When a certain metal surface is illuminated with light of frequency $$v$$, the stopping potential for photoelectric current is $$\mathrm{V}_0$$. When the same surface is illuminated by light of frequency $$\frac{v}{2}$$, the stopping potential is $$\frac{\mathrm{V}_0}{4}$$, the threshold frequency of photoelectric emission is
From a metallic surface photoelectric emission is observed for frequencies $$v_1$$ and $$v_2\left(v_1 > v_2\right)$$ of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $$1: \mathrm{x}$$. Hence the threshold frequency of the metallic surface is
The wave number of the last line of the Balmer series in the hydrogen spectrum will be $$\left(\right.$$ Rydberg's cons $$\left.\tan t, R=\frac{10^7}{\mathrm{~m}}\right)$$
Photoemission from metal surface takes place for frequencies '$$v_1$$' and '$$v_2$$' of incident rays $$\left(v_1>v_2\right)$$. Maximum kinetic energy of photoelectrons emitted is in the ratio $$1: \mathrm{K}$$. The threshold frequency of metallic surface is