In an optical instrument- microscope, the wavelength of light used are $\lambda_1 = 6800$ Å and $\lambda_2 = 5100$ Å. The ratio of the resolving power of microscope corresponding to '$\lambda_1$' to that for '$\lambda_2$' is

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when light of wavelength $\lambda_1$ and $\lambda_2$ respectively are incident on a metallic surface. If $\lambda_1 = 3\lambda_2$ then

The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$ then $\lambda_3$ is given by

If the difference between $(n + 1)^{th}$ Bohr radius and $n^{th}$ Bohr radius is given by $(n - 1)^{th}$ Bohr radius, then the value of n is