Question

Monochromatic light of frequency \( 8 \times 10^{14} \, \text{Hz} \) is incident on a metal surface whose threshold frequency is \( 5 \times 10^{14} \, \text{Hz} \). The stopping potential is approximately:
(Given \( h = 6.6 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

• A. 0.62 V
• B. 1.24 V
• C. 2.48 V
• D. 3.72 V

Hint:

To perform quick calculations in photoelectric effect questions, convert energies into electron-volts (eV). Since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \), dividing the energy in Joules by \( e \) directly gives the stopping potential in Volts.

Answer 1

The Correct Option is B

Step 1: Read the physics.
A photon hands its energy to an electron. Part of that energy ($h\nu_0$) is used just to escape the metal, and the leftover becomes kinetic energy that the stopping potential must cancel.
Step 2: Convert both frequencies into photon energies.
Incident photon energy $E = h\nu = 6.6\times10^{-34}\times 8\times10^{14}$ joules. Work function $W = h\nu_0 = 6.6\times10^{-34}\times 5\times10^{14}$ joules.
Step 3: Compute each in convenient units.
\[ E = 52.8\times10^{-20}\ \text{J}, \qquad W = 33.0\times10^{-20}\ \text{J} \]
Step 4: Subtract to get the kinetic energy.
\[ K_{\max} = E - W = 52.8\times10^{-20} - 33.0\times10^{-20} = 19.8\times10^{-20}\ \text{J} \]
Step 5: Turn energy into volts.
The stopping potential is just the kinetic energy measured in electron-volts, so divide by $e$: \[ V_0 = \frac{19.8\times10^{-20}}{1.6\times10^{-19}} \]
Step 6: Finish the arithmetic.
\[ V_0 = \frac{19.8\times10^{-20}}{16\times10^{-20}} = \frac{19.8}{16} \approx 1.24\ \text{V} \]
Notice this is the same as evaluating $\frac{h(\nu-\nu_0)}{e}$ directly, just organised energy first. \[ \boxed{V_0 \approx 1.24\ \text{V}} \]