Question

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10^?2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 280 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.

Answer 1

distance to travel=difference between the radii

=0.558-6.2*0.01 cm

=0.496 cm=496*10^(-5) m

potential difference=280 volts

then electric field=potential difference/distance

=5.6452*10^4 V/m

force on the electron=q*E

where q=magnitude charge on electron=1.6*10^(-19) C

E=electric field

acceleration=force/mass

=q*E/m

where m=mass of electron=9.1*10^(-31) kg

distance to be covered=d=496*10^(-5) m

using the formula:

final speed^2-initial speed^2=2*acceleration*distance

we get final speed=9.9228*10^6 m/s