For a clean metallic surface, it is known that its threshold frequency is `3.3 xx 10^14` Hz. Calculate the cutoff voltage for photoelectric emission if the light of frequency `8.2 xx 10^(14)` Hz is incident on the same metallic surface. (Given h = `6.63 xx 10^(-34)` J s)

Options

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

Explanation

To solve the problem, we need to calculate the cutoff voltage for photoelectric emission using the given values. Here’s the step-by-step solution:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of a certain frequency strikes a metallic surface, it can cause the emission of electrons. The energy of the incident photons must be greater than the work function (or the energy corresponding to the threshold frequency) of the metal for the electrons to be emitted.

### Step 2: Identify Given Values
- Threshold frequency (\( \nu_0 \)) = \( 3.3 \times 10^{14} \) Hz
- Incident frequency (\( \nu \)) = \( 8.2 \times 10^{14} \) Hz
- Planck's constant (\( h \)) = \( 6.63 \times 10^{-34} \) J·s
- Charge of an electron (\( e \)) = \( 1.6 \times 10^{-19} \) C

### Step 3: Calculate the Energy of the Incident Photons
The energy of a photon can be calculated using the formula:
\[ E = h \nu \]
For the incident photons:
\[ E_{\text{incident}} = h \nu = (6.63 \times 10^{-34} \, \text{J·s}) \times (8.2 \times 10^{14} \, \text{Hz}) \]
Calculating this gives:
\[ E_{\text{incident}} = 5.43 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the Energy Corresponding to the Threshold Frequency
Now, calculate the energy corresponding to the threshold frequency:
\[ E_{\text{threshold}} = h \nu_0 = (6.63 \times 10^{-34} \, \text{J·s}) \times (3.3 \times 10^{14} \, \text{Hz}) \]
Calculating this gives:
\[ E_{\text{threshold}} = 2.19 \times 10^{-19} \, \text{J} \]

### Step 5: Calculate the Kinetic Energy of the Emitted Electrons
The kinetic energy (KE) of the emitted electrons can be found by subtracting the threshold energy from the incident energy:
\[ KE = E_{\text{incident}} - E_{\text{threshold}} = (5.43 \times 10^{-19} \, \text{J}) - (2.19 \times 10^{-19} \, \text{J}) \]
Calculating this gives:
\[ KE = 3.24 \times 10^{-19} \, \text{J} \]

### Step 6: Relate Kinetic Energy to Cutoff Voltage
The kinetic energy of the emitted electrons is also given by:
\[ KE = eV \]
Where \( V \) is the cutoff voltage. Rearranging this gives:
\[ V = \frac{KE}{e} \]
Substituting the values:
\[ V = \frac{3.24 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \]
Calculating this gives:
\[ V = 2.025 \, \text{V} \]

### Final Answer
The cutoff voltage for photoelectric emission is approximately \( 2.03 \, \text{V} \).

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