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-by-Step Solution:

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

2. **Determine the frequencies of the light waves:**
- From the equation, we have two angular frequencies:
- \(\omega_1 = 10^{15} \, \text{s}^{-1}\)
- \(\omega_2 = 8 \times 10^{15} \, \text{s}^{-1}\)

3. **Calculate the corresponding frequencies (\(f\)) from angular frequencies (\(\omega\)):**
- The frequency \(f\) is related to angular frequency by the formula:
\[ f = \frac{\omega}{2\pi} \]
- For \(\omega_2\):
\[ f_2 = \frac{8 \times 10^{15}}{2\pi} \approx \frac{8 \times 10^{15}}{6.2832} \approx 1.27 \times 10^{15} \, \text{Hz} \]

4. **Calculate the energy of the photons using the frequency:**
- The energy \(E\) of a photon is given by:
\[ E = h f \]
- Where \(h\) (Planck's constant) is approximately \(6.626 \times 10^{-34} \, \text{J s}\).
- Thus, the energy corresponding to frequency \(f_2\) is:
\[ E_2 = h f_2 = 6.626 \times 10^{-34} \times 1.27 \times 10^{15} \approx 8.42 \times 10^{-19} \, \text{J} \]

5. **Convert the energy from Joules to electron volts:**
- To convert Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E_2 \text{ (in eV)} = \frac{8.42 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 5.26 \, \text{eV} \]

6. **Calculate the maximum kinetic energy of the photoelectrons:**
- The maximum kinetic energy (\(K_{\text{max}}\)) of the photoelectrons is given by:
\[ K_{\text{max}} = E - \phi \]
- Substituting the values:
\[ K_{\text{max}} = 5.26 \, \text{eV} - 2.0 \, \text{eV} = 3.26 \, \text{eV} \]

### Final Answer:
The maximum kinetic energy of the photoelectrons is approximately **3.26 eV**.