Question

What is the final velocity of an electron accelerating through a potential of \[{\text{1600 V}}\] if its initial velocity is zero?
A) \[{\text{2}}{\text{.37}} \times {\text{1}}{{\text{0}}^7}{\text{ m/s}}\]
B) \[{\text{2}}{\text{.37}} \times {\text{1}}{{\text{0}}^8}{\text{ m/s}}\]
C) \[{\text{2}}{\text{.37}} \times {\text{1}}{{\text{0}}^6}{\text{ m/s}}\]
D) \[{\text{2}}{\text{.37}} \times {\text{1}}{{\text{0}}^5}{\text{ m/s}}\]

Answer 1

Hint: Obtain the final velocity of the electron by equating the equations for the kinetic energy of the moving electron and the energy of the electron accelerated by applying a potential.

Complete step by step answer:
In the question, the potential through which an electron is accelerated is given. The initial velocity is zero. You have to find the final velocity of the electron.
When an electron is accelerated by applying a potential, the energy of the electron is given by the expression \[{\text{E = eV }}...{\text{ }}...\left( 1 \right)\].
Here, E is the energy of the electron, e is the charge on the electron and V is the potential through which the electron is accelerated.
When an electron is moving with a speed u, its kinetic energy is given by the expression \[{\text{E = }}\dfrac{1}{2}{\text{m}}{{\text{u}}^2}{\text{ }}...{\text{ }}...\left( 2 \right)\].
Here, E is the kinetic energy, m is the mass of the electron and u is the velocity of the electron.
But the energies represented by two equations (1) and (2) are the same.
\[{\text{eV = }}\dfrac{1}{2}{\text{m}}{{\text{u}}^2} \\
{{\text{u}}^2} = \dfrac{{{\text{2eV}}}}{{\text{m}}} \\
{\text{u}} = \sqrt {\dfrac{{{\text{2eV}}}}{{\text{m}}}} \\\]
Substitute values in the above expression and calculate the final velocity of the electron.
\[{\text{u}} = \sqrt {\dfrac{{{\text{2eV}}}}{{\text{m}}}} \\
= \sqrt {\dfrac{{{\text{2}} \times {\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 19}} \times {\text{1600}}}}{{9.1 \times {{10}^{ - 31}}}}} \\
= 2.37 \times {10^7}{\text{ m/s}} \\\]
Thus, the final velocity of an electron is \[2.37 \times {10^7}{\text{ m/s}}\].

Hence, the option A ) is the correct option.

Note: When an electric field is applied to the electron at rest, the electron is accelerated. The kinetic energy of the electron after acceleration is equal to the energy of the electron in presence of the applied electric field.