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 14.2 cm, and the voltage difference is 146 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.)
7.559×10^-1

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

Answer 1

Let velocity = β

V=voltage diff = 146 * 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 = 146 * 10^3 * 1.6 * 10^19 = 2.336* 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 = 2.336 * 10^-14  / 8.176 * 10^-14 + 1

Y = 1.285

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 (1-1/1.285^2) * 3* 10^8 m/s

β = 1.884 * 10^8 m/s or 0.628c , where c is velocity of light

Final velocity β of the electron using relativistic mechanics β = 1.884 * 10^8 m/s or 0.628c