The velocity of an electron so that its momentum is equal to that of a photon of wavelength $$660 \mathrm{~nm}$$ is

$$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $$\lambda_1$$ and $$\lambda_2$$ respectively are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$, then

The threshold frequency for a metal surface is '$$n_0$$'. A photo electric current '$$I$$' is produced when it is exposed to a light of frequency $$\left(\frac{11}{6}\right) \mathrm{n}_{\mathrm{o}}$$ and intensity $$\mathrm{I}_{\mathrm{n}}$$. If both the frequency and intensity are halved, the new photoelectric current '$$\mathrm{I}^1$$' will become:

The mass of a particle $$\mathrm{A}$$ is double that of the particle $$\mathrm{B}$$ and the kinetic energy of $$\mathrm{B}$$ is $$\frac{1}{8}$$th that of A then the ratio of the de- Broglie wavelength of A to that of B is: