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 cut-off voltage (stopping potential) is the negative potential required to stop the fastest moving photoelectrons.
Step 2: Formula Application:
$K_{max} = eV_0$
Step 3: Explanation:
If $V_0 = 1.5$ V, then the energy is $e \times 1.5$ V. In the unit of electron-volts (eV), this is simply numerically equal to the voltage. $K_{max} = 1.5$ eV.
Step 4: Final Answer:
The kinetic energy is 1.5 eV.