When light of frequencies ${\nu _1}$ and ${\nu _2}$ $({\nu _1} > {\nu _2})$ is incident on a metal surface,the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:k$. What is the threshold frequency of the metal?

When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

In a photoelectric experiment for $4000 \, \mathring{A}$ incident radiation,the potential difference to stop the ejection is $2 \, V$. If the incident light is changed to $3000 \, \mathring{A}$,then the potential required to stop the ejection of electrons will be

Consider the following statements regarding the photoelectric effect experiment:
$(I)$ Photoelectrons are emitted as soon as the metal is exposed to light.
$(II)$ There is a minimum frequency below which no photocurrent is observed.
$(III)$ The stopping potential is proportional to the frequency of light.
$(IV)$ The photocurrent varies linearly with the intensity of the light.
Which of the above statements indicate that light consists of quanta (photons) with energy proportional to frequency?

The threshold frequency of a photosensitive material is $v$. When photons of frequency $2v$ are incident on the material,photoelectrons are emitted with a maximum linear momentum $P$. To get photoelectrons with maximum linear momentum $2P$,the frequency of the incident photons is: (in $v$)

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