$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

When light is incident on a surface,photoelectrons are emitted. For these photoelectrons:

For intensity $I$ of a light of wavelength $5000 \, Å$, the photoelectron saturation current is $0.40 \, μA$ and the stopping potential is $1.36 \, V$. The work function of the metal is ........... $eV$.

For the photoelectric effect,which of the following statements are true?
$I$ The kinetic energies of the photoelectrons do not depend on the frequency of light.
$II$ The photoelectric effect will always occur for highly intense light.
$III$ The maximum kinetic energy of a photoelectron does not depend upon the intensity of the light.
$IV$ The escaping electron's kinetic energy is larger for a larger frequency.

When a photon of energy $4.25\, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A\, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70\, eV$ is ${T_B} = ({T_A} - 1.50)\, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

Why can wave theory not explain the change in the kinetic energy of an electron with a change in the frequency of incident light?