The work function of material (in eV), whose threshold frequency is` 6×10^(14)` Hz would be

Options

A. 5.48 eV
B. 3.48 eV
C. 2.48 eV
D. 8.48 eV

Answer

To find the work function of the material given its threshold frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between work function and threshold frequency**:
The work function (Φ) is related to the threshold frequency (ν₀) by the equation:
\[ \Phi = h \cdot \nu_0 \]
where \( h \) is Planck's constant.

2. **Identify the values**:
- Given threshold frequency, \( \nu_0 = 6 \times 10^{14} \) Hz.
- Planck's constant, \( h = 6.626 \times 10^{-34} \) J·s.

3. **Calculate the work function in Joules**:
Substitute the values into the equation:
\[ \Phi = (6.626 \times 10^{-34} \, \text{J·s}) \cdot (6 \times 10^{14} \, \text{Hz}) \]
\[ \Phi = 3.9756 \times 10^{-19} \, \text{J} \]

4. **Convert the work function from Joules to electron volts**:
To convert Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Phi \, (\text{in eV}) = \frac{3.9756 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
\[ \Phi \, (\text{in eV}) = 2.484 \, \text{eV} \]

5. **Final answer**:
The work function of the material is approximately \( 2.48 \, \text{eV} \).