When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential of ejected photoelectrons is $4.8 \, V$. If the same surface is illuminated by radiation of double the previous wavelength,then the stopping potential becomes $1.6 \, V$. The threshold wavelength of the metal is $... \lambda$.

Which of the following graphs shows the correct variation of maximum kinetic energy $(E)$ of photoelectrons with the intensity of incident radiation $(I)$?

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ eV \ m\right]$

When the energy of incident radiation is increased by $20 \%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5 \ eV$ to $0.8 \ eV$. The work function of the metal is (in $eV$)

Photons of wavelength $\lambda$ emitted by a source of power $P$ are incident on a photocell. If the current produced in the cell is $I$,then the percentage of incident photons which produce current in the photocell is: (where $h$ is Planck's constant and $c$ is the speed of light in vacuum)

Light from the Paschen series of a hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is: