When a metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

When photons of energy $8 \times 10^{-19} \ J$ are incident on a photosensitive material,the de Broglie wavelength of the photoelectrons emitted with maximum kinetic energy is $10 \ Å$. The work function of the photosensitive material is nearly (in $eV$)

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

The threshold wavelength of tungsten is $2300 \, \mathring{A}$. If ultraviolet light of wavelength $1800 \, \mathring{A}$ is incident on it,then the maximum kinetic energy of photoelectrons would be about ............ $eV$.

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to:

Explain: "For photoelectric emission, the frequency of incident radiation should be greater than the threshold frequency."