91Ó°ÊÓ

The separation between the plates,$\mathrm{d}=1.\text{0cm}$.

The potential difference, $\mathrm{Δ³Õ}=\text{625V}$.

According tothe law of conservation of energy can neither be created nor destroyed only converted from one form of energy to another.

Formulae:

The change in potential energy is define as,

$\mathrm{Δ\pm «}=\mathrm{Δ³Õ}×\mathrm{e}$

The kinetic energy is,

$\mathrm{Δ°­}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2$

Where, $\mathrm{K}$ is kinetic energy, $\mathrm{m}$ is mass, $U$ is potential energy, and $\mathrm{v}$ is velocity. and $\mathrm{V}$ is a potential difference.

Apply the Law of Conservation of Energy to get the initial speed${\mathrm{V}}_{\mathrm{i}}\text{m/s}$at the first plate,

$\begin{array}{c}({\mathrm{K}}_{\mathrm{f}}−{\mathrm{K}}_{\mathrm{i}})=−\mathrm{Δ\pm «}\\ =−\mathrm{Δ³Õ}×\mathrm{e}\end{array}$

Here,

The electron charge, $\mathrm{e}=1.6×{10}^{−19}\text{C}$

Substitute known values in the above equation.

$0−\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2=−625\text{V}×1.6×{10}^{−19}\text{C}$

Solving for ${\mathrm{v}}_{\mathrm{i}}$,

$\begin{array}{c}{\mathrm{v}}_{\mathrm{i}}=\sqrt{\frac{\left(625\text{V}\right)×(1.6×{10}^{−19}\text{C})×2}{9.1×{10}^{−31}\text{kg}}}\\ =\sqrt{\text{219.78}×{10}^{12}{\text{m}}^2/{\text{s}}^2}\\ =1.48×1{\text{0}}^{\text{7}}\text{m}/\text{s}\end{array}$.

Hence, the initial speed of the electron is $1.48×1{\text{0}}^{\text{7}}\text{m}/\text{s}$.