The work function of a photoelectric material is 3.3 eV. The thershold frequency will be equal to

Options

A. `8 xx 10^(4) Hz`
B. `8 xx 10^(56) Hz`
C. `8 xx 10^(10) Hz`
D. `8 xx 10^(14) Hz`

Answer

To find the threshold frequency (\( \nu_0 \)) of a photoelectric material given its work function (\( \phi \)), we can use the formula:

\[ \phi = h \nu_0 \]

where:
- \( \phi \) is the work function in joules,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( \nu_0 \) is the threshold frequency in Hz.

### Step 1: Convert the work function from eV to joules
The work function is given as \( 3.3 \, \text{eV} \). To convert this to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).

\[ \phi = 3.3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]

Calculating this gives:

\[ \phi = 3.3 \times 1.6 \times 10^{-19} = 5.28 \times 10^{-19} \, \text{J} \]

### Step 2: Rearrange the formula to solve for threshold frequency
We can rearrange the formula \( \phi = h \nu_0 \) to solve for \( \nu_0 \):

\[ \nu_0 = \frac{\phi}{h} \]

### Step 3: Substitute the values into the equation
Now we can substitute the values of \( \phi \) and \( h \):

\[ \nu_0 = \frac{5.28 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \]

### Step 4: Perform the calculation
Calculating this gives:

\[ \nu_0 = \frac{5.28 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 7.97 \times 10^{14} \, \text{Hz} \]

### Step 5: Round the result
We can round \( 7.97 \times 10^{14} \, \text{Hz} \) to \( 8.0 \times 10^{14} \, \text{Hz} \).

### Final Answer
The threshold frequency \( \nu_0 \) is approximately \( 8.0 \times 10^{14} \, \text{Hz} \).

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