a) What voltage will accelerate electrons to a speed of 6.00*10-7m/s? (b) Find the radius of curvature of the path of a proton accelerated through this potential in a 0.500-T field and compare this with the radius of curvature of an electron accelerated through the same potential.

a) V = 1.024*10∧-24V

b) Radius of curvature for electron = 6.825*10∧-18 m

c) For proton r = 2.92*10∧-16 m

Explanation:

1. The total energy for the electric force is conserved

k1+u1=k2+u2

k1=u2=0

∴ k2=u1

∴ u1=㏑V, k∨2=1/2 mv²

Rearranging

V=mv²/2e = (9.1*10∧-31) * (6*10∧-7) ²/2 (1.6*10∧-19) = 1.024*10∧-24V

2. The radius of curvature for an electron accelerated in that potential is given by r=mv/qB = (9.1*10∧-31) * (6*10∧-7) / (1.6*10∧-19) (0.5) = 6.825*10∧-18 m

3. For a proton accelerated in the same potential

ev=1/2mv² = √2 (1.6*10∧-19) (1.024*10∧-24V) / 1.67*10∧-27 = 1.401*10∧-8 ms-1

The radius of curvature for the proton is r=mv/qB

1.67*10∧-27*1.401*10∧-8 ms-1/1.6*10∧-19*0.5 = 2.92*10∧-16 m