Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$,respectively,are incident on a metallic surface. If $\lambda_{1} = 3 \lambda_{2}$,then:

The electric field associated with a light wave is given by $E = E_0 \sin [1.57 \times 10^7 (x - ct)]$,where $x$ is in meters and $t$ is in seconds. If this light is used to produce photoelectric emission from a metal surface with a work function of $1.9 \ eV$,what will be the stopping potential in $V$?

The work function of caesium is $2.14 \ eV$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$. (Answer in $nm$)

The work functions of metals are given as $Na = 1.92 \, eV$,$K = 2.15 \, eV$,$Mo = 4.17 \, eV$,and $Ni = 5.0 \, eV$. Which of these metals will not show photoelectric emission for radiation of wavelength $3300 \, \mathring{A}$ from a $He-Cd$ laser?

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:

$A$ photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_c$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de-Broglie wavelength is given by ($m$: mass of the electron,$h$: Planck's constant,and $c$: speed of light).