Given below are two statements:

Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.

Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.

In the light of above statements, choose the correct answer form the options given below

The difference between threshold wavelengths for two metal surfaces $$\mathrm{A}$$ and $$\mathrm{B}$$ having work function $$\phi_{A}=9 ~\mathrm{eV}$$ and $$\phi_{B}=4 \cdot 5 ~\mathrm{eV}$$ in $$\mathrm{nm}$$ is:

$$\{$$ Given, hc $$=1242 ~\mathrm{eV} \mathrm{nm}\}$$

A proton and an $$\alpha$$-particle are accelerated from rest by $$2 \mathrm{~V}$$ and $$4 \mathrm{~V}$$ potentials, respectively. The ratio of their de-Broglie wavelength is :

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :

(Assume $$m_{p}=m_{e} \times 1849$$ )