Answer:
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
