Question

The photoelectric cut-off voltage in a certain experiment is 1.5 V. The kinetic energy of photoelectrons emitted will be ______.

• A. 1.5 J
• B. 1.5 eV
• C. 2.4 eV
• D. 2.4 J

Hint:

To convert from \textbf{eV} to \textbf{Joules}, multiply by $1.6 \times 10^{-19}$. Here, $1.5\text{ eV} = 2.4 \times 10^{-19}\text{ J}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The stopping potential (cut-off voltage) is the negative potential applied to the collector plate that just stops the fastest-moving photoelectrons.
Step 2: Detailed Explanation:
The maximum kinetic energy ($K_{max}$) is equal to the work done by the stopping potential to stop the electron: $$K_{max} = e \times V_0$$ If $V_0 = 1.5\text{ V}$, then: $$K_{max} = e \times 1.5\text{ V} = 1.5\text{ eV}$$
Step 3: Final Answer:
The kinetic energy is 1.5 eV.