The work function of a metal is $1.6 \ eV$. What is the maximum wavelength of light in $\mathring{A}$ that can cause photoelectric emission from this metal? $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

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:

The threshold frequency of a photoelectric metal is $\nu_0$. If light of frequency $4\nu_0$ is incident on this metal,then the maximum kinetic energy of emitted electrons will be $.......h\nu_0$.

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]

The threshold frequency of cesium is $5.16 \times 10^{14} \ Hz$. Then its work function is . . . . . . $eV$.

Photoelectric emission is observed from a metallic surface for frequencies ${\nu _1}$ and ${\nu _2}$ of the incident light rays $({\nu _1} > {\nu _2})$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is