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 a material given its threshold frequency, we can use the formula that relates work function (φ) to threshold frequency (ν₀):

\[ \phi = h \cdot \nu_0 \]

where:
- \( \phi \) is the work function in Joules,
- \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \, \text{Joule second} \),
- \( \nu_0 \) is the threshold frequency in Hz.

Given:
- \( \nu_0 = 6 \times 10^{14} \, \text{Hz} \)

### Step 1: Calculate the work function in Joules
Substituting the values into the formula:

\[ \phi = h \cdot \nu_0 = (6.63 \times 10^{-34} \, \text{Joule second}) \cdot (6 \times 10^{14} \, \text{Hz}) \]

### Step 2: Perform the multiplication
Calculating the above expression:

\[ \phi = 6.63 \times 6 \times 10^{-34} \times 10^{14} \]
\[ = 39.78 \times 10^{-20} \, \text{Joules} \]
\[ = 3.978 \times 10^{-19} \, \text{Joules} \]

### Step 3: Convert Joules to electron volts
To convert Joules to electron volts, we use the conversion factor:

\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{Joules} \]

Now, we convert the work function:

\[ \phi \, \text{(in eV)} = \frac{3.978 \times 10^{-19} \, \text{Joules}}{1.6 \times 10^{-19} \, \text{Joules/eV}} \]

### Step 4: Perform the division
Calculating the above expression:

\[ \phi \, \text{(in eV)} = \frac{3.978}{1.6} \approx 2.48625 \, \text{eV} \]

### Step 5: Round the result
Rounding to two decimal places gives:

\[ \phi \approx 2.48 \, \text{eV} \]

Thus, the work function of the material is approximately **2.48 eV**.