The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

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

The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35 \text{ eV}$ by electromagnetic radiation whose electric component varies with time as: $E=a[1+\cos(2 \pi f_1 t)] \cos(2 \pi f_2 t)$ (where $a$ is a constant) is (given $f_1=3.6 \times 10^{15} \text{ Hz}$,$f_2=1.2 \times 10^{15} \text{ Hz}$ and Planck's constant $h=6.6 \times 10^{-34} \text{ Js}$) (in $\text{ eV}$)

The work functions of cesium $(Cs)$ and lithium $(Li)$ metals are $1.9 \ eV$ and $2.5 \ eV$,respectively. If we incident light of wavelength $550 \ nm$ on these two metal surfaces,then the photoelectric effect is possible for which case?

The work functions for metals $A, B$,and $C$ are $1.92 \ eV, 2.0 \ eV$,and $5 \ eV$ respectively. According to Einstein's photoelectric equation,which metal$(s)$ will emit photoelectrons when irradiated with light of wavelength $4100 \ \mathring A$?