Photons of energy $10 eV$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15} Hz$. The kinetic energy in $eV$ of the photoelectrons emitted is [Planck's constant $h = 6.63 \times 10^{-34} Js$]. (in $eV$)

The photoelectric threshold wavelength for a certain metal surface is $3600 \mathring A$. If the metal surface is irradiated by a wavelength of $1100 \mathring A$,the kinetic energy of the emitted photoelectrons is (in $\text{ eV}$)

Which one of the four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident radiation $(I)$ is correct?

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?

If the work function of a metal is $\phi$ and the frequency of the incident light is $\nu$,there is no emission of photoelectrons if:

In the case of photoelectric emission from a certain metal,the cutoff frequency is $v$. If radiation of frequency $2v$ is incident on the metal plate,the maximum possible velocity of the emitted electrons will be ($m=$ mass of electron).