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:

If the frequency of light in a photoelectric experiment is doubled,the stopping potential will

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

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 $\text{ eV}$)

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

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