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

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.