Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

The frequency of monochromatic light incident on an emitting surface is $f$. If the threshold frequency for the surface is $f_0$,then the maximum kinetic energy of the emitted photoelectrons is .......

Two photons having energies twice and thrice the work function of a metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively:

If the work function of a photometal is $6.825 \; eV$,its threshold wavelength will be ............. $\mathring A$ $(c = 3 \times 10^8 \; m/s)$.

Photoelectric emission is observed from a metallic surface for frequencies ${\nu _1}$ and ${\nu _2}$ of the incident light rays $({\nu _1} > {\nu _2})$. 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 work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)