The electric field at a point associated with a light wave is `E=100Vm^-1 sin [(3*0xx10^15 s^-1)t] sin [(6*0xx10^15s^-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 find the maximum kinetic energy of the photoelectrons when light falls on a metal surface, we can follow these steps:

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

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