Question

The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \text{ m/s}$. The value of ratio of charge (e) to mass (m) of the photoelectron is $1.8 \times 10^{11} \text{ C/kg}$. The value of stopping potential in volt is

• A. 2.00
• B. 2.25
• C. 2.50
• D. 3.00

Hint:

Stopping potential is independent of the number of electrons and depends only on their maximum speed.

Answer 1

The Correct Option is B

Step 1: Concept
Maximum kinetic energy $K_{max} = \frac{1}{2}mv^2 = eV_s$.

Step 2: Meaning
Stopping potential $V_s = \frac{v^2}{2(e/m)}$.

Step 3: Analysis
$V_s = \frac{(9 \times 10^5)^2}{2 \times 1.8 \times 10^{11}} = \frac{81 \times 10^{10}}{3.6 \times 10^{11}}$.
$V_s = \frac{81}{36} = 2.25 \text{ V}$.

Step 4: Conclusion
The stopping potential is $2.25 \text{ V}$. Final Answer: (B)