Light described at a palce by te equation `E=(100 V/m) [sinxx10^15 s ^(-1) t +sin (8xx 10^15 s^(-1) t]` falls on a metal surface having work function 2.0 eV. Calcualte the maximum kinetic energy of the photoelectrons.

Explanation

To solve the problem of calculating the maximum kinetic energy of photoelectrons when light falls on a metal surface, we can follow these steps:

### Step 1: Identify the given parameters
- The electric field equation of light is given as:
\[ E = (100 \, \text{V/m}) \left[ \sin(1 \times 10^{15} \, \text{s}^{-1} \, t) + \sin(8 \times 10^{15} \, \text{s}^{-1} \, t) \right] \]
- The work function (\( \phi \)) of the metal is given as \( 2.0 \, \text{eV} \).

### Step 2: Determine the frequencies of the light
From the electric field equation, we can see that there are two frequencies:
- \( \omega_1 = 1 \times 10^{15} \, \text{s}^{-1} \)
- \( \omega_2 = 8 \times 10^{15} \, \text{s}^{-1} \)

### Step 3: Calculate the frequency corresponding to the higher angular frequency
We will use the higher frequency \( \omega_2 \) to find the frequency \( f_2 \):
\[ f_2 = \frac{\omega_2}{2\pi} = \frac{8 \times 10^{15}}{2 \times 3.14} \approx 1.27 \times 10^{15} \, \text{Hz} \]

### Step 4: Calculate the energy of the photons
The energy of a photon can be calculated using the formula:
\[ E = h \cdot f \]
Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( f \) is the frequency we calculated.

Substituting the values:
\[ E = 6.626 \times 10^{-34} \times 1.27 \times 10^{15} \approx 8.41 \times 10^{-19} \, \text{J} \]

### Step 5: Convert the energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \text{ (in eV)} = \frac{8.41 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 5.26 \, \text{eV} \]

### Step 6: Calculate the maximum kinetic energy of the photoelectrons
Using the formula for the maximum kinetic energy of photoelectrons:
\[ K_{\text{max}} = E - \phi \]
Where \( \phi = 2.0 \, \text{eV} \):
\[ K_{\text{max}} = 5.26 \, \text{eV} - 2.0 \, \text{eV} = 3.26 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the photoelectrons is approximately \( 3.26 \, \text{eV} \).

---