(a) An electron is to be accelerated from 3.00 x 10⁶ m/s to 8.00 x 10⁶ m/s. Through what potential difference must the electron pass to accomplish this?

(b) Through what potential difference must the electron pass if it is to be slowed from 8.00 x 10⁶ m/s to a halt?

A positive charge creates positive potential, but if it is negative, then the potential will be negative.

Necessary data:

$$\begin{gathered} {\mathrm{q}}_{\mathrm{e}}=-1.0621\times {10}^{-19}\mathrm{C} \\ {\mathrm{m}}_{\mathrm{e}}=9.101\times {10}^{-31}\mathrm{kg} \\ {\mathrm{v}}_1=3\times {10}^6\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2=8\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Electric potential is equal to the potential energy per unit charge,

$$\begin{gathered} V=\frac{U}{q_0} \\ U=V{q}_0 \end{gathered}$$

By the Law of conservation of energy,

$$\begin{gathered} E_i=E_f \\ {U}_i+K_i=U_f+K_f \\ {V}_1{q}_0+\frac{1}{2}m{v}_1^2=V_2{q}_0+\frac{1}{2}m{v}_2^2 \\ {q}_0\left(V_1-V_2\right)=\frac{1}{2}m\left(v_2^2+v_1^2\right) \\ \left(V_1-V_2\right)=\frac{{\displaystyle \frac{1}{2}}m\left(v_2^2+v_1^2\right)}{q_0} \\ ∆V=\frac{{\displaystyle \frac{1}{2}}\left(9.101\times {10}^{-31}\right)\left({\left(8\times {10}^6\right)}^2-{\left(3\times {10}^6\right)}^2\right)}{-1.602\times {10}^{-19}} \\ ∆V=-156.355V \end{gathered}$$

Therefore, the potential difference required to accelerate the electron from V₁ to V₂ is -156.355 volts.

By the Law of conservation of energy,

$$\begin{gathered} E_i=E_f \\ {U}_i+K_i=U_f+K_f \\ {V}_1{q}_0+\frac{1}{2}m{v}_2^2=V{q}_0+\frac{1}{2}m{v}^2 \\ {q}_0\left(V_2-V\right)=\frac{1}{2}m\left(v^2+v_2^2\right) \\ \left(V_1-V_2\right)=\frac{{\displaystyle \frac{1}{2}m\left(v^2+v_2^2\right)}}{q_0} \\ ∆V=\frac{{\displaystyle \frac{1}{2}\left(9.101\times {10}^{-31}\right)\left({\left(0\right)}^2-{\left(8\times {10}^6\right)}^2\right)}}{-1.602\times {10}^{-19}} \\ ∆V=+181.94V \end{gathered}$$

Therefore, the potential difference required to deaccelerate the electron from V₂to 0 is 181.94 volts.