In the case of the photoelectric effect, on increasing the frequency of incident light:

The figure showing the correct relationship between the stopping potential $V_0$ and the frequency $\nu$ of light for potassium and tungsten is

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

When the wavelength of radiation falling on a metal is changed from $500 \, nm$ to $200 \, nm$,the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to $..... \, eV$.

If the work function of a metal is $3 \ eV$,then the threshold wavelength will be ............ $\mathring{A}$.

The work functions for metals $A$, $B$, and $C$ are $1.92 eV$, $2.0 eV$, and $5 eV$ respectively. The metal(s) which will emit photoelectrons for incident radiation of wavelength $4100 Å$ is/are $[h=6.63 \times 10^{-34} J s, e=1.6 \times 10^{-19} C, c=3 \times 10^8 m/s]$.