In a photoelectric cell, current stops when a negative potential of 0.5v is given to the collector w.r.t. emitter. The maximum K.E. of emitted electron is

Options

A. `0.8xx10^(-19)` joule
B. `0.8xx10^(-19)` erg
C. `0.5` joule
D. 0.5 erg

Answer

To solve the problem step by step, we will determine the maximum kinetic energy (K.E.) of the emitted electrons in a photoelectric cell when a stopping potential is applied.

### Step 1: Understand the Concept of Stopping Potential
When a negative potential is applied to the collector in a photoelectric cell, it creates an electric field that opposes the motion of the emitted electrons. The potential at which the current stops is known as the stopping potential (V₀).

### Step 2: Identify the Given Data
From the problem, we know:
- Stopping potential (V₀) = 0.5 V

### Step 3: Relate Stopping Potential to Maximum Kinetic Energy
The maximum kinetic energy (K.E.) of the emitted electrons can be expressed in terms of the stopping potential using the formula:
\[ K.E. = e \cdot V₀ \]
where \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs).

### Step 4: Substitute the Values into the Formula
Now we can calculate the maximum kinetic energy:
\[ K.E. = e \cdot V₀ = (1.6 \times 10^{-19} \text{ C}) \cdot (0.5 \text{ V}) \]

### Step 5: Perform the Calculation
Calculating the above expression:
\[ K.E. = 1.6 \times 10^{-19} \cdot 0.5 = 0.8 \times 10^{-19} \text{ J} \]

### Step 6: Convert to Electron Volts
Since the question asks for the kinetic energy in electron volts (eV), we can express it as:
\[ K.E. = 0.5 \text{ eV} \]
This is because 1 eV is defined as the energy gained by an electron when it is accelerated through a potential difference of 1 volt.

### Final Answer
Thus, the maximum kinetic energy of the emitted electron is:
\[ K.E. = 0.5 \text{ eV} \]

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