The electron gun in a TV picture tube accelerates electrons between two parallel
ID: 2047629 • Letter: T
Question
The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.1 { m cm} apart with a 27 { m kV} potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.
With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: the exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big.
Express your answer using two significant figures.
Vexit = m/s
Explanation / Answer
The electrons accelerated by this gun are not relativistic. You need electrons well over 100 keVs in order to have significant relativistic corrections. The speed of the electrons accelerated by this gun is just KE = 1/2 mV^2 solving for V V = sqrt( 2 KE / m) substituting values we get V = sqrt(2*27E3 *1.6E-19/9.11E-31) = 9.74E+07 m/s where I used the conversion 1 eV = 1.6E-19 Joules. The value for the speed is less than 1/3 the speed of light so a relativistic correction is really not necessary. If you insist on treating these electrons as relativistic, then you have to use the following equation for the kinetic energy KE = mc^2 - moc^2 = mo c^2( 1/sqrt(1-V^2/c^2) - 1) solving for V we get V = c sqrt(1 - 1/ (KE / mo c^2 +1)^2) substituting values V = 3E8*sqrt(1 - 1/ (27E3*1.6E-19 / (9.11E-31*3E8*3E8) +1)^2) V = 9.37E+07 m/s as you can see, the relativistic value for the speed is just about the same quantity as the classical so relativistic corrections are not necessary. ah! the electric field strength is E = 27/1.5E-2 = 1800 kV/m
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