$$\mathrm{K}_1$$ 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:

The difference in energy levels of an electron at two excited levels is $$13.75 \mathrm{~eV}$$. If it makes a transition from the higher energy level to the lower energy level then what will be the wave length of the emitted radiation? [given $$h=6.6 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} \mathrm{~s}^{-1} ; c=3 \times 10^8 \mathrm{~ms}^{-1} ; 1 \mathrm{~eV}=1.6 \times 10^{-19} \mathrm{~J}$$]