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 $nm$)

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

When a certain metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the 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 $\text{ eV}$)

Consider a metal exposed to light of wavelength $600 \, nm$. The maximum kinetic energy of the emitted electron doubles when light of wavelength $400 \, nm$ is used. Find the work function of the metal in $eV$. (in $, eV$)

When the wavelength of incident radiation is changed from $400 \ nm$ to $310 \ nm$,the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metal is ........ $eV$.