Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface.(Mass of the electron `9.11xx10^(-31)kg`)`
1. Calculate the value of work function is eV.

Options

A. 1.77
B. 3.27
C. 5.69
D. 2.32

Explanation

To calculate the work function of silver in electron volts, we will follow these steps:

### Step 1: Convert the wavelength of the threshold (λ₀) from Angstroms to meters.
Given:
- λ₀ = 3800 Å = 3800 × 10⁻¹⁰ m

### Step 2: Calculate the frequency (f₀) corresponding to the threshold wavelength (λ₀).
Using the formula:
\[ f = \frac{c}{\lambda} \]
where:
- c = speed of light = \( 3 \times 10^8 \) m/s

Substituting the values:
\[ f₀ = \frac{3 \times 10^8 \text{ m/s}}{3800 \times 10^{-10} \text{ m}} \]

### Step 3: Calculate the work function (W) using the frequency (f₀).
The work function can be calculated using the formula:
\[ W = h f₀ \]
where:
- h = Planck's constant = \( 6.63 \times 10^{-34} \) J·s

Substituting the frequency calculated in Step 2 into the equation:
\[ W = 6.63 \times 10^{-34} \text{ J·s} \times f₀ \]

### Step 4: Convert the work function from joules to electron volts.
To convert joules to electron volts, use the conversion factor:
1 eV = \( 1.6 \times 10^{-19} \) J

Thus,
\[ W \text{ (in eV)} = \frac{W \text{ (in J)}}{1.6 \times 10^{-19}} \]

### Step 5: Calculate the value of the work function in eV.
Now, we will perform the calculations step by step.

1. **Calculate frequency (f₀)**:
\[ f₀ = \frac{3 \times 10^8}{3800 \times 10^{-10}} = \frac{3 \times 10^8}{3.8 \times 10^{-7}} \approx 7.89 \times 10^{14} \text{ Hz} \]

2. **Calculate work function (W) in joules**:
\[ W = 6.63 \times 10^{-34} \times 7.89 \times 10^{14} \approx 5.22 \times 10^{-19} \text{ J} \]

3. **Convert work function to eV**:
\[ W \text{ (in eV)} = \frac{5.22 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.26 \text{ eV} \]

Thus, the work function of silver is approximately **3.26 eV**.