The photoelectric work function for a metal is $2.4 \ eV$. Among the four wavelengths,the wavelength of light for which photoemission does not take place is: (in $nm$)

In a photoelectric experiment,if the wavelength of incident radiation is reduced from $6000 \ \mathring{A}$ to $4000 \ \mathring{A}$ while keeping the intensity of radiation constant,then:

The work function of a metal is $2 \ eV$. If a radiation of wavelength $3000 \ \text{Å}$ is incident on it,the maximum kinetic energy of the emitted photoelectrons is (Planck's constant $h=6.6 \times 10^{-34} \ \text{Js}$; velocity of light $c=3 \times 10^8 \ \text{m/s}$; $1 \ \text{eV}=1.6 \times 10^{-19} \ \text{J}$).

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

The frequency of the incident light falling on a photosensitive metal plate is doubled. The kinetic energy of the emitted photoelectrons is:

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