From a metallic surface,photoelectric emission is observed for frequencies $v_1$ and $v_2$ $(v_1 > v_2)$ of the incident light. The maximum kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence,the threshold frequency of the metallic surface is

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons, moving perpendicular to the field, are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$, then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. 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

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is [$h=$ Planck's constant,$m=$ mass of electron and $c=$ speed of light]

The variation of stopping potential $V_{0}$ with frequency $\nu$ of incident radiation for a given photosensitive material is a straight line [frequency $\nu$ of incident radiation is greater than threshold frequency $\nu_{0}$]. The slope of this line is . . . . . . .