.11*10-31 kg)(4*107 m/s)2 = 72.88*10-17 J = [(72.88*10-17) /(1.6*10-19)] eV = 4555 eV so each electron uses the voltage V = 4555 V to its acceleration and get velocity v = 4*107 m/s. electric field E = voltage/ separation = (4555)/(5*10-3) = 911*103 V/m c) power P = IV = (50)(4555) = 2.27*105 WRelated Questions
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