Answer:
Approximately [tex]1.8 \times 10^{-18}\; {\rm J}[/tex].
Explanation:
Since the speed of this electron is significantly lower than the speed of light (around [tex]0.015[/tex],) the kinetic energy of this electron can be approximated with the equation:
[tex]\displaystyle (\text{KE}) = \frac{1}{2}\, m\, v^{2}[/tex],
Where:
Substitute in the values and evaluate to obtain:
[tex]\begin{aligned} (\text{KE}) &= \frac{1}{2}\, m\, v^{2} \\ &\approx \frac{1}{2}\, \left(9.109 \times 10^{-31}\; {\rm kg})\, (2.0 \times 10^{6}\; {\rm m\cdot s^{-1}})^{2} \\ &\approx 1.82 \times 10^{-18}\; {\rm J}\end{aligned}[/tex].
In other words, the kinetic energy of this electron would be approximately [tex]1.82 \times 10^{-18}\; {\rm J}[/tex].
