A photon of energy `8 eV` is incident on metal surface of threshold frequency `1.6 xx 10^(15) Hz`, The maximum kinetic energy of the photoelectrons emitted ( in `eV`) (Take `h = 6 xx 10^(-34) Js`).

Options

A. `1.6`
B. `6`
C. `2`
D. `1.2`

Explanation

To find the maximum kinetic energy of the photoelectrons emitted when a photon of energy 8 eV is incident on a metal surface with a threshold frequency of \(1.6 \times 10^{15} \, \text{Hz}\), we can follow these steps:

### Step 1: Calculate the Work Function
The work function (\( \phi \)) of the metal can be calculated using the formula:
\[ \phi = h \nu_0 \]
where:
- \( h = 6 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( \nu_0 = 1.6 \times 10^{15} \, \text{Hz} \) (threshold frequency)

Substituting the values:
\[ \phi = (6 \times 10^{-34} \, \text{Js}) \times (1.6 \times 10^{15} \, \text{Hz}) \]

### Step 2: Convert Work Function to Electron Volts
To convert the work function from Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ \phi \text{ (in eV)} = \frac{\phi \text{ (in J)}}{1.6 \times 10^{-19}} \]

### Step 3: Calculate the Maximum Kinetic Energy
Using Einstein's photoelectric equation:
\[ K.E. = E - \phi \]
where:
- \( E = 8 \, \text{eV} \) (energy of the incident photon)
- \( \phi \) is the work function calculated in eV.

Substituting the values:
\[ K.E. = 8 \, \text{eV} - \phi \]

### Step 4: Final Calculation
After calculating \( \phi \) in eV, substitute it back into the equation to find the maximum kinetic energy \( K.E. \).

### Detailed Calculation
1. Calculate the work function:
\[ \phi = (6 \times 10^{-34}) \times (1.6 \times 10^{15}) = 9.6 \times 10^{-19} \, \text{J} \]
2. Convert to eV:
\[ \phi = \frac{9.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 6 \, \text{eV} \]
3. Calculate the maximum kinetic energy:
\[ K.E. = 8 \, \text{eV} - 6 \, \text{eV} = 2 \, \text{eV} \]

Thus, the maximum kinetic energy of the photoelectrons emitted is \( \boxed{2 \, \text{eV}} \).