Identify the correct statement if the electron has an initial speed of $4.0 \times {10^6}m/\sec $ is brought to rest by an electric field. It is given that the mass of an electron is $9 \times {10^{ - 31}}kg$ and charge of an electron is $1.6 \times {10^{ - 19}}C$.
(A) The electron moves from the region of low potential to high potential through a potential difference of $11.4\mu V$
(B) The electron moves from the region of high potential to low potential through a potential difference of $11.4\mu V$
(C) The electron moves from the region of low potential to high potential through a potential difference of $45V$
(D) The electron moves from the region of high potential to low potential through a potential difference of $45V$

Hint: The charge which has high concentration of charges is known as higher potential while the charge which has low concentration of charges is known as lower potential.
Formula used The formula used for the calculation of potential difference of the electron is –
$qV = \dfrac{1}{2}m{v^2}$
where, $V$ is the potential difference
$q$ is the charge of particle
$m$ is the mass of particle
$v$ is the velocity of the particle

Complete step by step solution:
When the two bodies are charged and are in contact with each other, the charge starts flowing from one conductor to the other body. The electric condition which tells us the flow of charge from one conductor to the other in contact is known as electric potential. The positively charged conductor has more positive potential than the earth. A negatively charged conductor has negative potential less than that of the earth.
In an electric circuit, when the current flows between two points we consider the charge between those points, so it is not important to know the potential on both the points. It is sufficient to know the potential difference between those points. Therefore, the potential difference between the two points is equal to the work done in moving a unit positive charge from one point to the other.
The charge which has a high concentration of charges is known as higher potential while the charge which has low concentration of charges is known as lower potential.
Now, we have to find the potential difference for the given particle which is an electron. So, let $q$ be the charge of particle $m$ be the mass of particle $v$ be the velocity of the particle
Therefore, according to the question, it is given that –
$q = 1.6 \times {10^{ - 19}}C$
$m = 9 \times {10^{ - 31}}kg$
$v = 4.0 \times {10^6}m/\sec $
Now, we have to use the formula –
$
  qV = \dfrac{1}{2}m{v^2} \\
  \therefore V = \dfrac{1}{2}\dfrac{m}{q}{v^2} \\
 $
Putting the values in the above formula, we get –
$ \Rightarrow V = \dfrac{1}{2} \times \dfrac{{9 \times {{10}^{ - 31}} \times {{\left( {4.0 \times {{10}^6}} \right)}^2}}}{{1.6 \times {{10}^{ - 19}}}}$
Doing further calculations, we get –
$ \Rightarrow V = 45V$
Since it has been stopped it will travel from high potential to low potential.

Hence, the correct option is (D).

Note: We know that direction of electric field is from high potential to lower potential and the direction of the electron is in the direction of electric field, so it flows from higher potential to low potential.