The electric field associated with a light wave at a point is given...

The given electric field of the light wave is:

$E = 100 \sin(3.0 \times 10^{15} t) \sin(6.0 \times 10^{15} t)$

Using the trigonometric identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$, the electric field can be written as:

$E = 50 [\cos(3.0 \times 10^{15} t) - \cos(9.0 \times 10^{15} t)]$

This indicates that the light wave is a superposition of two waves with angular frequencies $\omega_1 = 3.0 \times 10^{15} \text{ s}^{-1}$ and $\omega_2 = 9.0 \times 10^{15} \text{ s}^{-1}$.

The maximum kinetic energy of the emitted photoelectrons is determined by the photon with the higher frequency, which is $\omega_2 = 9.0 \times 10^{15} \text{ s}^{-1}$.

The energy of this incident photon is:

$E_p = \dfrac{h \omega_2}{2\pi} = \dfrac{6.63 \times 10^{-34} \times 9.0 \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} \text{ eV}$

$E_p \approx \dfrac{59.67 \times 10^{-19}}{10.05 \times 10^{-19}} \text{ eV} \approx 5.93 \text{ eV}$

Using Einstein's photoelectric equation, the maximum kinetic energy is:

K_{\text\max} = E_p - \phi

K_{\text\max} = 5.93 \text{ eV} - 2.0 \text{ eV} = 3.93 \text{ eV}