The threshold wavelength for photoelectric emission from a material is $5200 \, \mathring{A}$. Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

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 $V$?

Write the slope of the $eV_0 - \nu$ graph.

$A$ photosensitive surface made of cesium on tungsten is irradiated by monochromatic radiation of wavelength $640.2 \ nm$ $(1 \ nm = 10^{-9} \ m)$ from a neon bulb. The measured stopping potential is $0.54 \ V$. If the source is replaced by another source and the same photocell is irradiated by a line of $427.2 \ nm$,what will be the new stopping potential in $V$?

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-