The frequency and work function of an incident photon are $\nu$ and $\phi_0$. If $\nu_0$ is the threshold frequency,then the necessary condition for the emission of photoelectrons is:

The graph of stopping potential $(V_0)$ versus the frequency $(f)$ of incident radiation for the photoelectric effect is ..........

Light of wavelength $1824 \mathring A$,incident on the surface of a metal,produces photo-electrons with maximum energy $5.3 \ eV$. When light of wavelength $1216 \mathring A$ is used,the maximum energy of photoelectrons is $8.7 \ eV$. The work function of the metal surface is .............. $eV$.

The threshold wavelength for photoelectric effect from a metal is $200 \ nm$. The maximum kinetic energy of the emitted photoelectrons when radiation of wavelength $100 \ nm$ is incident on the metal is .......... $eV$.

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

For two different photosensitive materials having work functions $\phi$ and $2 \phi$ respectively,which are illuminated with light of sufficient energy to emit electrons,if the graph of stopping potential $(V_s)$ versus frequency $(\nu)$ is drawn for these two materials,what is the ratio of the slopes of the graphs for these two materials?