If electric field of EM wave is given by

60 [ sin(3 × 10¹⁴ t) + sin(12 × 10¹⁴ t) ] at x=0 falls on a photosensitive material having work function 2.8 eV. Find maximum kinetic energy of ejected electrons.

To find the maximum kinetic energy of the ejected electrons, we first analyze the given electric field equation of the electromagnetic wave falling on the photosensitive material:
$E=60\left[\sin \right(3\times {10}^{14}t)+\sin (12\times {10}^{14}t\left)\right]$
This electromagnetic wave consists of two different frequency components.

The general form of a sinusoidal component of an electric field is given by:
$E\left(t\right)=E_0\sin \left(\omega t\right)$
where $\omega$ is the angular frequency.

From the given expression, the two angular frequencies are:
${\omega }_1=3\times {10}^{14}\text{rad/s}$
${\omega }_2=12\times {10}^{14}\text{rad/s}$

According to Einstein's photoelectric equation, the maximum kinetic energy ($K_{\max }$) of photoelectrons depends on the frequency of the incident light. To find the maximum kinetic energy, we must use the maximum frequency (or maximum angular frequency) component, since energy of a photon $E=\hslash \omega$ increases with frequency.
Therefore, we select the higher angular frequency:
$\omega =12\times {10}^{14}\text{rad/s}$

The energy of the corresponding photon ($E_p$) is given by:
$E_p=h\nu =\frac{h}{2\pi }\omega =\hslash \omega$
Let us calculate this energy in electron-volts (eV).
Planck's constant $h\approx 6.63\times {10}^{-34}\text{J·s}$.
Thus, the energy of the photon in Joules is:
$E_p=\frac{6.63\times {10}^{-34}}{2\times 3.1416}\times 12\times {10}^{14}\text{J}$
$E_p\approx 1.055\times {10}^{-34}\times 12\times {10}^{14}\text{J}$
$E_p\approx 1.266\times {10}^{-19}\text{J}$

Convert this energy to eV by dividing by the elementary charge $e\approx 1.6\times {10}^{-19}\text{C}$:
$E_p=\frac{1.266\times {10}^{-19}}{1.6\times {10}^{-19}}\text{eV}\approx 4.96\text{eV}$
(Alternatively, using $h\approx 4.136\times {10}^{-15}\text{eV·s}$, we have:
$E_p=\frac{4.136\times {10}^{-15}}{2\pi }\times 12\times {10}^{14}\text{eV}\approx 0.658\times {10}^{-15}\times 12\times {10}^{14}\text{eV}\approx 7.9\times 0.628\text{eV}\approx 4.96\text{eV}$)

The work function ($\Phi$) of the photosensitive material is given as:
$\Phi =2.8\text{eV}$

Using Einstein's photoelectric equation:
$K_{\max }=E_p-\Phi$
$K_{\max }=4.96\text{eV}-2.8\text{eV}=2.16\text{eV}$

Thus, the maximum kinetic energy of the ejected electrons is 2.16 eV.