To solve the problem, we will use the photoelectric effect equation derived from Einstein's theory. The equation relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted electrons.
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Threshold frequency, \( \nu_0 = 3.3 \times 10^{14} \, \text{Hz} \)
- Incident light frequency, \( \nu = 8.2 \times 10^{14} \, \text{Hz} \)
- Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{Js} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)
2. **Calculate the Work Function (\( \phi \)):**
The work function can be calculated using the threshold frequency:
\[
\phi = h \nu_0 = 6.62 \times 10^{-34} \, \text{Js} \times 3.3 \times 10^{14} \, \text{Hz}
\]
\[
\phi = 2.185 \times 10^{-19} \, \text{J}
\]
3. **Calculate the Energy of the Incident Photons (\( E \)):**
The energy of the incident photons can be calculated using the incident frequency:
\[
E = h \nu = 6.62 \times 10^{-34} \, \text{Js} \times 8.2 \times 10^{14} \, \text{Hz}
\]
\[
E = 5.4264 \times 10^{-19} \, \text{J}
\]
4. **Calculate the Kinetic Energy of the Emitted Electrons:**
According to the photoelectric effect, the kinetic energy (\( KE \)) of the emitted electrons is given by:
\[
KE = E - \phi
\]
Substituting the values:
\[
KE = 5.4264 \times 10^{-19} \, \text{J} - 2.185 \times 10^{-19} \, \text{J}
\]
\[
KE = 3.2414 \times 10^{-19} \, \text{J}
\]
5. **Relate Kinetic Energy to Stopping Potential (\( V_s \)):**
The kinetic energy of the emitted electrons can also be expressed in terms of stopping potential:
\[
KE = e V_s
\]
Therefore,
\[
V_s = \frac{KE}{e} = \frac{3.2414 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}}
\]
\[
V_s = 2.026 \, \text{V}
\]
6. **Final Result:**
Rounding to two significant figures, the cut-off voltage for photoelectric emission is approximately:
\[
V_s \approx 2 \, \text{V}
\]
