The electric field at a point associated with a light wave is `E=(100Vm^-1) sin [(3*0x10^15 s^-1)t] sin [(6*0x10^15s^-1)t].` If this light falls on a metal surface having a work function of 2*0 eV, what will be the maximum kinetic energy of the photoelectrons?

Explanation

To solve the problem step by step, we will follow the outlined approach based on the given electric field equation and the photoelectric effect principles.

### Step 1: Analyze the Electric Field Equation
The electric field associated with the light wave is given as:
\[ E = (100 \, \text{V/m}) \sin[(3 \times 10^{15} \, \text{s}^{-1})t] \sin[(6 \times 10^{15} \, \text{s}^{-1})t] \]

### Step 2: Use the Product-to-Sum Formula
We can simplify the product of the two sine functions using the identity:
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