Answer
The fraction of the electron's initial kinetic energy that will be lost is $~~0.113$
Work Step by Step
We can find the magnitude of the force on the electron:
$F = q~E$
$F = (1.6\times 10^{-19}~C)(1.00\times 10^3~N/C)$
$F = 1.6\times 10^{-16}~N$
We can find the magnitude of deceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{1.6\times 10^{-16}~N}{9.109\times 10^{-31}~kg}$
$a = 1.7565\times 10^{14}~m/s^2$
We can find the electron's velocity after decelerating over a distance of $8.00~mm$:
$v_f^2 = v_0^2+2ax$
$v_f = \sqrt{v_0^2+2ax}$
$v_f = \sqrt{(5.00\times 10^6~m/s)^2+(2)(-1.7565\times 10^{14}~m/s^2)(8.0\times 10^{-3}~m)}$
$v_f = 4.71\times 10^6~m/s$
We can find the fraction of kinetic energy that remains:
$\frac{K_f}{K_0} = \frac{\frac{1}{2}m(4.71\times 10^6~m/s)^2}{\frac{1}{2}m(5.00\times 10^6~m/s)^2}$
$\frac{K_f}{K_0} = \frac{(4.71\times 10^6~m/s)^2}{(5.00\times 10^6~m/s)^2}$
$\frac{K_f}{K_0} = 0.887$
The fraction of the electron's initial kinetic energy that will be lost is $~~0.113$
