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 16.5 cm, and the voltage difference is 149 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

Let velocity = β
V=voltage diff = 149 * 10^3 V
KE = kinetic energy
mo = Rest mass of Electron = 9.11×10-31 kg
Rest energy of Electron (mo* c^2) = 511Kev
1ev = 1.6 * 10^-19 J
Rest energy of Electron (mo* c^2) = 511 *10^3 * 1.6 * 10^-19 J = 8.176 * 10^-14 J

We know Kinetic Energy Gained by Electron inside a parallel Plate capacitor = V * Charge on electron = 149 * 10^3 * 1.6 * 10^19 = 2.384 * 10^-14 J

Kinetc energy of electron is given by -
KE = (Y-1)moc^2
Where Y = 1/ sqrt(1 - (β/c)²) -----------1
Solving for Y

2.384 * 10^-14  = (Y-1) * 8.176 * 10^-14
Y = 2.384 * 10^-14  / 8.176 * 10^-14 + 1
Y = 1.291
Substituting Value of Y in Equation 1

Y = 1/ sqrt(1 - (β/c)²)
1 - (β/c)² = 1/y²
(β/c)² = 1- 1/y²
β = sqrt( 1- 1/y²) c
β = sqrt (1-1/1.291^2) * 3* 10^8 m/s
β = 1.897 * 10^8 m/s or 0.632c , where c is velocity of light

Final velocity β of the electron using relativistic mechanics β = 1.897 * 10^8 m/s or 0.632c