Monochromatic photons of energy 3.3 eV are incident on a photo sensitive surface of work function 2.4 eV. Calculate the threshold frequency of incident radiation.

Options

A. `6.82 xx 10^(14) Hz`
B. `5.82 xx 10^(14) Hz`
C. `7.82 xx 10^(14) Hz`
D. `10.82 xx 10^(14) Hz`

Answer

To solve the problem, we need to calculate the threshold frequency of the incident radiation based on the given work function. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Threshold Frequency
The threshold frequency (\( \nu_0 \)) is the minimum frequency of incident light required to eject electrons from a photosensitive surface. It is related to the work function (\( \phi \)) of the material by the equation:
\[ \phi = h \nu_0 \]
where \( h \) is Planck's constant.

### Step 2: Identify Given Values
- Work function (\( \phi \)) = 2.4 eV
- Planck's constant (\( h \)) = \( 6.63 \times 10^{-34} \) J·s
- To convert eV to Joules, we use the conversion factor: \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \).

### Step 3: Convert Work Function to Joules
Convert the work function from electron volts to joules:
\[ \phi = 2.4 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.84 \times 10^{-19} \text{ J} \]

### Step 4: Rearrange the Threshold Frequency Formula
We can rearrange the equation for threshold frequency:
\[ \nu_0 = \frac{\phi}{h} \]

### Step 5: Substitute the Values
Substituting the values we have:
\[ \nu_0 = \frac{3.84 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J·s}} \]

### Step 6: Calculate the Threshold Frequency
Now, perform the calculation:
\[ \nu_0 = \frac{3.84}{6.63} \times 10^{15} \text{ Hz} \approx 0.579 \times 10^{15} \text{ Hz} \approx 5.79 \times 10^{14} \text{ Hz} \]

### Final Answer
Thus, the threshold frequency of the incident radiation is approximately:
\[ \nu_0 \approx 5.79 \times 10^{14} \text{ Hz} \]

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