Question

Stopping potential is given by 1.8 eV, then what will be the maximum kinetic energy of photoelectrons?

Hint:

The relationship is direct and simple: If stopping potential $V_0 = x$ Volts, then max Kinetic Energy $K_{max} = x$ electron-Volts (eV).

Answer 1

Step 1: Understanding the Concept:
In the photoelectric effect, stopping potential (\(V_0\)) is the minimum negative (retarding) potential applied to the anode with respect to the cathode that completely stops even the most energetic photoelectrons from reaching the anode.
Step 2: Key Formula or Approach:
The work done by the stopping potential field against the most energetic electron exactly equals its maximum kinetic energy.
\[ K_{max} = e V_0 \]
Where:
\(K_{max}\) = maximum kinetic energy
\(e\) = elementary charge
\(V_0\) = stopping potential in volts
Step 3: Detailed Explanation:
Given: Stopping potential conceptually implies the energy barrier is 1.8 eV, or \(V_0 = 1.8 \text{ V}\).
Using the formula:
\[ K_{max} = e V_0 \]
Substitute \(V_0 = 1.8 \text{ V}\):
\[ K_{max} = e \times (1.8 \text{ V}) \]
\[ K_{max} = 1.8 \text{ eV} \]
The numerical value of the maximum kinetic energy in electron-volts (eV) is exactly the same as the numerical value of the stopping potential in Volts (V).
Since the question states the value directly as 1.8 eV, it essentially directly provided the maximum kinetic energy.
Step 4: Final Answer:
The maximum kinetic energy of the photoelectrons is 1.8 eV.