When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are $$\mathrm{v}_1$$ and $$\mathrm{v}_2$$ respectively. The ratio $$\mathrm{v}_1: \mathrm{v}_2$$ is

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

If the kinetic energy of a free electron doubles, it's de Broglie wavelength ($$\lambda$$) changes by a factor