# An electron with an initial speed of 6.00 x10^5 m/s is brought to rest by an electric field. Did the electron move into a region of higher potential or lower potential? What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?

This article aims to find an electron’s stopping potential difference and initial kinetic energy. The negative potential of the collector plate at which the photoelectric current becomes zero is called the stopping potential or threshold potential. The stopping potential is the value of the retarding potential difference between the two plates that is sufficient to stop the most efficient photoelectrons from being emitted. It is marked Vo.

1. The stopping potential does not depend on the intensity of the incident radiation. As the intensity increases, the value of the saturation current increases, while the stopping potential remains unchanged.
2. The stopping potential depends on the frequency of given radiation intensity.

Kinetic energy

In physics, an object’s kinetic energy is its energy as a result of its motion. It is the work required to accelerate a body of a given mass from rest to its given speed. After a body has acquired this energy during its acceleration, it maintains this kinetic energy unless its velocity changes. The body does the same amount of work in decelerating from its current speed to a state of rest.

Formula for the kinetic energy with mass $m$ and speed $v$ is given as:

$K.E=\dfrac{1}{2}mv^{2}$

Given Data:

The amount of charge is given as:

$e=1.602\times 10^{-19}C$

Mass of the electron is:

$m=9.11\times 10^{-31}kg$

Part(a)

The electron moves to a region of the lower potential because it must move in the opposite direction of the force to rest.

Part(b)

The stopping potential difference for the electron is:

$\dfrac{mv^{2}}{2}=-q\Delta V$

$\Delta V=\dfrac{mv^{2}}{2e}$

Plug the values:

$\Delta V=\dfrac{(9.11\times 10^{-31}kg)(6.00\times 10^{5}\dfrac{m}{s})^{2}}{2(1.602\times 10^{-19}C)}$

$=102.4\times10^{-2}V$

$=1.02 V$

Part(c)

Initial kinetic energy of the electron is given as:

$\Delta K=\dfrac{mv^{2}}{2}$

$=\dfrac{(9.11\times 10^{-31}kg)(6.00\times 10^{5}\dfrac{m}{s})^{2}}{2}$

$=1.64\times 10^{-19}J$

$=1.64\times 10^{-19}J(\dfrac{1eV}{1.602\times 10^{-19}J})$

$=1.02eV$

The kinetic energy of electrons in electron volt is $\Delta K=1.02eV$

## Numerical Result

1. Electron moves in the region of lower potential.
2. The stopping potential difference for the electron is $\Delta V=1.02 V$
3. The kinetic energy of the electron is $\Delta K=1.02eV$

## Example

Electron with initial velocity $10 \times 10^{5}\dfrac{m}{s}$ is brought to rest by an electric field.

1. Did the electron move to a region of higher potential or lower potential?
2. What potential difference stopped the electron?
3. Calculate the initial kinetic energy of the electron in electron volts?

Solution

Given Data:

The amount of charge is given as:

$e=1.602\times 10^{-19}C$

Mass of the electron is:

$m=9.11\times 10^{-31}kg$

Part(a)

Electron moves to a region of lower potential because it must move in the opposite direction of the force to rest.

Part(b)

The stopping potential difference for the electron is:

$\dfrac{mv^{2}}{2}=-q\Delta V$

$\Delta V=\dfrac{mv^{2}}{2e}$

Plug the values:

$\Delta V=\dfrac{(9.11\times 10^{-31}kg)(10\times 10^{5}\dfrac{m}{s})^{2}}{2(1.602\times 10^{-19}C)}$

$=2.84 V$

Part(c)

Initial kinetic energy of the electron is:

$\Delta K=\dfrac{mv^{2}}{2}$

$=\dfrac{(9.11\times 10^{-31}kg)(10\times 10^{5}\dfrac{m}{s})^{2}}{2}$

$=4.55\times 10^{-19}J$

$=4.55\times 10^{-19}J(\dfrac{1eV}{1.602\times 10^{-19}J})$

$=2.84eV$

The kinetic energy of electrons in electron volt is $\Delta K=2.84eV$

1. Electron moves in the region of lower potential.
2. The stopping potential difference for the electron is $\Delta V=2.84 V$
3. The kinetic energy of the electron is $\Delta K=2.84eV$