Yellow light of wavelength $557 \ nm$ is incident on a cesium surface. When the cathode-anode voltage drops below $0.25 \ V$,no photoelectrons flow in the circuit. What is the threshold wavelength of the cesium surface in $nm$?

The threshold wavelength of a metal is $400 \ nm$. The maximum kinetic energy of the emitted photoelectrons is $1.5 \ eV$. Find the wavelength of the incident photon in $\mathring{A}$.

The retarding potential necessary to stop the emission of photoelectrons,when a target material of work function $1.24 eV$ is irradiated with light of wavelength $4.36 \times 10^{-7} m$ is (in $eV$)

Light of wavelength $4000 \,\mathring A$ falls on a photosensitive metal and a negative $2 \,V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{-34} \,Js, \,e = 1.6 \times 10^{-19} \,C, \,c = 3 \times 10^8 \,m/s)$.

The work function for a certain metal is $4.2 \; eV$. Will this metal give photoelectric emission for incident radiation of wavelength $330 \; nm$?

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be: