Answer:
[tex]1.33\cdot 10^7 m/s[/tex]
Explanation:
For a charged particle accelerated by an electric field, the kinetic energy gained by the particle is equal to the decrease in electric potential energy of the particle; therefore:
[tex]K_f-K_i = -q\Delta V[/tex]
where
[tex]K_f[/tex] is the final kinetic energy
[tex]K_i[/tex] is the initial kinetic energy
q is the charge of the particle
[tex]\Delta V[/tex] is the potential difference
In this problem,
[tex]q=-1.6\cdot 10^{-19}C[/tex] is the charge of the electron
[tex]\Delta V=-500 V-(-1000 V)=500 V[/tex]
The electron starts from rest, so its initial kinetic energy is
[tex]K_i=0[/tex]
Therefore,
[tex]K_f=-(-1.6\cdot 10^{-19})(500)=8\cdot 10^{-17}J[/tex]
We can write the final kinetic energy of the electron as
[tex]K_f=\frac{1}{2}mv^2[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron mass
v is the final speed
And solving for v,
[tex]v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(8\cdot 10^{-17})}{9.11\cdot 10^{-31}}}=1.33\cdot 10^7 m/s[/tex]
