Solution:
Solution
Standard form will be like $$E=E_0[sin(\omega _1t) +sin (\omega _2 t)]$$
so we have $$\omega _1 =2\times 10^{15}rad/s$$ and $$\omega _2 =6.28\times 10^{15}rad/s$$
corresponding frequency will be $$\nu _1=\omega _1/2\pi=\dfrac{2\times 10^{15}}{6.28}$$
and $$\nu _2=\omega _2/2\pi=\dfrac{6.28\times 10^{15}}{6.28}=10^{15}Hz$$
We can see that $$\nu _2 $$ is $$higher $$ so it will provide $$maximum $$ kinetic energy.
Energy of the $$\nu_2 $$ photon will be $$E=h\nu _2=6.6\times 10^{-34}Js\times 10^{15} Hz=6.6\times 10^{-19} Joule=\dfrac{6.6\times 10^{-19}Joule}{1.6\times 10^{-19}J/eV}$$
or $$E=4.125eV$$
Consumed energy in work function will be $$W=1.5eV$$
So remaining energy that is maximum kinetic energy will be $$KE=E-W=(4.125-1.5)eV=2.625eV$$