Answer:
![1.33\cdot 10^7 m/s](https://img.qammunity.org/2021/formulas/chemistry/high-school/8azwh0v80mka8qp99n2x4t6qorybev2df7.png)
Step-by-step 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:
![K_f-K_i = -q\Delta V](https://img.qammunity.org/2021/formulas/chemistry/high-school/907cowuo40ntpbzsantmmcyz0fzloc6lmz.png)
where
is the final kinetic energy
is the initial kinetic energy
q is the charge of the particle
is the potential difference
In this problem,
is the charge of the electron
![\Delta V=-500 V-(-1000 V)=500 V](https://img.qammunity.org/2021/formulas/chemistry/high-school/alfu2v55jxnbwv7sjz3vxhqcr2qgdc2c6k.png)
The electron starts from rest, so its initial kinetic energy is
![K_i=0](https://img.qammunity.org/2021/formulas/physics/high-school/z21cpewl35vqba01s8tg3emapcsjmaq0hx.png)
Therefore,
![K_f=-(-1.6\cdot 10^(-19))(500)=8\cdot 10^(-17)J](https://img.qammunity.org/2021/formulas/chemistry/high-school/mqxskdp9c116gtey1ubmv3fylzv9e0alql.png)
We can write the final kinetic energy of the electron as
![K_f=(1)/(2)mv^2](https://img.qammunity.org/2021/formulas/physics/college/9fbdshn8186g6e26o9z0p6dcfzyofnpxlj.png)
where
is the electron mass
v is the final speed
And solving for v,
![v=\sqrt{(2K_f)/(m)}=\sqrt{(2(8\cdot 10^(-17)))/(9.11\cdot 10^(-31))}=1.33\cdot 10^7 m/s](https://img.qammunity.org/2021/formulas/chemistry/high-school/mqey686te68ynegjst93k2ot49nyg8mwz8.png)