The kinetic energy of the photoelectrons increases by $0.52 eV$ when the wavelength of incident light is changed from $500 \,nm$ to another wavelength. The new wavelength is approximately: (in $\,nm$)

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\lambda/2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is three times that in the first case,the work function of the surface is

Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$,respectively,are incident on a metallic surface. If $\lambda_{1} = 3 \lambda_{2}$,then:

Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is

Light of wavelength $5000 \ \mathring A$ and intensity $4.68 \ mW/cm^2$ is incident on a photosensitive surface. If only $5\%$ of the incident photons produce photoelectrons,what is the number of photoelectrons emitted per unit area per unit time?