The photoelectric effect was successfully explained first by

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

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for the photoelectric current is $V_s$. When the same surface is illuminated by a light of frequency $v/2$,the stopping potential is $V_s/4$. The threshold frequency of photoelectric emission is

Energy of the incident photons on the photosensitive metal surface is $3W$ and then $5W$ where '$W$' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?