Question

A.) An electron is to be accelerated from a velocity of 3.00×106 m/s to a velocity of 8.50×106 m/s . Through what potential difference must the electron pass to accomplish this?

V(initial)-V(final) = ______ V

B.) Through what potential difference must the electron pass if it is to be slowed from 8.50×10⁶ m/s to a halt?

V(initial)-V(final) = ______ V

Answer 1

electron aquires kinetic energy as it accelerates through potential difference.

Work done by potential difference = change in kinetic energy

eV = final kinetic energy - initial kinetic energy

so

(a)

putting appropriate values of m (mass of electron) = 9.11E-31 kg, e (charge on electron) = (-1.6e-19) C and v1= 3E6 m/s, v2 = 8.5E6 m/s:

we get potential diff V = -180.06 volts

(b)

in above formula put V1 = 8.5E6 m/s and v2 = 0 m/s

required potential diff. V = 205 volts