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(a) What voltage will accelerate electrons to a speed of 6.00×107m/s6.00 \times 10 ^ { - 7 } \mathrm { m } / \mathrm { 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.

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a) Since the electric force is a conservative force, the total energy is conserved:

K1+U1=K2+U2K_1 + U_1 = K_2+U_2

K1=U2=0\because K_1 = U_2 = 0

K2=U1\therefore K_2= U_1

U1=eV\because U_1 =eV and K2=12mv2K_2 = \dfrac{1}{2}mv^2

eV=12mv2\therefore eV = \dfrac{1}{2}mv^2

Rearranging:

V=mv22e=(9.1×1031)(6×107)22(1.6×1019)=1.024×1024VV=\dfrac{mv^2}{2e}=\dfrac{(9.1 \times 10^{-31})(6 \times 10^{-7})^2}{2(1.6 \times 10^{-19})}= 1.024 \times 10^{-24} \, \, \mathrm{V}

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