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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**.

- 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.** - 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:

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

**Expert Answer**

**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**

- Electron moves in the region of lower potential.
- The stopping potential difference for the electron is \[\Delta V=1.02 V\]
- 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.**

**Did the electron move to a region of higher potential or lower potential?****What potential difference stopped the electron?****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 $\Delta K=2.84eV$

- Electron moves in the region of lower potential.
- The
**stopping potential difference**for the electron is \[\Delta V=2.84 V\] - The
**kinetic energy**of the electron is \[\Delta K=2.84eV \]