Question

An electron is accelerated inside a parallel plate capacitor. The electron leaves the negative plate with a negligible initial velocity and then after the acceleration it hits the positive plate with a final velocity β. The distance between the plates is 12.4 cm, and the voltage difference is 122 kV. Determine the final velocity β of the electron using classical mechanics. (The rest mass of the electron is 9.11×10^-31 kg, the rest energy of the electron is 511 keV.)

What is the final velocity β of the electron if you use relativistic mechanics?

Answer 1

using classical mechanics,

Workdone by elctric field = gain in kinetic energy of the electron

q*delta_V = (1/2)*m*beta^2

beta^2 = 2*q*delta_V/m

beta = sqrt(2*q*delta_V/m)

= sqrt(2*1.6*10^-19*122*10^3/(9.11*10^-31))

= 2.07*10^8 m/s <<<<<<<<------------------------------Answer

Using relativistic mechanics

relativistic kinetic energy, KE = Total energy - rest mass energy

KE = m*c^2 - mo*c^2

q*delta_V = mo*c^2/(sqrt(1 - (beta/c)^2) - mo*c^2

e*121k V = 511 keV/sqrt(1 - (beta/c)^2) - 511 keV

121 keV = 511 keV/sqrt(1 - (beta/c)^2) - 511 keV

121 = 511/sqrt(1 - (beta/c)^2) - 511

121 + 511 = 511/sqrt(1 - (beta/c)^2)

sqrt(1 - (beta/c)^2) = 511/(121 + 511)

sqrt(1 - (beta/c)^2) = 0.8085

1 - (beta/c)^2 = 0.8085^2

(beta/c)^2 = 1 - 0.8085^2

beta/c = sqrt(1 - 0.8085^2)

beta/c = 0.5885

beta = c*0.5885

= 3*10^8*0.05885

= 1.76*10^8 m/s <<<<<<<<<<----------------------Answer