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$(D)$ The work done by the electric field on the electron is equal to the change in its kinetic energy.According to the work-energy theorem, $W = \Del

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The electron moves from a region of higher potential to lower potential through a potential difference of $45 \, V$.

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$(D)$ The work done by the electric field on the electron is equal to the change in its kinetic energy.According to the work-energy theorem, $W = \Delta K = K_f - K_i$.Since the electron is brought to rest, $K_f = 0$, so $W = -K_i = -\frac{1}{2}mv^2$.The work done by an electric field is $W = -qV$, where $V$ is the potential difference.Equating the two, $-qV = -\frac{1}{2}mv^2$, which gives $V = \frac{mv^2}{2q}$.Substituting the given values: $V = \frac{9 	imes 10^{-31} 	imes (4.0 	imes 10^6)^2}{2 	imes 1.6 	imes 10^{-19}}$.$V = \frac{9 	imes 10^{-31} 	imes 16 	imes 10^{12}}{3.2 	imes 10^{-19}} = \frac{144 	imes 10^{-19}}{3.2 	imes 10^{-19}} = 45 \, V$.Since an electron is negatively charged, it gains potential energy when moving from higher potential to lower potential, which results in a loss of kinetic energy. Thus, it moves from a region of higher potential to lower potential.

An electron with an initial speed of $4.0 \times 10^6 \, m/s$ is brought to rest by an electric field. The mass and charge of an electron are $9 \times 10^{-31} \, kg$ and $1.6 \times 10^{-19} \, C$, respectively. Identify the correct statement.

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