In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

For a photoelectric cell,the graph showing the variation of stopping potential $(V_o)$ with frequency $(\nu)$ of incident light is best represented by:

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $\lambda_1$ and $K_2$ be the maximum kinetic energy corresponding to wavelength $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

$A$ ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely $1$, $2$ and $3$. The threshold wavelengths of these photoelectric cells are $\lambda_1$, $\lambda_2$ and $\lambda_3$, respectively, and the magnitudes of the stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and the threshold wavelengths is $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 >> \lambda$. The correct option is:

The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$,the maximum velocity of the ejected photoelectrons will be:

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$,then the threshold frequency of the metallic surface is