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 11.0 cm, and the voltage difference is 147 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

here,

Assuming Notations :

v = velocity = β
V = voltage diff
KE = kinetic energy
Y = gamma
m = mass
q = charge
1 ev = 1.6*10^-19 J

By Classical Method Final Velocity of electron will be
v = sqrt(-(2*q*V)/(m) )
v = sqrt(-2( -1.6*10^-19*147000 ) / (9.11*10^-31 ) )
v = 2.27*10^8 m/s

as 1 m/s = 3.335* 10^-9 c

V = 0.757 c = β

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
By Relativistic Method Final Velocity of electron will be :
if speeds higher than > 1/10c (10% speed of light), we have to use relativity for accuracy.

as, KE = (Y-1)m*c^2

solving for gamma, Y
Y = (KE / m*c^2) + 1
Y = ( (511000 * 1.60*10^-19 ) / ( (9.11*10^-31)(3*10^8)^2 ) + 1
Y = 1.997

Therefore Velocity will be ,
v = sqrt( 1- (1/Y^2) )*c
v = sqrt( 1- (1/(1.997)^2) )*(3*10^8)
v = 2.59*10^8 m/s

as 1 m/s = 3.335* 10^-9 c

V = 0.864 c = β