The threshold frequency of a certain metal is `3.3xx10^(14)Hz`. If light of frequency `8.2xx10^(14)Hz` is incident on the metal, predict the cut off voltage for photoelectric emission. Given Planck's constant, `h=6.62xx10^(-34)Js`.

Options

A. 2 V
B. 4 V
C. 6 V
D. 8 V

Explanation

To find the cut-off voltage (stopping potential) for photoelectric emission, we can use Einstein's photoelectric equation. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Threshold frequency of the metal, \( \nu_0 = 3.3 \times 10^{14} \, \text{Hz} \)
- Frequency of incident light, \( \nu = 8.2 \times 10^{14} \, \text{Hz} \)
- Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{Js} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Use Einstein's Photoelectric Equation
Einstein's photoelectric equation relates the kinetic energy of emitted electrons to the energy of the incident photons and the work function of the metal:
\[ KE = E - \phi_0 \]
Where:
- \( KE \) is the kinetic energy of the emitted electrons,
- \( E \) is the energy of the incident photons,
- \( \phi_0 \) is the work function of the metal.

### Step 3: Express Energy in Terms of Frequency
The energy of the incident photons can be expressed as:
\[ E = h \nu \]
And the work function can be expressed as:
\[ \phi_0 = h \nu_0 \]

### Step 4: Substitute into the Equation
Substituting the expressions for \( E \) and \( \phi_0 \) into the photoelectric equation gives:
\[ KE = h \nu - h \nu_0 \]

### Step 5: Relate Kinetic Energy to Stopping Potential
The kinetic energy can also be expressed in terms of stopping potential \( V_0 \):
\[ KE = e V_0 \]
Thus, we have:
\[ e V_0 = h \nu - h \nu_0 \]

### Step 6: Solve for Stopping Potential \( V_0 \)
Rearranging the equation to solve for \( V_0 \):
\[ V_0 = \frac{h}{e} (\nu - \nu_0) \]

### Step 7: Substitute the Values
Now substitute the known values:
\[ V_0 = \frac{6.62 \times 10^{-34} \, \text{Js}}{1.6 \times 10^{-19} \, \text{C}} \left(8.2 \times 10^{14} \, \text{Hz} - 3.3 \times 10^{14} \, \text{Hz}\right) \]

### Step 8: Calculate the Difference in Frequencies
Calculate \( \nu - \nu_0 \):
\[ \nu - \nu_0 = 8.2 \times 10^{14} - 3.3 \times 10^{14} = 4.9 \times 10^{14} \, \text{Hz} \]

### Step 9: Substitute Back into the Equation
Now substitute this value back into the equation for \( V_0 \):
\[ V_0 = \frac{6.62 \times 10^{-34}}{1.6 \times 10^{-19}} \times 4.9 \times 10^{14} \]

### Step 10: Perform the Calculation
Calculating the right-hand side:
\[ V_0 = \frac{6.62 \times 4.9}{1.6} \times 10^{-34 + 14 + 19} \]
\[ V_0 = \frac{32.438}{1.6} \times 10^{-1} \]
\[ V_0 \approx 20.27375 \times 10^{-1} \approx 2.027375 \, \text{V} \]

### Final Answer
Thus, the cut-off voltage (stopping potential) is approximately:
\[ V_0 \approx 2 \, \text{V} \]